I have created a page for updation of record. I want to pass the id of student from one page to another. I am trying to send it through window.location but it is not working. In ajax I tried to navigate to other page but didn't succeed in that too. How can i pass the data and receive on other page without showing in query string?
ajax code is
var id = $(this).data('username');
$.ajax({
var id = $(this).data('username');
$.ajax({type: "POST",
cache: false,
data: id,
url: "Update.php",
success: function(dto)
{
//but I do not require this return call I just
// want to pass the data to update.php
}
});
//this is the code where the button is being clicked
<table class="table table-condensed" >
<thead style="background-color:#665851" align="center">
`<tr>
<td style="color:white">Roll No</td>
<td style="color:white">Name</td>
<td style="color:white">Department</td>
<td style="color:white">Click To Update</td>
</tr>
</thead>
<tbody style="background-color:whitesmoke; border:initial" id="tblBody" align="center">
<?php
$database="firstdatabase"; //database name
$con = mysqli_connect("localhost","root" ,"");//for wamp 3rd field is balnk
if (!$con)
{ die('Could not connect: ' . mysql_error());
}
mysqli_select_db($con,$database );
$state = "SELECT rollno ,name, dept FROM student ;";
$result = mysqli_query($con,$state);
$output = 1;
$outputDisplay = "";
$noRows = mysqli_affected_rows($result);
if($result)
{
$num = mysqli_affected_rows($con);
//$row = mysqli_fetch_array($result,MYSQLI_NUM);
while ($row = mysqli_fetch_array($result))
{
$r = $row['rollno'];
$n = $row['name'];
$d = $row['dept'];
$outputDisplay .= "<tr><td>".$r."</td><td>".$n."</td><td>".$d."</td><td align='right'>
<button type='button' name='theButton' value='Detail' class='btn' id='theButton' data-username='$r'> <img src='edit.jpg'/></button>
</td>
</tr>";
}
}
else
{
$outputDisplay .= "<br /><font color =red> MYSql Error No: ".mysqli_errno();
$outputDisplay .= "<br /> My SQl Error: ".mysqli_error();
}
?>
<?php
print $outputDisplay;
?>
</tbody>
</table>
If both pages are at same domain you can use localStorage, storage event to pass data between html documents
At second page
window.addEventListener("storage", function(e) {
// do stuff at `storage` event
var id = localStorage.getItem("id");
});
at first page
// do stuff, set `localStorage.id`
localStorage.setItem("id", "abc");
When you use window.location then your page go to another page. ajax work on active page. You can not use.
Generally you can use sessions for this $_SESSION variable to store it into the session, or you can pass that value via get parameter. And afterwards get that parameter with $_GET
Or $_POST parameter if you want to submit form.
you can try using cookies
Set the cookie
<?php
setcookie("name","value",time()+$int);
/*name is your cookie's name
value is cookie's value
$int is time of cookie expires*/
?>
Get the coockie
<?php
echo $_COOKIE["your cookie name"];
?>
Populate a <form method="post" [...]> in the first page with the information needed; you can change their aspect with CSS as desired.
When the <form> is send you only need a PHP script/page that uses $_POST to fill the new page.
Easier than AJAX if you try to navigate from the first page to the second.
If you already have a form and wanna post that to an update script, you could just add the student id as an hidden form element example:
<input type="hidden" name="student_id" value="<?php echo $student_id; ?>">
Else if you want to redirect from another page to the update page, with a student id, the best way will probably be a $_GET variable.
So the URL would look something like this: http://domain.com/update.php?student_id=1
And then your update.php will include a simple check like this.
if(!empty($_GET['student_id'])) {
$student_id = $_GET['student_id'];
// Ready to update
} else {
// Throw 404 error, or redirect to an create page
}
Related
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.
I'm learning PHP and SQL and as exercise I'm working on a page that is actually something like admin panel for a website that lists movies. I'm using lampp and phpmyadmin where I have created a simple database that contains two tables, movie list and users list.
Because I'm beginner and my code is probably messy, I'm describing what I tried to achieve. There's login.php page where the only functionality is typing username and password. If info matches info from SQL table, user proceeds to adminpanel.php.
This page should load a list of movies and create a table with that data. At the end of each row I want two buttons, edit and delete. What I'm trying to achieve is to delete current row where delete button is clicked, for delete button. Edit button should show hidden form just for the row where button was clicked. This form would contain button that actually updates data in SQL table after filling form and clicking the button. (I haven't added function that shows form yet, I care about buttons much more) Form for adding movies at the end of the file works.
Here's adminpanel.php
<html>
<head>
<script src="https://code.jquery.com/jquery-3.3.1.js"
integrity="sha256-2Kok7MbOyxpgUVvAk/HJ2jigOSYS2auK4Pfzbm7uH60="
crossorigin="anonymous">
</script>
<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/core.js"></script>
<script type="text/javascript" src="changes.js"></script>
<script type="text/javascript" src="https://stackpath.bootstrapcdn.com/bootstrap/4.1.3/css/bootstrap.min.css"></script>
<style type="text/css">
*{text-align: center;}
.skriveni_input{
display: none;
};
</style>
</head>
<?php
require_once('connection.php');
if(!isset($_POST['btnlogin'])){
exit;
}
$username = $_POST['username'];
$password = $_POST['password'];
$query = "SELECT usrname,password FROM usrs WHERE usrname='$username' AND password='$password' ";
$res = mysqli_query($conn,$query);
$rows = mysqli_num_rows($res);
if($rows == 1){
echo "Welcome ".$_POST['username']."<br><br>";
} else {
echo "<script>
alert('Wrong login info');
window.location.href='login.php';
</script>";
exit;
}
$query = "SELECT * FROM movies";
$result = $conn->query($query);
echo "<table align = center cellspacing = 0 border = 0;><thead><tr><th>Name</th><th>Year</th><th>Genre</th></tr></thead><tbody>";
while ($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo '<td id="row_id" style="display:none;" value="'.$row["movie_id"].'">'.$row["movie_id"].'</td>';
echo '<td>'.$row["name"].'</td>';
echo '<td>'.$row["year"].'</td>';
echo '<td>'.$row["genre"].'</td>';
echo '<td><input type="submit" name="edit" value="edit" data-index="' . $row['movie_id'] . '" class="btnedit" id="btnedit"></input></td>';
echo '<td><input type="submit" name="delete" value="delete" class="btndlt" id="btndlt"></input></td>';
echo "</tr>";
echo "<tr>
<td><input type='text' class='hidden_input' id='hidden_name" . $row['movie_id'] . "'placeholder='hidden name'></input></td>
<td><input type='text' class='hidden_input' id='hidden_year" . $row['movie_id'] . "'placeholder='hidden year'></input></td>
<td><input type='text' class='hidden_input' id='hidden_genre" . $row['movie_id'] . "'placeholder='hidden genre'></input></td>
</tr>";
}
echo "</tbody></table>";
?>
<h3>Add movie form: </h3>
<form action="" method="POST">
<label for="movie_name">Movie name : </label>
<input type="text" name="movie_name" id="movie_name">
<br><br>
<label for="movie_year">Year: </label>
<input type="text" name="movie_year" id="movie_year">
<br><br>
<label for="movie_genre">Genre: </label>
<input type="text" name="movie_genre" id="movie_genre">
<br><br>
<input type="submit" name="submit_movie" id="submit_movie" value="Submit">
</form>
</html>
Here's my javascript file with ajax calls:
$(document).ready(function(e){
$('#submit_movie').click(function(e){
e.preventDefault();
var movie_name = $('#movie_name').val();
var movie_year = $('#movie_year').val();
var movie_genre = $('#movie_genre').val();
$.ajax({
type: 'POST',
data: {movie_name:movie_name, movie_year:movie_year, movie_genre:movie_genre},
url: "insert.php",
success: function(result){
alert('Movie ' + movie_name + ' (' + movie_year + ')' +' added successfully.');
document.location.reload();
}
})
});
$('.btnedit').click(function(e){
var id = $(this).parent().prev().prev().prev().prev().html();
alert(id);
//unfinished function
})
$('.btndlt').click(function(e){
var id = $(this).parent().prev().prev().prev().prev().prev().html();
e.preventDefault();
$.ajax({
type: 'POST',
data: {id:id},
url: 'delete_row.php',
success: function(result){
alert('Successfully deleted.');
document.location.reload();
}
})
})
});
Here's php page for adding a movie, insert.php (this one works, posting it just for more information) :
<?php
require_once('connection.php');
if($_REQUEST['movie_name']){
$name = $_REQUEST['movie_name'];
$year = $_REQUEST['movie_year'];
$genre = $_REQUEST['movie_genre'];
$sql = "INSERT INTO movies(name, year, genre) VALUES ('$name','$year','$genre')";
$query = mysqli_query($conn, $sql);
}
?>
Here's delete_row.php file for deleting entry with delete button:
<?php
require_once('connection.php');
$id = $_REQUEST['id'];
if(isset($_REQUEST['delete'])){
$sql = "DELETE FROM `movies` WHERE movie_id = $id";
$query = mysqli_query($conn, $sql);
}
?>
As you can probably see I was all over the place with php and ajax because I tried to implement multiple solutions or mix them to solve the problem.
At this stage when I click delete button I get alert message that says erasing is successful and adminpanel.php reloads with list of movies. However the movie is still there and in SQL database.
When I tried to debug delete_row.php I found out that index "id" is undefined every time even though I think I'm passing it with ajax call.
Edit
I should've said that security is not my concern right now, I do this exercise just for functionalities I described. :) Security is my next step, I am aware this code is not secure at all.
When I tried to debug delete_row.php I found out that index "id" is
undefined every time even though I think I'm passing it with ajax
call.
The reason this happens is probably because you're accessing delete_row.php directly through the browser, and because the form is not submitted (it will later through ajax) the $_REQUEST variable will always be undefined.
When debugging $_REQUEST (or $_POST) variables in the future, you should use Postman where you can actually request that php file sending your own POST arguments.
On your specific code, the query will never run because of this line:
if(isset($_REQUEST['delete']))
Which is checking for a delete variable that was never sent in the first place, hence will always resolve false
Use this code instead on delete_row.php:
<?php
require_once('connection.php');
if(isset($_REQUEST['id'])){
$id = $_REQUEST['id'];
$sql = "DELETE FROM `movies` WHERE movie_id = $id";
$query = mysqli_query($conn, $sql);
}
?>
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.
i have a dynamic table that has a list of records, basically for all rows of the table there is a link to click, that passes the id of that row clicked to the next page,
my question is how can i save and display the number of times each link have been clicked ?
here is my code
<table width="100%" >
<tr bgcolor="#FF3399" style="color:#FFF">
<td><h3><strong>Topic</strong></h3></td>
<td><h3>Author</h3></td>
<td><h3>Date</h3></td>
<td><strong>Replies</strong></td>
<td><strong>Views</strong></td>
</tr>
<?php do { ?>
<tr bgcolor="#009900" style="color:#FFF">
<td><h4><a style="color:#FFF" onclick="spinn();" data-ajax="false" href="send.php?id=<?php echo $row_forum['id']; ?>"><strong><?php echo $row_forum['Topic']; ?></strong></a></h4></td>
<td bgcolor="#009900"><h4><?php echo $row_forum['Author']; ?></h4></td>
<td><h4><?php echo date("g : i a, j/F/Y,",strtotime($row_forum['Date'])); ?></h4></td>
<td> <?php echo $row_forum['Replies']; ?></td>
<td><?php echo $row_forum['Views']; ?></td>
</tr>
<?php } while ($row_forum = mysql_fetch_assoc($forum)); ?>
</table>
On the page that is the target of your link, simply invoke a method which will increase your count field in the database. You can put code like this in your target page:
<?php
function increaseCount() {
// ... connect to db
mysql_query('UPDATE table SET count=count+1 WHERE id='.$id);
}
increaseCount($_GET['id']);
?>
Please note, that the above code is very trivial and dangerous and only shows the concept. It does not protect you from SQL Injection at all. Doesn't even check, if the id parameter in the address exists.
Also, don't use methods like mysql_query - they're depracated.
Alternatively, try using Ajax to change the behavior of link's click event - send an ajax request to PHP script which will increment your count field in the database, and then follow your link's href :)
XmlHttpRequest
jQuery.ajax
Edit
$topicId = $_GET['id']; // you need to change 'id' to the name of your ID-parameter in the URL
$viewsIncrementQuery = "UPDATE `topic` SET `Views` = `Views` + 1 WHERE `id` = " . $topicID;
$incremented = mysql_query($viewsIncrementQuery, $epl) or die(mysql_error());
You need to put that code somewhere close to where you extract data from the database, so probably after the $query_lin = sprintf( (...) part.
If it throws an error - show it's message.
I am doing a jquery refresh on a div containing a mysql query. The query however needs to grab a variable from the URL. It works, but as soon as it does the refresh it stops grabbing the variable and thus the query doesn't work.
Here is my refresh code:
<script type="text/javascript">
var auto_refresh = setInterval(
function ()
{
$('#records').load('user_records.php').fadeIn("slow");
}, 30000); // refresh every 30 seconds
</script>
and then my basic query...
$username = $_GET['user'];
echo $username; //for testing
echo "<div id='records'><h1 align='center'>Today's Transfers</h1>
<table cellspacing='0' cellpadding='0'>
<tr>
<th> Customer Name</th><th>Phone Number</th><th>Disposition</th><th>DID</th><th>Date Called</th>
</tr>
";
$sql = "SELECT * FROM users WHERE username = '$username'";
$result = mysqli_query($link, $sql, MYSQLI_STORE_RESULT);
while($row = $result->fetch_assoc()){
$user_id = $row['id'];
}
followed by echoing my results, etc, etc.
The query gets the username and works when the page load, but when it does the refresh it stops getting ther $username variable, even though it is still in the url. Is there a way to keep getting the variable each time? Thanks for any help!
You don't seem to pass the user to the server. If you can access the username in javascript somewhere you can do the following.
$('#records').load('user_records.php?user='+username).fadeIn("slow");