subtract when clicked button php

Question

i have a button, i want it every time i click the button it'll subtract from my credit on database (MySQL). i tried everything that already answered on another question but still unable to solved my problem.

here's my code

<input type="submit" name="submit" value="Check" class="button_x" style="width:100px" >
<?php
if($_SERVER['REQUEST_METHOD'] == "POST" and isset($_POST['submit']))
{
    func();
}
function func()
{
    $query = "UPDATE users SET user_credit = user_credit - 2";
    $result = mysqli_query($conn, $query);
}
?>

if im not using the if function, it work! but it would subtract the credit when i click the sub menu as well. or should i add 'where' at my query?

thank you.

@jon stirling do i really have to provide an 3v4l... oh. oh dear. — Ne Ma, Oct 23 '18 at 14:58
Note: The object-oriented interface to `mysqli` is significantly less verbose, making code easier to read and audit, and is not easily confused with the obsolete `mysql_query` interface. Before you get too invested in the procedural style it’s worth switching over. Example: `$db = new mysqli(…)` and `$db->prepare("…")` The procedural interface is an artifact from the PHP 4 era when `mysqli` API was introduced and should not be used in new code. — tadman, Oct 23 '18 at 15:13

GrumpyCrouton · Accepted Answer · 2018-10-23T14:57:11.263

1

I think your problem is you are not actually doing a POST, your form is probably using GET.

Another thing is you don't really need to check $_SERVER['REQUEST_METHOD'] at all. Try this code:

if(isset($_GET['submit']))
{
    func();
}

If the above works, then my assumption was correct, you'll either need to use $_GET or change your form to use $_POST.

You can change your form to use $_POST by adding method="POST" to your form element around your button.

<form method="POST">
    <input type="submit" name="submit" value="Check" class="button_x" style="width:100px">
</form>

If the above doesn't work, I suspect your $_SERVER['REQUEST_METHOD'] variable is empty somehow (Could be a something with your sever), in which case, the following should work:

if(isset($_POST['submit']))
{
    func();
}

Another way you could do this is to not care if it's POST or GET, by doing it this way:

if(isset($_REQUEST['submit']))
{
    func();
}

edited Oct 23 '18 at 14:57

answered Oct 23 '18 at 14:45

GrumpyCrouton

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@JonStirling Because it's not PHP's syntax? – Sam Oct 23 '18 at 14:48
@Samuel Yes it is? – Jonnix Oct 23 '18 at 14:48
@JonStirling Is that valid syntax? I've never seen it in practice, but now looking at the manual it appears it is correct. Could possibly be `$_SERVER['REQUEST_METHOD']` being empty maybe? – GrumpyCrouton Oct 23 '18 at 14:48
http://php.net/manual/en/language.operators.logical.php same as `&&` but with different precedence I think? – Jonnix Oct 23 '18 at 14:48
@JonStirling yes you are right... It's valid logic... My bad – Sam Oct 23 '18 at 14:49
The OP didn't post `form` code, so the method could be `GET` – AnTrakS Oct 23 '18 at 14:49
2

@D.Dimitrov could be... but at this point it is all but speculations. – Sam Oct 23 '18 at 14:51
@Samuel I agree it's all speculation, but that is really the only reasoning I can think of this would happen, unless it's an issue with `$_SERVER['REQUEST_METHOD']` being empty, but my answer would solve that as well. – GrumpyCrouton Oct 23 '18 at 14:52

score 0 · Answer 2 · answered Oct 23 '18 at 14:52

Have you put it inside a <form> tag? Like this:

<form action="" method="post">
    <input type="submit" name="submit" value="Check" class="button_x" style="width:100px">
</form>
<?php
if($_SERVER['REQUEST_METHOD'] == "POST" && isset($_POST['submit']))
{
    func();
}
function func()
{
    $query = "UPDATE users SET user_credit = user_credit - 2";    
    $result = mysqli_query($conn, $query);
}

Fabian · Answer 3 · 2018-10-23T15:02:29.687

Have you checked to make sure that the $conn variable you are using is accessible inside the function? It's difficult to tell since I can't see the rest of your code, but if $conn is created outside the function it will not be reachable from inside. A solution to this would be to either create it again inside the function or to pass it in as a parameter like such:

func($conn);

function func($conn)
{
    $query = "UPDATE users SET user_credit = user_credit - 2";
    $result = mysqli_query($conn, $query);
}

Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?

I missed that part. I guess the code we see in func() is incomplete since this apparently works despite the scope issues. — Fabian, Oct 24 '18 at 06:28

subtract when clicked button php

3 Answers3