global variable empty if function is called using variable

Question

Why this doesn't work

$(document).on('click', '.title', function(){
    let fn = 'abc';
    $.post('common.php', {fn}, function(data) {
        console.log(data);
    });
});

common.php

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

$cols = '323';

function abc() {
    global $db;  // this works (db connection);
    global $cols;
    echo $cols;  //  doesn't work result is empty
    echo '323';  // this works
}

There is no logic - some global variables work (for example $db connection) and some don't work.

Any help?

It work in 3v4l.org/p630p But you should use ; after $db — Mohammad
– Mohammad, Commented Sep 27, 2018 at 6:29
@Mohammad, that's not my question. Pls read the title - if function is called using variable. Call abc() using if (isset($_POST['fn'])) {$_POST['fn']();} Typo corrected. — user7461846
– user7461846, Commented Sep 27, 2018 at 6:32
Because when you calling function in first line, $cols doesn't defined. But if you define variable before function call, it work correctly. 3v4l.org/8cUNj — Mohammad
– Mohammad, Commented Sep 27, 2018 at 6:35
it's working here link . are you sure that your function is executed? make sure that your isset($_POST['fn']) is true. — user4587108
– user4587108, Commented Sep 27, 2018 at 6:36

Mohammad · Accepted Answer · 2018-09-27 06:40:22Z

1

Because when you calling function in first line

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

$cols variable doesn't defined. But if you define variable before function call, it work correctly.

$cols = '323';
if (isset($_POST['fn'])) {$_POST['fn']();}

function abc() {...

Check result in demo

answered Sep 27, 2018 at 6:40

Mohammad

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