I have a MYSQL Table called users.
I also have a column called online_status.
On my page I want a user to be able to toggle their status as 'Online' or 'Offline' and have this updated in the database when they click on the div using Ajax, without refreshing the page.
Here's my PHP/HTML code:
<?php if ($profile['online_status'] == "Online") {
$status = "Offline";
}else{
$status = "Online";
} ?>
<div id="one"><li class="far fa-circle" onClick="UpdateRecord(<? echo $profile['online_status']; ?>);"/></li><? echo 'Show as ' .$status; ?></div>
My Ajax:
<script type="text/javascript" src="/js/jquery.js"></script>
<script>
function UpdateRecord(id)
{
jQuery.ajax({
type: "POST",
url: "update_status.php",
data: 'id='+id,
cache: false,
success: function(response)
{
alert("Record successfully updated");
}
});
}
</script>
update_status.php
<?php
$var = #$_POST['id'] ;
$sql = "UPDATE users SET online_status = 'Offline' WHERE user_id = 1";
$result = mysqli_query($conn,$sql) or die(mysqli_error($conn));
//added for testing
echo 'var = '.$var;
?>
I am currently getting no alert, nothing is being updated in my database either. Please can someone help me improve/fix the code to get it to work? Also, if there's a way of eradicating the need for the update_status.php file and have the ajax self post then this would be preferred.
Thank you in advance.
From what i see, the reason why no alert pops up nor nothing gets updated is because of the onclick() on button you have. Add quotes around the parameter to the update function. As you have it, javascript sees the parameter as a javascript variable as $profile['online_status']; is a string.
If you had debugged your code, you should see an error pointing towards the onclick() line
Change this
onClick="UpdateRecord(<? echo $profile['online_status']; ?>);"
To
onClick="UpdateRecord('<? echo $profile['online_status']; ?>');"
Also you are hardcoding the where clause in your update statement. You should be using the $_POST['id'] variable via prepared statements
pass data to PHP file
data: { id: id },
add a database connection to your PHP file
<?php
$var = $_POST['id'] ;
$sql = "UPDATE users SET online_status = 'Offline' WHERE user_id = '$var'";
$result = mysqli_query($conn,$sql) or die(mysqli_error($conn));
?>
If you still see any errors then press F12 and go to network tab, then click on that div, network tab will record your ajax file returns, you can check there on by selecting your php file's response, hope it helps
I have to transfer data from one div to another, I am using AJAX to do this.
<script type="text/javascript" src="lib/jquery-1.6.2.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#aq").click(function(){
var name1 = $("#n1").val();
$.ajax({
type: "POST",
url: "risultato.php"
data: "name1=" + name1 ,
dataType: "html",
success: function(msg)
{
$("#risultato1").html(msg);
},
error: function()
{
alert("Chiamata fallita, si prega di riprovare...");
}
});
});
});
</script>';
<form name="modulo1'.$dationennx['id'].'">
<input type="hidden" name="name1" value="'.$dati['id'].'"
id="n1'.$dationennx['id'].'">
<a href="javascript:rispondithread(\'homeq\');"
id="aq">'.stripslashes($dationennx['oggetto']).'</a><br>
</form>
<script>
function rispondithread(h) {
$("#rispondithreadforum").attr("style", "display:block;");
}
</script>`
I am fetching the data from my table from the 'risultato.php' page, which i want to use to show a textarea on my main page with the fetched data.
<?php
$nome = $_POST['name1'];
$query = "SELECT * FROM login2.podcast
WHERE login2.podcast.id = '$nome'
ORDER BY login2.podcast.data DESC";
$dati = mysql_query($query);
while($ris = mysql_fetch_array($dati) ){
echo'
<textarea class="form-control textareaabc" readonly tabindex="8">'.stripslashes($ris['testo']).'</textarea>';
}
?>
It doesn't work if i try to fetch the data using mysql_query, but it does when i try echoing the post data in the page.
$nome = $_POST['name1'];
echo $nome
This writes the '$nome' variable in my main page.
$nome = $_POST['name1'];
echo'<input type="text" value="'.$nome.'" name="nome">';
i don't understand this. why it doesn't work? what's wrong?
It is highly likely that there is no data for the '$nome' variable in your table
Make sure that you are actually receiving data from the database, does it print out the textarea? If not, you do not have any id in your podcast table table matching the '$nome' variable.
Testing your code
Try checking if you actually get anything back when you print something out in that page, could you possibly be pointing to the wrong page?
other
Overall, i would recommend using PDO or atleast mysqli, MySQL is no longer supported since PHP 7 and deprecated since PHP 5. See: PHP.net documentation about mysql extension
sorry, i forget to include the connection to the database into the file risultato.php :P
thanks to everybody
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 a total AJAX noob, so please forgive me, but this is what I'm trying to do...
I have a php form that submits the information via ajax to a parser file. I need to get a few ids from that form to the parser file so I can use them in my sql update. I'll try to keep my code simple but give enough info so someone can answer.
My form is being generated via a foreach loop that iterates through a list of teams and grabs their various characteristics. For simplicity, let's say the main thing I need to get to the parser file is that team_id.
I'm not sure if I need to add
<input type="hidden" name="team_id" value="<?=$team->id ?>">
or
<tr data-teamid="<?=$team->id; ?>">
or something like that to my form....but either way, it gets passed through this AJAX file...
<script type="text/javascript">
function updateNames() {
jQuery('#form-message, #form-errors').html("");
var post_data = jQuery('form[name="update_names"]').serialize();
$.ajax({
url: 'parsers/update_names.php',
method: 'POST',
data : post_data,
success: function(resp) {
if(resp == 'success'){
jQuery('#form-message').html("Names and Scores have been Updated!");
}else{
jQuery('#form-errors').html(resp);
}
}
});
return false; // <--- important, prevents the link's href (hash in this example) from executing.
}
jQuery(document).ready(function() {
$(".linkToClick").click(updateNames);
});
</script>
And is making it to my parser file, which looks like this...
require_once '../core/init.php';
$db = DB::getInstance();
$errors = [];
// $camp_id = Input::get('camp_id');
$camp_id = 18;
//Find the Teams that Belong to the Camp
$sql = "SELECT * FROM teams WHERE camp_id = $camp_id";
$teamsQ = $db->query($sql);
$all_teams = $teamsQ->results();
//validation and sanitization removed for simplicity.
if(empty($errors)){
$fields = [];
foreach($_POST as $k => $v){
if($k != 'camp_id'){
$fields[$k] = Input::get($k);
}
}
$db->update('teams',$all_teams->id,$fields);
echo 'success';
}else{
echo display_errors($errors);
}
SO. The main question I have is how do I get that camp_id and team_id into the parser file so I can use them to update my database?
A secondary question is this...is the fact that the form is being generated by a foreach loop going to make it difficult for the ajax to know which field to update?
So, how would I get that camp_id to
$sql = "SELECT * FROM teams WHERE camp_id = $camp_id";
And the team_id to
$db->update('teams',$all_teams->id,$fields);
I tried to break this down to the simplest form and it's still not getting to the function. This code...
<form name="update_names" method="post">
<input type="hidden" name="team_id" value="<?=$teams->id ?>">
<button onclick="updateNames();return false;" class="btn btn-large btn-primary pull-right">test</button>
<script type="text/javascript">
function updateNames() {
alert('test');
}
</script>
Gives me... Uncaught ReferenceError: updateNames is not defined
The jQuery .serialize() method uses the name attribute of an element to assign a variable name. It ignores the element's id, any classes and any other attribute. So, this is the correct format if using .serialize():
<input type="hidden" name="team_id" value="<?=$team->id ?>">
Looking at your ajax code, your parser file would be called parsers/update_names.php.
To verify that the desired field is getting to your parser file, add this to the top for a temporary test:
<?php
$tid = $_POST['team_id'];
echo 'Returning: ' .$tid;
die();
and temporarily modify the ajax code block to:
$.ajax({
url: 'parsers/update_names.php',
method: 'POST',
data : post_data,
success: function(resp) {
alert(resp);
{
});
return false;
If the ajax processor file (your "parser") receives the team_id data, then you will get that data returned to you in an alert box.
Thus, you can now determine:
1. That you are receiving the team_id information;
2. That the ajax back-and-forth communications are working
Note that you also can install FirePHP and echo text to the browser's console from the php processor file.
I am trying to send data from Javascript to a php script. I am new to PHP,javascript and JSON etc. I am tryin to use jQuery Upvote plugin. I am reading the documentation from readme.md here. However, I find it too vague for my understanding.
I have added the following code in HTML body. The plugin is working fine in front-end.
<div id="topic" class="upvote">
<a class="upvote"></a>
<span class="count">0</span>
<a class="downvote"></a>
<a class="star"></a>
</div>
<div id="output"> </div>
<script language="javascript">
var callback = function(data) {
$.ajax({
url: '/vote',
type: 'post',
data: { id: data.id, up: data.upvoted, down: data.downvoted, star: data.starred }
});
};
$('#topic').upvote({id: 123, callback: callback});
</script>
Can someone please tell how should I write the php counterpart so as to get the information about change of state of plugin? I think it is related to json maybe but I am stuck here..
EDIT: As per the suggestion, I have modified a bit of code and wrote PHP script.
In above code I have changed URL to my PHP script url: 'voter.php'.
This is the PHP code (in brief) :
<?php require_once('Connections/conn.php'); ?>
mysql_select_db($database_conn, $conn);
if ($_POST['up']!=0){
$query_categories = "UPDATE user SET score=1 WHERE u_id =5";
$categories = mysql_query($query_categories, $conn) or die(mysql_error());
}
else if ($_POST['down']!=0) {
$query_categories = "UPDATE user SET score=-1 WHERE u_id =5";
$categories = mysql_query($query_categories, $conn) or die(mysql_error());
}
The database connection is working fine, but still the database values are not updated by above code..
Your script sends AJAX Post request to the php script. You can access those post variables like this, using $_POST[] array:
$id = $_POST['id'];
$up = $_POST['up'];
$down = $_POST['down'];
$star = $_POST['star'];
Then you can populate the mysql table, etc.