AJAX function on button click not running PHP after the first time

Question

I have the following function on a button:

$(function() {
    $("#submit").click(function(){
        $.ajaxSetup({ cache: false });
        $.ajax({
        url:'crfile.php',
        type:'GET',
    });
};

It calls the following PHP file:

if (session_status() == PHP_SESSION_ACTIVE){
    session_destroy();
}
session_start();

$extension = ".txt";
$_SESSION['fname'] = substr(md5(rand()), 0, 7).$extension;
fopen("temp/" . $_SESSION['fname'], 'w');
$fname = $_SESSION['fname'];

It creates a file with a random name in the temp/ folder. Both are working properly except when I click the #submit button a second time without reloading the page. It reruns the function, however it doesn't generate a new filename in the PHP script.

I thought it was cache or sessions, however neither is the case. I am using PHP 5.5.

Can someone please advise?

YONGSOO KIM · Accepted Answer · 2015-07-06 01:07:23Z

1

At first, you need to check your ajax function. trying Use the method 'post', or use on('click',function(){ .....}) for your code.

Secondly, you need to make sure your ajax function really works. use 'print' and 'exit' for debugging.

if (session_status() == PHP_SESSION_ACTIVE){
    session_destroy();
}
session_start();

$extension = ".txt";
$_SESSION['fname'] = substr(md5(rand()), 0, 7).$extension;
//debugging command
print  $_SESSION['fname'];
print  'see this?';
exit;


fopen("temp/" . $_SESSION['fname'], 'w');
$fname = $_SESSION['fname'];

this way you will know whether your ajax work or not. You can use debugging command at the end of the code above.

Try those codes and let me know if you need more help.

answered Jul 6, 2015 at 1:07

YONGSOO KIM

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