Why am I not able to update my sql database using AJAX?

Question

$(document).ready(function(){
    $('.button').click(function(){
        var clickBtnValue = $(this).val();
        var ajaxurl = 'functions/delivered.php',
        data =  {'action': clickBtnValue};
        $.post(ajaxurl, data, function (response) {
            // Response div goes here.
            alert("action performed successfully");
        });
    });
});

This is the ajax code and I am trying to update my database

functions/delivered.php:

<?php
$db=new mysqli("localhost","root","","restaurant")
if(isset($_POST['onum'])){
    $onum=$_POST['onum'];
    $query="UPDATE `revenue` SET `payment`='y' WHERE onum=$onum";
    $db->query($query);
}
?>

Please describe what's not working. What are you expecting to happen? What actually happened? What error messages are you getting? Please see [**How do I ask a good question?**](https://stackoverflow.com/help/how-to-ask) and [**What topics can I ask about here?**](https://stackoverflow.com/help/on-topic). — Alex Howansky, Jan 16 '20 at 17:58
Your post request only has one key on it, `action`. There is no `onum` key included. — Taplar, Jan 16 '20 at 17:58
Also please note that your code is vulnerable to [**SQL injection**](https://en.wikipedia.org/wiki/SQL_injection) attacks. Instead of building queries with string concatenation, always use [**prepared statements**](https://secure.php.net/manual/en/pdo.prepare.php) with [**bound parameters**](https://secure.php.net/manual/en/pdostatement.bindparam.php). See [**this page**](https://phptherightway.com/#databases) and [**this post**](https://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) for some good examples. — Alex Howansky, Jan 16 '20 at 17:58
`$db=new mysqli("localhost","root","","restaurant")` < you didn't close that off with a `;` and should have thrown you a parse error notice for it. — Funk Forty Niner, Jan 16 '20 at 18:03

score 2 · Answer 1 · answered Jan 16 '20 at 18:04

2

You're not sending an onum parameter in the AJAX call.

$(document).ready(function(){
    $('.button').click(function(){
        var clickBtnValue = $(this).val();
        var onumValue = $("#onum").val();
        var ajaxurl = 'functions/delivered.php',
        data =  {'action': clickBtnValue, onum: onumValue};
        $.post(ajaxurl, data, function (response) {
            // Response div goes here.
            alert("action performed successfully");
        });
    });
});

Replace $("#onum") with the actual selector for the input that contains the value you want to send to PHP.

answered Jan 16 '20 at 18:04

Barmar

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You missed something for the "php" side of things. See [my comment](https://stackoverflow.com/questions/59775331/why-am-i-not-able-to-update-my-sql-database-using-ajax#comment105694774_59775331) about that. – Funk Forty Niner Jan 16 '20 at 18:04
1

Could just be a copying error, your comment seems to be sufficient for that typo. – Barmar Jan 16 '20 at 18:08
Could be, but we've seen more than our share of those. – Funk Forty Niner Jan 16 '20 at 18:09
Either way, the comment is enough. Putting it in the answer doesn't add any value. – Barmar Jan 16 '20 at 18:10
Ok. Well, let's see what the OP has to say. They seem to not be responding or have logged out. – Funk Forty Niner Jan 16 '20 at 18:13
I really don't care. Typo-squashing is for comments. – Barmar Jan 16 '20 at 18:14
I tried putting action instead of onum but it still doesn't work – Xen Jan 18 '20 at 07:14
Add the HTML to the question. – Barmar Jan 18 '20 at 15:53

Why am I not able to update my sql database using AJAX?

1 Answers1