mysql_query executes automatically but not when onClick

Question

I'm a newbie in php and I need help with this. The query runs every time on page load and not when I click my button.

<input type="button" name="button" onclick="<?php 
                  $temp_id = $row_RecStudent['stu_id'];
                  mysql_query("UPDATE tbl_studentdetail SET stu_queue = 0 WHERE stu_id = $temp_id"); ?>" value="Unqueue" />

This is so frustrated because I run this button in a loop but every time the button loads it automatically run the mysql_query inside not when I click it.

SOLVED! Check on AnthonyB's answer.

you can't attach a PHP code to a javascript onclick event. instead the onclick can trigger an ajax call to run the update — Mohamed Yasin
– Mohamed Yasin, Commented Nov 29, 2016 at 7:06
that is not how this works: all the PHP is executed before anything is sent to the browser, and all the javascript is executed after everything is sent to the browser. you should read some tutorials about how webservers work. and you shouldn't keep on using mysql-functions, they are deprecated and in PHP7 removed. use mysqli or PDO — Franz Gleichmann
– Franz Gleichmann, Commented Nov 29, 2016 at 7:13
Thanks for the replies. Indeed I knew nothing about this previously and now I've learnt a lot more in these hours compared to last few days. — Kenny
– Kenny, Commented Nov 29, 2016 at 9:14

e4c5 · Accepted Answer · 2016-11-29 07:05:15Z

3

The query runs every time on page load and not when I click my button.

Which is because onclick events are usually executed in the browser with client side scripting, which invariably means javascript. What you really need to have is

 <input type="button" name="button" onclick="js_function()" value="Unqueue" />

Where js_function is a javascript that will make an ajax request to the page or simply cause the form to be submitted. What you are doing is making onclick equal to the result of the PHP code fragment, which obviously happens at the server side and has absolutely nothing at all to do with the user clicking on the button.

answered Nov 29, 2016 at 7:05

e4c5

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3 Comments

Anthony Over a year ago

PHP is a server side language. That means it is executed once, just before return HTML to client. JavaScript is here a client side language. That means it is executed by the browser (not by the server). In your case, PHP execute the query "UPDATE ...", and then return html. What you want, as said by @e4c5, is make an Ajax request to do this. An AJAX request will send an HTTP request to your server, and your server will execute update query.

e4c5 Over a year ago

thank you @AnthonyB for the comment. e4c6 is the Caro-Kann while my preferencce is for the Sicilian

Anthony Over a year ago

I did'nt understand chess reference, but I fixed my comment. You could add example of AJAX request (perhaps using jQuery) with the server side code, in order to your answer be more useful.

Anthony · Accepted Answer · 2016-11-29 08:10:57Z

1

You should use an AJAX request, because there is no redirection nor reloading. To use the exemple below, you must include jQuery

<!-- data-id is the id you got before -->    
<input type="button" data-id="<?php echo $row_RecStudent['stu_id']; ?>" name="button" class="button-on-click" value="Unqueue" />

<script type="text/javascript">
    $(function() {
        //On click on this kind of button
        $(".button-on-click").on('click', function() {
            var id = $(this).attr('data-id');
            //Send AJAX request on page youraction.php?id=X where X is data-id attribute's value
            //the ID used in mysql query
            $.ajax({
                url: "youraction.php",
                method: "GET",
                data: {
                    id: id
                }
            }).done(function() {
                //Once it is done
                console.log("OK!");
            });
        });
    });
</script>

On youraction.php

<?php
//Cast $_GET['id'] to integer to prevent SQL injection.
//If there is no id given, stop
if (empty($_GET['id'])) {
    exit();
}
$id = (int)$_GET['id'];
mysql_query("UPDATE tbl_studentdetail SET stu_queue = 0 WHERE stu_id = $id");

Be careful, mysql_* functions are deprecated. Instead, use mysqli_* functions or PDO. And you could prefere use prepared statement for security reason.

edited Nov 29, 2016 at 8:10

answered Nov 29, 2016 at 7:35

Anthony

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8 Comments

Kenny Over a year ago

I appreciate this so much. But there an issue at line data: {

Anthony Over a year ago

Indeed I fixed it. There were errors because of parenthesis and a coma in data. Now that should work.

e4c5 Over a year ago

good answer +1 from me for including an ajax sample and for the warning about mysql_*

Anthony Over a year ago

Thanks @e4c5! I was inspired by your answer.

Kenny Over a year ago

Your code is quite clear but I still don't manage to get it to work. I've been rechecking but when I click on button just nothing happen.

|

Indrasis Datta · Accepted Answer · 2016-11-29 07:08:17Z

0

Please note that you're supposed to mention a Javascript function for onClick event. You can't write PHP script here.

One way to overcome this problem is create a separate PHP file which will contain the query.

// page1.php

<a href="action.php?stu_id=<?php echo $row_RecStudent['stu_id'] ?>">Unqueue</a>

// action.php

<?php 
    $temp_id = $_GET['stu_id'];
    mysql_query("UPDATE tbl_studentdetail SET stu_queue = 0 WHERE stu_id = $temp_id"); 
    header("Location:page1.php");
?>"

answered Nov 29, 2016 at 7:08

Indrasis Datta

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1 Comment

Anthony Over a year ago

Using a link is not a bad idea, instead of using an AJAX request. But it cause a reloading of the page. If this is not a problem, you can use it. But some times, a reloading is a problem.

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