cannot get php global variables using ajax

Question

if(!isset($_GET['t'])){$table = 'notes';}
else{$table = $_GET['t'];}

$cats = $table . '_cats';

include 'test.php';

test.php

test(); // works fine

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

function test(){
    global $cats, $table;
    echo $cats;
    echo $table;
}

js

$(document).on('click', '.atitle', function(){
    $.post('test.php', {fn: 'test'}, function(data){
        console.log(data); // empty
    });
});

I'm expecting$cats and $table written in console

Any help?

are you confirmed to received test in $_POST['fn'] ? please check first — prasanth
– prasanth, Commented Mar 28, 2020 at 5:00
@prasanth - yes because echo 'lorem' inside fn test - works — qadenza
– qadenza, Commented Mar 28, 2020 at 5:03
What is the file name where you written include 'test.php'; ? — Niklesh Raut
– Niklesh Raut, Commented Mar 28, 2020 at 5:07

Niklesh Raut · Accepted Answer · 2020-03-28 05:27:54Z

2

As your first file name is a_notes.php then call a_notes.php from ajax. You are defining $cats, $table in a_notes.php not in test.php.

$(document).on('click', '.atitle', function(){
    $.post('a_notes.php', {fn: 'test'}, function(data){
        console.log(data);
    },'json');// set dataType as json
});

Also in test.php, As you are calling a function test() then echo returned value of function.

echo test();

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

function test(){
    global $cats, $table;
    return json_encode(["cats"=>$cats,"table"=>$table]);
}

edited Mar 28, 2020 at 5:27

answered Mar 28, 2020 at 5:11

Niklesh Raut

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2 Comments

qadenza Over a year ago

but test.php is included in a_notes.php. I tried to define cats and table inside test.php - the same result. a_notes.php is the main file.

Niklesh Raut Over a year ago

@qadenza: Call $.post('a_notes.php'

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