PHP: variable not working inside of function?

Question

echo $path; //working
function createList($retval) {
    echo $path; //not working
    print "<form method='POST' action='' enctype='multipart/form-data'>";
    foreach ($retval as $value) {
            print "<input type='checkbox' name='deletefiles[]' id='$value' value='$value'>$value<br>";
    }
    print "<input class='submit' name='deleteBtn' type='submit' value='Datei(en) löschen'>";
    print "</form>";    
}

what am I doing wrong? why is $path printed correctly outside of the createList function, but it's not accessible inside the function?

Where is $path defined? Is it defined as global $path?

Marius Schulz
– Marius Schulz

2010-06-14 21:44:42 +00:00
Commented Jun 14, 2010 at 21:44 — Marius Schulz
– Marius Schulz, Commented Jun 14, 2010 at 21:44

Joshua Pinter · Accepted Answer · 2020-09-25 14:14:45Z

52

Because it's not defined in the function.

There are a few ways to go about this:

1) Use what Alex said by telling the function it is a global variable:

echo $path; // working

function createList($retval) {
  global $path;

  echo $path; // working
}

2) Define it as a constant:

define(PATH, "/my/test/path"); // You can put this in an include file as well.
  
echo PATH; // working

function createList($retval) {
  echo PATH; // working
}

3) Pass it into the function if it's specific to that function:

echo $path; // working

function createList($retval, $path) {
  echo $path; // working
}

Based on how the function really works, one of those will do ya.

edited Sep 25, 2020 at 14:14

answered Jun 14, 2010 at 21:54

Joshua Pinter

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Comments

Alex Ciminian · Accepted Answer · 2010-06-14 21:45:34Z

10

If you want it to work, you should use global $path in the function, so it looks outside the function scope.

Please note that global variables are sent from hell :).

answered Jun 14, 2010 at 21:45

Alex Ciminian

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Comments

Crozin · Accepted Answer · 2010-06-14 21:46:26Z

2

Cause $path inside createList() and outside it (in global scope) are two different variables. Read more about variable scope in PHP.

answered Jun 14, 2010 at 21:46

Crozin

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Comments

Aif · Accepted Answer · 2010-06-14 21:45:45Z

1

you must use the global modifier.

echo $path;
function createList($retval) {
    global path;
    echo $path; // works now :)

answered Jun 14, 2010 at 21:45

Aif

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Comments

thetaiko · Accepted Answer · 2010-06-14 21:49:56Z

0

As an alternative to using a global variable, just pass $path in. Of course, if you don't need the variable inside the function, don't bother!

echo $path;
function createList($retval, $path) {
    echo $path;
    print "<form method='POST' action='' enctype='multipart/form-data'>";
    foreach ($retval as $value) {
            print "<input type='checkbox' name='deletefiles[]' id='$value' value='$value'>$value<br>";
    }
    print "<input class='submit' name='deleteBtn' type='submit' value='Datei(en) löschen'>";
    print "</form>";    
}

answered Jun 14, 2010 at 21:49

thetaiko

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Comments

dippas · Accepted Answer · 2022-07-18 13:50:44Z

0

Just use $GLOBALS['your variable name'] and use it anywhere in your function

edited Jul 18, 2022 at 13:50

dippas

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answered Apr 5, 2022 at 10:41

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Because it's not defined in the function.

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