How to call included php file with AJAX?

How to call included php file with AJAX? - javascript

I am not sure if this is best approach, but how to make AJAX call with included file? Let me make working simple example.
I have, main index.php file, with following content, where I want to access all data returned from included file where magic happens. Data is array type in reality.
<?php
require_once('magic.php');
?>
<html>
<body>
<form action="<?=$_SERVER['PHP_SELF'];?>" method="POST">
<input type="text" name="variable" id="variable" />
<input type="submit" value="Do magic" name="submit" />
</form>
<?php
echo "<pre>";
print_r($data);
echo "</pre>";
?>
</body>
</html>
And "magic.php" file, where all the magic happens.
<?php
$variable = $_POST['variable'];
function doMagic($variable) {
# do some magic with passed variable ...
# and return data
return $data;
}
$data = doMagic($variable);
# now return $data variable so main file can use it
return $data;
?>
What I want to do is to call this function doMagic() with AJAX, but all logic is already included and available in global space, and return $data to modal or pop-up window.
What is best approach, because I am already passing variable to included file?

Use ajax like the following
$('form').submit(function(){
$.ajax({url: "magic.php",
data: {variable : $'#variable').val()},
success:function(data){console.log(data);});
});
and in magic.php, use json_encode
<?php
$variable = $_POST['variable'];
function doMagic($variable) {
# do some magic with passed variable ...
# and return data
return $data;
}
$data = doMagic($variable);
# now return $data variable so main file can use it
return json_encode($data);
?>

Since you have jQuery as a tag...
You could do something like this:
In your index.php:
<input type="text" id="variable">
<div id="#myButton"></div>
<div id="#response"></div>
Then have a Javascript file (or include in your index.php):
$('#myButton').click(function(){
var myVariable = $('#variable').val();
$.post( "magic.php",{variable: myVariable}, function(data) {
$( "#response" ).html( data );
});
});
This basically means, when you click the item that has the id of myButton, it'll grab the value of what's inside of the input with the id of variable, POST that via jQuery.post() to your magic.php file, and return the data. Then, it replaces the html of the div with the id of response with the data it returns.
Just make sure to include the jQuery library

Related

How can I get Javascript variable using PHP Code? [duplicate]

I want to pass JavaScript variables to PHP using a hidden input in a form.
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
Here is the code:
<script type="text/javascript">
// View what the user has chosen
function func_load3(name) {
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("You have chosen: " + oSelectBox.options[iChoice].text);
//document.write(oSelectBox.options[iChoice].text);
var sa = oSelectBox.options[iChoice].text;
document.getElementById("hidden1").value = sa;
}
</script>
<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<?php
$salarieid = $_POST['hidden1'];
$query = "select * from salarie where salarieid = ".$salarieid;
echo $query;
$result = mysql_query($query);
?>
<table>
Code for displaying the query result.
</table>

You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<DOCTYPE html>
<html>
<head>
<title>My Test Form</title>
</head>
<body>
<form method="POST">
<p>Please, choose the salary id to proceed result:</p>
<p>
<label for="salarieids">SalarieID:</label>
<?php
$query = "SELECT * FROM salarie";
$result = mysql_query($query);
if ($result) :
?>
<select id="salarieids" name="salarieid">
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
}
?>
</select>
<?php endif ?>
</p>
<p>
<input type="submit" value="Sumbit my choice"/>
</p>
</form>
<?php if isset($_POST['salaried']) : ?>
<?php
$query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
$result = mysql_query($query);
if ($result) :
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
echo '</tr>';
}
?>
</table>
<?php endif?>
<?php endif ?>
</body>
</html>

Just save it in a cookie:
$(document).ready(function () {
createCookie("height", $(window).height(), "10");
});
function createCookie(name, value, days) {
var expires;
if (days) {
var date = new Date();
date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
expires = "; expires=" + date.toGMTString();
}
else {
expires = "";
}
document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}
And then read it with PHP:
<?PHP
$_COOKIE["height"];
?>
It's not a pretty solution, but it works.

There are several ways of passing variables from JavaScript to PHP (not the current page, of course).
You could:
Send the information in a form as stated here (will result in a page refresh)
Pass it in Ajax (several posts are on here about that) (without a page refresh)
Make an HTTP request via an XMLHttpRequest request (without a page refresh) like this:
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var PageToSendTo = "nowitworks.php?";
var MyVariable = "variableData";
var VariablePlaceholder = "variableName=";
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable;
xmlhttp.open("GET", UrlToSend, false);
xmlhttp.send();
I'm sure this could be made to look fancier and loop through all the variables and whatnot - but I've kept it basic as to make it easier to understand for the novices.

Here is the Working example: Get javascript variable value on the same page in php.
<script>
var p1 = "success";
</script>
<?php
echo "<script>document.writeln(p1);</script>";
?>

Here's how I did it (I needed to insert a local timezone into PHP:
<?php
ob_start();
?>
<script type="text/javascript">
var d = new Date();
document.write(d.getTimezoneOffset());
</script>
<?php
$offset = ob_get_clean();
print_r($offset);

When your page first loads the PHP code first runs and sets the complete layout of your webpage. After the page layout, it sets the JavaScript load up.
Now JavaScript directly interacts with DOM and can manipulate the layout but PHP can't - it needs to refresh the page. The only way is to refresh your page to and pass the parameters in the page URL so that you can get the data via PHP.
So, we use AJAX to get Javascript to interact with PHP without a page reload. AJAX can also be used as an API. One more thing if you have already declared the variable in PHP before the page loads then you can use it with your Javascript example.
<?php $myname= "syed ali";?>
<script>
var username = "<?php echo $myname;?>";
alert(username);
</script>
The above code is correct and it will work, but the code below is totally wrong and it will never work.
<script>
var username = "syed ali";
var <?php $myname;?> = username;
alert(myname);
</script>
Pass value from JavaScript to PHP via AJAX
This is the most secure way to do it, because HTML content can be edited via developer tools and the user can manipulate the data. So, it is better to use AJAX if you want security over that variable. If you are a newbie to AJAX, please learn AJAX it is very simple.
The best and most secure way to pass JavaScript variable into PHP is via AJAX
Simple AJAX example
var mydata = 55;
var myname = "syed ali";
var userdata = {'id':mydata,'name':myname};
$.ajax({
type: "POST",
url: "YOUR PHP URL HERE",
data:userdata,
success: function(data){
console.log(data);
}
});
PASS value from JavaScript to PHP via hidden fields
Otherwise, you can create a hidden HTML input inside your form. like
<input type="hidden" id="mydata">
then via jQuery or javaScript pass the value to the hidden field. like
<script>
var myvalue = 55;
$("#mydata").val(myvalue);
</script>
Now when you submit the form you can get the value in PHP.

I was trying to figure this out myself and then realized that the problem is that this is kind of a backwards way of looking at the situation. Rather than trying to pass things from JavaScript to php, maybe it's best to go the other way around, in most cases. PHP code executes on the server and creates the html code (and possibly java script as well). Then the browser loads the page and executes the html and java script.
It seems like the sensible way to approach situations like this is to use the PHP to create the JavaScript and the html you want and then to use the JavaScript in the page to do whatever PHP can't do. It seems like this would give you the benefits of both PHP and JavaScript in a fairly simple and straight forward way.
One thing I've done that gives the appearance of passing things to PHP from your page on the fly is using the html image tag to call on PHP code. Something like this:
<img src="pic.php">
The PHP code in pic.php would actually create html code before your web page was even loaded, but that html code is basically called upon on the fly. The php code here can be used to create a picture on your page, but it can have any commands you like besides that in it. Maybe it changes the contents of some files on your server, etc. The upside of this is that the php code can be executed from html and I assume JavaScript, but the down side is that the only output it can put on your page is an image. You also have the option of passing variables to the php code through parameters in the url. Page counters will use this technique in many cases.

PHP runs on the server before the page is sent to the user, JavaScript is run on the user's computer once it is received, so the PHP script has already executed.
If you want to pass a JavaScript value to a PHP script, you'd have to do an XMLHttpRequest to send the data back to the server.
Here's a previous question that you can follow for more information: Ajax Tutorial
Now if you just need to pass a form value to the server, you can also just do a normal form post, that does the same thing, but the whole page has to be refreshed.
<?php
if(isset($_POST))
{
print_r($_POST);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="data" value="1" />
<input type="submit" value="Submit" />
</form>
Clicking submit will submit the page, and print out the submitted data.

We can easily pass values even on same/ different pages using the cookies shown in the code as follows (In my case, I'm using it with facebook integration) -
function statusChangeCallback(response) {
console.log('statusChangeCallback');
if (response.status === 'connected') {
// Logged into your app and Facebook.
FB.api('/me?fields=id,first_name,last_name,email', function (result) {
document.cookie = "fbdata = " + result.id + "," + result.first_name + "," + result.last_name + "," + result.email;
console.log(document.cookie);
});
}
}
And I've accessed it (in any file) using -
<?php
if(isset($_COOKIE['fbdata'])) {
echo "welcome ".$_COOKIE['fbdata'];
}
?>

Your code has a few things wrong with it.
You define a JavaScript function, func_load3(), but do not call it.
Your function is defined in the wrong place. When it is defined in your page, the HTML objects it refers to have not yet been loaded. Most JavaScript code checks whether the document is fully loaded before executing, or you can just move your code past the elements it refers to in the page.
Your form has no means to submit it. It needs a submit button.
You do not check whether your form has been submitted.
It is possible to set a JavaScript variable in a hidden variable in a form, then submit it, and read the value back in PHP. Here is a simple example that shows this:
<?php
if (isset($_POST['hidden1'])) {
echo "You submitted {$_POST['hidden1']}";
die;
}
echo <<<HTML
<form name="myform" action="{$_SERVER['PHP_SELF']}" method="post" id="myform">
<input type="submit" name="submit" value="Test this mess!" />
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<script type="text/javascript">
document.getElementById("hidden1").value = "This is an example";
</script>
HTML;
?>

You can use JQuery Ajax and POST method:
var obj;
$(document).ready(function(){
$("#button1").click(function(){
var username=$("#username").val();
var password=$("#password").val();
$.ajax({
url: "addperson.php",
type: "POST",
async: false,
data: {
username: username,
password: password
}
})
.done (function(data, textStatus, jqXHR) {
obj = JSON.parse(data);
})
.fail (function(jqXHR, textStatus, errorThrown) {
})
.always (function(jqXHROrData, textStatus, jqXHROrErrorThrown) {
});
});
});
To take a response back from the php script JSON parse the the respone in .done() method.
Here is the php script you can modify to your needs:
<?php
$username1 = isset($_POST["username"]) ? $_POST["username"] : '';
$password1 = isset($_POST["password"]) ? $_POST["password"] : '';
$servername = "xxxxx";
$username = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (username, password)
VALUES ('$username1', '$password1' )";
;
if ($conn->query($sql) === TRUE) {
echo json_encode(array('success' => 1));
} else{
echo json_encode(array('success' => 0));
}
$conn->close();
?>

Is your function, which sets the hidden form value, being called? It is not in this example. You should have no problem modifying a hidden value before posting the form back to the server.

May be you could use jquery serialize() method so that everything will be at one go.
var data=$('#myForm').serialize();
//this way you could get the hidden value as well in the server side.

This obviously solution was not mentioned earlier. You can also use cookies to pass data from the browser back to the server.
Just set a cookie with the data you want to pass to PHP using javascript in the browser.
Then, simply read this cookie on the PHP side.

We cannot pass JavaScript variable values to the PHP code directly... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
So it's better to use the AJAX to parse the JavaScript value into the php Code.
Or alternatively we can make this done with the help of COOKIES in our code.
Thanks & Cheers.

Use the + sign to concatenate your javascript variable into your php function call.
<script>
var JSvar = "success";
var JSnewVar = "<?=myphpFunction('" + JSvar + "');?>";
</script>`
Notice the = sign is there twice.

How to query filter by month from calendar with startTime value? [duplicate]

I want to pass JavaScript variables to PHP using a hidden input in a form.
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
Here is the code:
<script type="text/javascript">
// View what the user has chosen
function func_load3(name) {
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("You have chosen: " + oSelectBox.options[iChoice].text);
//document.write(oSelectBox.options[iChoice].text);
var sa = oSelectBox.options[iChoice].text;
document.getElementById("hidden1").value = sa;
}
</script>
<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<?php
$salarieid = $_POST['hidden1'];
$query = "select * from salarie where salarieid = ".$salarieid;
echo $query;
$result = mysql_query($query);
?>
<table>
Code for displaying the query result.
</table>

You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<DOCTYPE html>
<html>
<head>
<title>My Test Form</title>
</head>
<body>
<form method="POST">
<p>Please, choose the salary id to proceed result:</p>
<p>
<label for="salarieids">SalarieID:</label>
<?php
$query = "SELECT * FROM salarie";
$result = mysql_query($query);
if ($result) :
?>
<select id="salarieids" name="salarieid">
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
}
?>
</select>
<?php endif ?>
</p>
<p>
<input type="submit" value="Sumbit my choice"/>
</p>
</form>
<?php if isset($_POST['salaried']) : ?>
<?php
$query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
$result = mysql_query($query);
if ($result) :
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
echo '</tr>';
}
?>
</table>
<?php endif?>
<?php endif ?>
</body>
</html>

Just save it in a cookie:
$(document).ready(function () {
createCookie("height", $(window).height(), "10");
});
function createCookie(name, value, days) {
var expires;
if (days) {
var date = new Date();
date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
expires = "; expires=" + date.toGMTString();
}
else {
expires = "";
}
document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}
And then read it with PHP:
<?PHP
$_COOKIE["height"];
?>
It's not a pretty solution, but it works.

There are several ways of passing variables from JavaScript to PHP (not the current page, of course).
You could:
Send the information in a form as stated here (will result in a page refresh)
Pass it in Ajax (several posts are on here about that) (without a page refresh)
Make an HTTP request via an XMLHttpRequest request (without a page refresh) like this:
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var PageToSendTo = "nowitworks.php?";
var MyVariable = "variableData";
var VariablePlaceholder = "variableName=";
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable;
xmlhttp.open("GET", UrlToSend, false);
xmlhttp.send();
I'm sure this could be made to look fancier and loop through all the variables and whatnot - but I've kept it basic as to make it easier to understand for the novices.

Here is the Working example: Get javascript variable value on the same page in php.
<script>
var p1 = "success";
</script>
<?php
echo "<script>document.writeln(p1);</script>";
?>

Here's how I did it (I needed to insert a local timezone into PHP:
<?php
ob_start();
?>
<script type="text/javascript">
var d = new Date();
document.write(d.getTimezoneOffset());
</script>
<?php
$offset = ob_get_clean();
print_r($offset);

When your page first loads the PHP code first runs and sets the complete layout of your webpage. After the page layout, it sets the JavaScript load up.
Now JavaScript directly interacts with DOM and can manipulate the layout but PHP can't - it needs to refresh the page. The only way is to refresh your page to and pass the parameters in the page URL so that you can get the data via PHP.
So, we use AJAX to get Javascript to interact with PHP without a page reload. AJAX can also be used as an API. One more thing if you have already declared the variable in PHP before the page loads then you can use it with your Javascript example.
<?php $myname= "syed ali";?>
<script>
var username = "<?php echo $myname;?>";
alert(username);
</script>
The above code is correct and it will work, but the code below is totally wrong and it will never work.
<script>
var username = "syed ali";
var <?php $myname;?> = username;
alert(myname);
</script>
Pass value from JavaScript to PHP via AJAX
This is the most secure way to do it, because HTML content can be edited via developer tools and the user can manipulate the data. So, it is better to use AJAX if you want security over that variable. If you are a newbie to AJAX, please learn AJAX it is very simple.
The best and most secure way to pass JavaScript variable into PHP is via AJAX
Simple AJAX example
var mydata = 55;
var myname = "syed ali";
var userdata = {'id':mydata,'name':myname};
$.ajax({
type: "POST",
url: "YOUR PHP URL HERE",
data:userdata,
success: function(data){
console.log(data);
}
});
PASS value from JavaScript to PHP via hidden fields
Otherwise, you can create a hidden HTML input inside your form. like
<input type="hidden" id="mydata">
then via jQuery or javaScript pass the value to the hidden field. like
<script>
var myvalue = 55;
$("#mydata").val(myvalue);
</script>
Now when you submit the form you can get the value in PHP.

PHP runs on the server before the page is sent to the user, JavaScript is run on the user's computer once it is received, so the PHP script has already executed.
If you want to pass a JavaScript value to a PHP script, you'd have to do an XMLHttpRequest to send the data back to the server.
Here's a previous question that you can follow for more information: Ajax Tutorial
Now if you just need to pass a form value to the server, you can also just do a normal form post, that does the same thing, but the whole page has to be refreshed.
<?php
if(isset($_POST))
{
print_r($_POST);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="data" value="1" />
<input type="submit" value="Submit" />
</form>
Clicking submit will submit the page, and print out the submitted data.

We can easily pass values even on same/ different pages using the cookies shown in the code as follows (In my case, I'm using it with facebook integration) -
function statusChangeCallback(response) {
console.log('statusChangeCallback');
if (response.status === 'connected') {
// Logged into your app and Facebook.
FB.api('/me?fields=id,first_name,last_name,email', function (result) {
document.cookie = "fbdata = " + result.id + "," + result.first_name + "," + result.last_name + "," + result.email;
console.log(document.cookie);
});
}
}
And I've accessed it (in any file) using -
<?php
if(isset($_COOKIE['fbdata'])) {
echo "welcome ".$_COOKIE['fbdata'];
}
?>

Your code has a few things wrong with it.
You define a JavaScript function, func_load3(), but do not call it.
Your function is defined in the wrong place. When it is defined in your page, the HTML objects it refers to have not yet been loaded. Most JavaScript code checks whether the document is fully loaded before executing, or you can just move your code past the elements it refers to in the page.
Your form has no means to submit it. It needs a submit button.
You do not check whether your form has been submitted.
It is possible to set a JavaScript variable in a hidden variable in a form, then submit it, and read the value back in PHP. Here is a simple example that shows this:
<?php
if (isset($_POST['hidden1'])) {
echo "You submitted {$_POST['hidden1']}";
die;
}
echo <<<HTML
<form name="myform" action="{$_SERVER['PHP_SELF']}" method="post" id="myform">
<input type="submit" name="submit" value="Test this mess!" />
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<script type="text/javascript">
document.getElementById("hidden1").value = "This is an example";
</script>
HTML;
?>

You can use JQuery Ajax and POST method:
var obj;
$(document).ready(function(){
$("#button1").click(function(){
var username=$("#username").val();
var password=$("#password").val();
$.ajax({
url: "addperson.php",
type: "POST",
async: false,
data: {
username: username,
password: password
}
})
.done (function(data, textStatus, jqXHR) {
obj = JSON.parse(data);
})
.fail (function(jqXHR, textStatus, errorThrown) {
})
.always (function(jqXHROrData, textStatus, jqXHROrErrorThrown) {
});
});
});
To take a response back from the php script JSON parse the the respone in .done() method.
Here is the php script you can modify to your needs:
<?php
$username1 = isset($_POST["username"]) ? $_POST["username"] : '';
$password1 = isset($_POST["password"]) ? $_POST["password"] : '';
$servername = "xxxxx";
$username = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (username, password)
VALUES ('$username1', '$password1' )";
;
if ($conn->query($sql) === TRUE) {
echo json_encode(array('success' => 1));
} else{
echo json_encode(array('success' => 0));
}
$conn->close();
?>

Is your function, which sets the hidden form value, being called? It is not in this example. You should have no problem modifying a hidden value before posting the form back to the server.

May be you could use jquery serialize() method so that everything will be at one go.
var data=$('#myForm').serialize();
//this way you could get the hidden value as well in the server side.

This obviously solution was not mentioned earlier. You can also use cookies to pass data from the browser back to the server.
Just set a cookie with the data you want to pass to PHP using javascript in the browser.
Then, simply read this cookie on the PHP side.

We cannot pass JavaScript variable values to the PHP code directly... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
So it's better to use the AJAX to parse the JavaScript value into the php Code.
Or alternatively we can make this done with the help of COOKIES in our code.
Thanks & Cheers.

Use the + sign to concatenate your javascript variable into your php function call.
<script>
var JSvar = "success";
var JSnewVar = "<?=myphpFunction('" + JSvar + "');?>";
</script>`
Notice the = sign is there twice.

Using javascript var in php to get wordpress attachment [duplicate]

I want to pass JavaScript variables to PHP using a hidden input in a form.
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
Here is the code:
<script type="text/javascript">
// View what the user has chosen
function func_load3(name) {
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("You have chosen: " + oSelectBox.options[iChoice].text);
//document.write(oSelectBox.options[iChoice].text);
var sa = oSelectBox.options[iChoice].text;
document.getElementById("hidden1").value = sa;
}
</script>
<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<?php
$salarieid = $_POST['hidden1'];
$query = "select * from salarie where salarieid = ".$salarieid;
echo $query;
$result = mysql_query($query);
?>
<table>
Code for displaying the query result.
</table>

You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<DOCTYPE html>
<html>
<head>
<title>My Test Form</title>
</head>
<body>
<form method="POST">
<p>Please, choose the salary id to proceed result:</p>
<p>
<label for="salarieids">SalarieID:</label>
<?php
$query = "SELECT * FROM salarie";
$result = mysql_query($query);
if ($result) :
?>
<select id="salarieids" name="salarieid">
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
}
?>
</select>
<?php endif ?>
</p>
<p>
<input type="submit" value="Sumbit my choice"/>
</p>
</form>
<?php if isset($_POST['salaried']) : ?>
<?php
$query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
$result = mysql_query($query);
if ($result) :
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
echo '</tr>';
}
?>
</table>
<?php endif?>
<?php endif ?>
</body>
</html>

Just save it in a cookie:
$(document).ready(function () {
createCookie("height", $(window).height(), "10");
});
function createCookie(name, value, days) {
var expires;
if (days) {
var date = new Date();
date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
expires = "; expires=" + date.toGMTString();
}
else {
expires = "";
}
document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}
And then read it with PHP:
<?PHP
$_COOKIE["height"];
?>
It's not a pretty solution, but it works.

There are several ways of passing variables from JavaScript to PHP (not the current page, of course).
You could:
Send the information in a form as stated here (will result in a page refresh)
Pass it in Ajax (several posts are on here about that) (without a page refresh)
Make an HTTP request via an XMLHttpRequest request (without a page refresh) like this:
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var PageToSendTo = "nowitworks.php?";
var MyVariable = "variableData";
var VariablePlaceholder = "variableName=";
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable;
xmlhttp.open("GET", UrlToSend, false);
xmlhttp.send();
I'm sure this could be made to look fancier and loop through all the variables and whatnot - but I've kept it basic as to make it easier to understand for the novices.

Here is the Working example: Get javascript variable value on the same page in php.
<script>
var p1 = "success";
</script>
<?php
echo "<script>document.writeln(p1);</script>";
?>

Here's how I did it (I needed to insert a local timezone into PHP:
<?php
ob_start();
?>
<script type="text/javascript">
var d = new Date();
document.write(d.getTimezoneOffset());
</script>
<?php
$offset = ob_get_clean();
print_r($offset);

When your page first loads the PHP code first runs and sets the complete layout of your webpage. After the page layout, it sets the JavaScript load up.
Now JavaScript directly interacts with DOM and can manipulate the layout but PHP can't - it needs to refresh the page. The only way is to refresh your page to and pass the parameters in the page URL so that you can get the data via PHP.
So, we use AJAX to get Javascript to interact with PHP without a page reload. AJAX can also be used as an API. One more thing if you have already declared the variable in PHP before the page loads then you can use it with your Javascript example.
<?php $myname= "syed ali";?>
<script>
var username = "<?php echo $myname;?>";
alert(username);
</script>
The above code is correct and it will work, but the code below is totally wrong and it will never work.
<script>
var username = "syed ali";
var <?php $myname;?> = username;
alert(myname);
</script>
Pass value from JavaScript to PHP via AJAX
This is the most secure way to do it, because HTML content can be edited via developer tools and the user can manipulate the data. So, it is better to use AJAX if you want security over that variable. If you are a newbie to AJAX, please learn AJAX it is very simple.
The best and most secure way to pass JavaScript variable into PHP is via AJAX
Simple AJAX example
var mydata = 55;
var myname = "syed ali";
var userdata = {'id':mydata,'name':myname};
$.ajax({
type: "POST",
url: "YOUR PHP URL HERE",
data:userdata,
success: function(data){
console.log(data);
}
});
PASS value from JavaScript to PHP via hidden fields
Otherwise, you can create a hidden HTML input inside your form. like
<input type="hidden" id="mydata">
then via jQuery or javaScript pass the value to the hidden field. like
<script>
var myvalue = 55;
$("#mydata").val(myvalue);
</script>
Now when you submit the form you can get the value in PHP.

PHP runs on the server before the page is sent to the user, JavaScript is run on the user's computer once it is received, so the PHP script has already executed.
If you want to pass a JavaScript value to a PHP script, you'd have to do an XMLHttpRequest to send the data back to the server.
Here's a previous question that you can follow for more information: Ajax Tutorial
Now if you just need to pass a form value to the server, you can also just do a normal form post, that does the same thing, but the whole page has to be refreshed.
<?php
if(isset($_POST))
{
print_r($_POST);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="data" value="1" />
<input type="submit" value="Submit" />
</form>
Clicking submit will submit the page, and print out the submitted data.

We can easily pass values even on same/ different pages using the cookies shown in the code as follows (In my case, I'm using it with facebook integration) -
function statusChangeCallback(response) {
console.log('statusChangeCallback');
if (response.status === 'connected') {
// Logged into your app and Facebook.
FB.api('/me?fields=id,first_name,last_name,email', function (result) {
document.cookie = "fbdata = " + result.id + "," + result.first_name + "," + result.last_name + "," + result.email;
console.log(document.cookie);
});
}
}
And I've accessed it (in any file) using -
<?php
if(isset($_COOKIE['fbdata'])) {
echo "welcome ".$_COOKIE['fbdata'];
}
?>

Your code has a few things wrong with it.
You define a JavaScript function, func_load3(), but do not call it.
Your function is defined in the wrong place. When it is defined in your page, the HTML objects it refers to have not yet been loaded. Most JavaScript code checks whether the document is fully loaded before executing, or you can just move your code past the elements it refers to in the page.
Your form has no means to submit it. It needs a submit button.
You do not check whether your form has been submitted.
It is possible to set a JavaScript variable in a hidden variable in a form, then submit it, and read the value back in PHP. Here is a simple example that shows this:
<?php
if (isset($_POST['hidden1'])) {
echo "You submitted {$_POST['hidden1']}";
die;
}
echo <<<HTML
<form name="myform" action="{$_SERVER['PHP_SELF']}" method="post" id="myform">
<input type="submit" name="submit" value="Test this mess!" />
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<script type="text/javascript">
document.getElementById("hidden1").value = "This is an example";
</script>
HTML;
?>

You can use JQuery Ajax and POST method:
var obj;
$(document).ready(function(){
$("#button1").click(function(){
var username=$("#username").val();
var password=$("#password").val();
$.ajax({
url: "addperson.php",
type: "POST",
async: false,
data: {
username: username,
password: password
}
})
.done (function(data, textStatus, jqXHR) {
obj = JSON.parse(data);
})
.fail (function(jqXHR, textStatus, errorThrown) {
})
.always (function(jqXHROrData, textStatus, jqXHROrErrorThrown) {
});
});
});
To take a response back from the php script JSON parse the the respone in .done() method.
Here is the php script you can modify to your needs:
<?php
$username1 = isset($_POST["username"]) ? $_POST["username"] : '';
$password1 = isset($_POST["password"]) ? $_POST["password"] : '';
$servername = "xxxxx";
$username = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (username, password)
VALUES ('$username1', '$password1' )";
;
if ($conn->query($sql) === TRUE) {
echo json_encode(array('success' => 1));
} else{
echo json_encode(array('success' => 0));
}
$conn->close();
?>

Is your function, which sets the hidden form value, being called? It is not in this example. You should have no problem modifying a hidden value before posting the form back to the server.

May be you could use jquery serialize() method so that everything will be at one go.
var data=$('#myForm').serialize();
//this way you could get the hidden value as well in the server side.

This obviously solution was not mentioned earlier. You can also use cookies to pass data from the browser back to the server.
Just set a cookie with the data you want to pass to PHP using javascript in the browser.
Then, simply read this cookie on the PHP side.

We cannot pass JavaScript variable values to the PHP code directly... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
So it's better to use the AJAX to parse the JavaScript value into the php Code.
Or alternatively we can make this done with the help of COOKIES in our code.
Thanks & Cheers.

Use the + sign to concatenate your javascript variable into your php function call.
<script>
var JSvar = "success";
var JSnewVar = "<?=myphpFunction('" + JSvar + "');?>";
</script>`
Notice the = sign is there twice.

Passing data from JavaScript Promise to MySQL database [PHP] [duplicate]

I want to pass JavaScript variables to PHP using a hidden input in a form.
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
Here is the code:
<script type="text/javascript">
// View what the user has chosen
function func_load3(name) {
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("You have chosen: " + oSelectBox.options[iChoice].text);
//document.write(oSelectBox.options[iChoice].text);
var sa = oSelectBox.options[iChoice].text;
document.getElementById("hidden1").value = sa;
}
</script>
<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<?php
$salarieid = $_POST['hidden1'];
$query = "select * from salarie where salarieid = ".$salarieid;
echo $query;
$result = mysql_query($query);
?>
<table>
Code for displaying the query result.
</table>

You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<DOCTYPE html>
<html>
<head>
<title>My Test Form</title>
</head>
<body>
<form method="POST">
<p>Please, choose the salary id to proceed result:</p>
<p>
<label for="salarieids">SalarieID:</label>
<?php
$query = "SELECT * FROM salarie";
$result = mysql_query($query);
if ($result) :
?>
<select id="salarieids" name="salarieid">
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
}
?>
</select>
<?php endif ?>
</p>
<p>
<input type="submit" value="Sumbit my choice"/>
</p>
</form>
<?php if isset($_POST['salaried']) : ?>
<?php
$query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
$result = mysql_query($query);
if ($result) :
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
echo '</tr>';
}
?>
</table>
<?php endif?>
<?php endif ?>
</body>
</html>

Just save it in a cookie:
$(document).ready(function () {
createCookie("height", $(window).height(), "10");
});
function createCookie(name, value, days) {
var expires;
if (days) {
var date = new Date();
date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
expires = "; expires=" + date.toGMTString();
}
else {
expires = "";
}
document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}
And then read it with PHP:
<?PHP
$_COOKIE["height"];
?>
It's not a pretty solution, but it works.

There are several ways of passing variables from JavaScript to PHP (not the current page, of course).
You could:
Send the information in a form as stated here (will result in a page refresh)
Pass it in Ajax (several posts are on here about that) (without a page refresh)
Make an HTTP request via an XMLHttpRequest request (without a page refresh) like this:
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var PageToSendTo = "nowitworks.php?";
var MyVariable = "variableData";
var VariablePlaceholder = "variableName=";
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable;
xmlhttp.open("GET", UrlToSend, false);
xmlhttp.send();
I'm sure this could be made to look fancier and loop through all the variables and whatnot - but I've kept it basic as to make it easier to understand for the novices.

Here is the Working example: Get javascript variable value on the same page in php.
<script>
var p1 = "success";
</script>
<?php
echo "<script>document.writeln(p1);</script>";
?>

Here's how I did it (I needed to insert a local timezone into PHP:
<?php
ob_start();
?>
<script type="text/javascript">
var d = new Date();
document.write(d.getTimezoneOffset());
</script>
<?php
$offset = ob_get_clean();
print_r($offset);

When your page first loads the PHP code first runs and sets the complete layout of your webpage. After the page layout, it sets the JavaScript load up.
Now JavaScript directly interacts with DOM and can manipulate the layout but PHP can't - it needs to refresh the page. The only way is to refresh your page to and pass the parameters in the page URL so that you can get the data via PHP.
So, we use AJAX to get Javascript to interact with PHP without a page reload. AJAX can also be used as an API. One more thing if you have already declared the variable in PHP before the page loads then you can use it with your Javascript example.
<?php $myname= "syed ali";?>
<script>
var username = "<?php echo $myname;?>";
alert(username);
</script>
The above code is correct and it will work, but the code below is totally wrong and it will never work.
<script>
var username = "syed ali";
var <?php $myname;?> = username;
alert(myname);
</script>
Pass value from JavaScript to PHP via AJAX
This is the most secure way to do it, because HTML content can be edited via developer tools and the user can manipulate the data. So, it is better to use AJAX if you want security over that variable. If you are a newbie to AJAX, please learn AJAX it is very simple.
The best and most secure way to pass JavaScript variable into PHP is via AJAX
Simple AJAX example
var mydata = 55;
var myname = "syed ali";
var userdata = {'id':mydata,'name':myname};
$.ajax({
type: "POST",
url: "YOUR PHP URL HERE",
data:userdata,
success: function(data){
console.log(data);
}
});
PASS value from JavaScript to PHP via hidden fields
Otherwise, you can create a hidden HTML input inside your form. like
<input type="hidden" id="mydata">
then via jQuery or javaScript pass the value to the hidden field. like
<script>
var myvalue = 55;
$("#mydata").val(myvalue);
</script>
Now when you submit the form you can get the value in PHP.

PHP runs on the server before the page is sent to the user, JavaScript is run on the user's computer once it is received, so the PHP script has already executed.
If you want to pass a JavaScript value to a PHP script, you'd have to do an XMLHttpRequest to send the data back to the server.
Here's a previous question that you can follow for more information: Ajax Tutorial
Now if you just need to pass a form value to the server, you can also just do a normal form post, that does the same thing, but the whole page has to be refreshed.
<?php
if(isset($_POST))
{
print_r($_POST);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="data" value="1" />
<input type="submit" value="Submit" />
</form>
Clicking submit will submit the page, and print out the submitted data.

We can easily pass values even on same/ different pages using the cookies shown in the code as follows (In my case, I'm using it with facebook integration) -
function statusChangeCallback(response) {
console.log('statusChangeCallback');
if (response.status === 'connected') {
// Logged into your app and Facebook.
FB.api('/me?fields=id,first_name,last_name,email', function (result) {
document.cookie = "fbdata = " + result.id + "," + result.first_name + "," + result.last_name + "," + result.email;
console.log(document.cookie);
});
}
}
And I've accessed it (in any file) using -
<?php
if(isset($_COOKIE['fbdata'])) {
echo "welcome ".$_COOKIE['fbdata'];
}
?>

Your code has a few things wrong with it.
You define a JavaScript function, func_load3(), but do not call it.
Your function is defined in the wrong place. When it is defined in your page, the HTML objects it refers to have not yet been loaded. Most JavaScript code checks whether the document is fully loaded before executing, or you can just move your code past the elements it refers to in the page.
Your form has no means to submit it. It needs a submit button.
You do not check whether your form has been submitted.
It is possible to set a JavaScript variable in a hidden variable in a form, then submit it, and read the value back in PHP. Here is a simple example that shows this:
<?php
if (isset($_POST['hidden1'])) {
echo "You submitted {$_POST['hidden1']}";
die;
}
echo <<<HTML
<form name="myform" action="{$_SERVER['PHP_SELF']}" method="post" id="myform">
<input type="submit" name="submit" value="Test this mess!" />
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<script type="text/javascript">
document.getElementById("hidden1").value = "This is an example";
</script>
HTML;
?>

You can use JQuery Ajax and POST method:
var obj;
$(document).ready(function(){
$("#button1").click(function(){
var username=$("#username").val();
var password=$("#password").val();
$.ajax({
url: "addperson.php",
type: "POST",
async: false,
data: {
username: username,
password: password
}
})
.done (function(data, textStatus, jqXHR) {
obj = JSON.parse(data);
})
.fail (function(jqXHR, textStatus, errorThrown) {
})
.always (function(jqXHROrData, textStatus, jqXHROrErrorThrown) {
});
});
});
To take a response back from the php script JSON parse the the respone in .done() method.
Here is the php script you can modify to your needs:
<?php
$username1 = isset($_POST["username"]) ? $_POST["username"] : '';
$password1 = isset($_POST["password"]) ? $_POST["password"] : '';
$servername = "xxxxx";
$username = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (username, password)
VALUES ('$username1', '$password1' )";
;
if ($conn->query($sql) === TRUE) {
echo json_encode(array('success' => 1));
} else{
echo json_encode(array('success' => 0));
}
$conn->close();
?>

Is your function, which sets the hidden form value, being called? It is not in this example. You should have no problem modifying a hidden value before posting the form back to the server.

May be you could use jquery serialize() method so that everything will be at one go.
var data=$('#myForm').serialize();
//this way you could get the hidden value as well in the server side.

This obviously solution was not mentioned earlier. You can also use cookies to pass data from the browser back to the server.
Just set a cookie with the data you want to pass to PHP using javascript in the browser.
Then, simply read this cookie on the PHP side.

We cannot pass JavaScript variable values to the PHP code directly... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
So it's better to use the AJAX to parse the JavaScript value into the php Code.
Or alternatively we can make this done with the help of COOKIES in our code.
Thanks & Cheers.

Use the + sign to concatenate your javascript variable into your php function call.
<script>
var JSvar = "success";
var JSnewVar = "<?=myphpFunction('" + JSvar + "');?>";
</script>`
Notice the = sign is there twice.

read java script variable in php [duplicate]

I want to pass JavaScript variables to PHP using a hidden input in a form.
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
Here is the code:
<script type="text/javascript">
// View what the user has chosen
function func_load3(name) {
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("You have chosen: " + oSelectBox.options[iChoice].text);
//document.write(oSelectBox.options[iChoice].text);
var sa = oSelectBox.options[iChoice].text;
document.getElementById("hidden1").value = sa;
}
</script>
<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<?php
$salarieid = $_POST['hidden1'];
$query = "select * from salarie where salarieid = ".$salarieid;
echo $query;
$result = mysql_query($query);
?>
<table>
Code for displaying the query result.
</table>

You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<DOCTYPE html>
<html>
<head>
<title>My Test Form</title>
</head>
<body>
<form method="POST">
<p>Please, choose the salary id to proceed result:</p>
<p>
<label for="salarieids">SalarieID:</label>
<?php
$query = "SELECT * FROM salarie";
$result = mysql_query($query);
if ($result) :
?>
<select id="salarieids" name="salarieid">
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
}
?>
</select>
<?php endif ?>
</p>
<p>
<input type="submit" value="Sumbit my choice"/>
</p>
</form>
<?php if isset($_POST['salaried']) : ?>
<?php
$query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
$result = mysql_query($query);
if ($result) :
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
echo '</tr>';
}
?>
</table>
<?php endif?>
<?php endif ?>
</body>
</html>

Just save it in a cookie:
$(document).ready(function () {
createCookie("height", $(window).height(), "10");
});
function createCookie(name, value, days) {
var expires;
if (days) {
var date = new Date();
date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
expires = "; expires=" + date.toGMTString();
}
else {
expires = "";
}
document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}
And then read it with PHP:
<?PHP
$_COOKIE["height"];
?>
It's not a pretty solution, but it works.

There are several ways of passing variables from JavaScript to PHP (not the current page, of course).
You could:
Send the information in a form as stated here (will result in a page refresh)
Pass it in Ajax (several posts are on here about that) (without a page refresh)
Make an HTTP request via an XMLHttpRequest request (without a page refresh) like this:
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var PageToSendTo = "nowitworks.php?";
var MyVariable = "variableData";
var VariablePlaceholder = "variableName=";
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable;
xmlhttp.open("GET", UrlToSend, false);
xmlhttp.send();
I'm sure this could be made to look fancier and loop through all the variables and whatnot - but I've kept it basic as to make it easier to understand for the novices.

Here is the Working example: Get javascript variable value on the same page in php.
<script>
var p1 = "success";
</script>
<?php
echo "<script>document.writeln(p1);</script>";
?>

Here's how I did it (I needed to insert a local timezone into PHP:
<?php
ob_start();
?>
<script type="text/javascript">
var d = new Date();
document.write(d.getTimezoneOffset());
</script>
<?php
$offset = ob_get_clean();
print_r($offset);

When your page first loads the PHP code first runs and sets the complete layout of your webpage. After the page layout, it sets the JavaScript load up.
Now JavaScript directly interacts with DOM and can manipulate the layout but PHP can't - it needs to refresh the page. The only way is to refresh your page to and pass the parameters in the page URL so that you can get the data via PHP.
So, we use AJAX to get Javascript to interact with PHP without a page reload. AJAX can also be used as an API. One more thing if you have already declared the variable in PHP before the page loads then you can use it with your Javascript example.
<?php $myname= "syed ali";?>
<script>
var username = "<?php echo $myname;?>";
alert(username);
</script>
The above code is correct and it will work, but the code below is totally wrong and it will never work.
<script>
var username = "syed ali";
var <?php $myname;?> = username;
alert(myname);
</script>
Pass value from JavaScript to PHP via AJAX
This is the most secure way to do it, because HTML content can be edited via developer tools and the user can manipulate the data. So, it is better to use AJAX if you want security over that variable. If you are a newbie to AJAX, please learn AJAX it is very simple.
The best and most secure way to pass JavaScript variable into PHP is via AJAX
Simple AJAX example
var mydata = 55;
var myname = "syed ali";
var userdata = {'id':mydata,'name':myname};
$.ajax({
type: "POST",
url: "YOUR PHP URL HERE",
data:userdata,
success: function(data){
console.log(data);
}
});
PASS value from JavaScript to PHP via hidden fields
Otherwise, you can create a hidden HTML input inside your form. like
<input type="hidden" id="mydata">
then via jQuery or javaScript pass the value to the hidden field. like
<script>
var myvalue = 55;
$("#mydata").val(myvalue);
</script>
Now when you submit the form you can get the value in PHP.

PHP runs on the server before the page is sent to the user, JavaScript is run on the user's computer once it is received, so the PHP script has already executed.
If you want to pass a JavaScript value to a PHP script, you'd have to do an XMLHttpRequest to send the data back to the server.
Here's a previous question that you can follow for more information: Ajax Tutorial
Now if you just need to pass a form value to the server, you can also just do a normal form post, that does the same thing, but the whole page has to be refreshed.
<?php
if(isset($_POST))
{
print_r($_POST);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="data" value="1" />
<input type="submit" value="Submit" />
</form>
Clicking submit will submit the page, and print out the submitted data.

We can easily pass values even on same/ different pages using the cookies shown in the code as follows (In my case, I'm using it with facebook integration) -
function statusChangeCallback(response) {
console.log('statusChangeCallback');
if (response.status === 'connected') {
// Logged into your app and Facebook.
FB.api('/me?fields=id,first_name,last_name,email', function (result) {
document.cookie = "fbdata = " + result.id + "," + result.first_name + "," + result.last_name + "," + result.email;
console.log(document.cookie);
});
}
}
And I've accessed it (in any file) using -
<?php
if(isset($_COOKIE['fbdata'])) {
echo "welcome ".$_COOKIE['fbdata'];
}
?>

Your code has a few things wrong with it.
You define a JavaScript function, func_load3(), but do not call it.
Your function is defined in the wrong place. When it is defined in your page, the HTML objects it refers to have not yet been loaded. Most JavaScript code checks whether the document is fully loaded before executing, or you can just move your code past the elements it refers to in the page.
Your form has no means to submit it. It needs a submit button.
You do not check whether your form has been submitted.
It is possible to set a JavaScript variable in a hidden variable in a form, then submit it, and read the value back in PHP. Here is a simple example that shows this:
<?php
if (isset($_POST['hidden1'])) {
echo "You submitted {$_POST['hidden1']}";
die;
}
echo <<<HTML
<form name="myform" action="{$_SERVER['PHP_SELF']}" method="post" id="myform">
<input type="submit" name="submit" value="Test this mess!" />
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<script type="text/javascript">
document.getElementById("hidden1").value = "This is an example";
</script>
HTML;
?>

You can use JQuery Ajax and POST method:
var obj;
$(document).ready(function(){
$("#button1").click(function(){
var username=$("#username").val();
var password=$("#password").val();
$.ajax({
url: "addperson.php",
type: "POST",
async: false,
data: {
username: username,
password: password
}
})
.done (function(data, textStatus, jqXHR) {
obj = JSON.parse(data);
})
.fail (function(jqXHR, textStatus, errorThrown) {
})
.always (function(jqXHROrData, textStatus, jqXHROrErrorThrown) {
});
});
});
To take a response back from the php script JSON parse the the respone in .done() method.
Here is the php script you can modify to your needs:
<?php
$username1 = isset($_POST["username"]) ? $_POST["username"] : '';
$password1 = isset($_POST["password"]) ? $_POST["password"] : '';
$servername = "xxxxx";
$username = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (username, password)
VALUES ('$username1', '$password1' )";
;
if ($conn->query($sql) === TRUE) {
echo json_encode(array('success' => 1));
} else{
echo json_encode(array('success' => 0));
}
$conn->close();
?>

Is your function, which sets the hidden form value, being called? It is not in this example. You should have no problem modifying a hidden value before posting the form back to the server.

May be you could use jquery serialize() method so that everything will be at one go.
var data=$('#myForm').serialize();
//this way you could get the hidden value as well in the server side.

This obviously solution was not mentioned earlier. You can also use cookies to pass data from the browser back to the server.
Just set a cookie with the data you want to pass to PHP using javascript in the browser.
Then, simply read this cookie on the PHP side.

We cannot pass JavaScript variable values to the PHP code directly... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
So it's better to use the AJAX to parse the JavaScript value into the php Code.
Or alternatively we can make this done with the help of COOKIES in our code.
Thanks & Cheers.

Use the + sign to concatenate your javascript variable into your php function call.
<script>
var JSvar = "success";
var JSnewVar = "<?=myphpFunction('" + JSvar + "');?>";
</script>`
Notice the = sign is there twice.

We Keep Coding

JavaScript is the programming language of the Web.