Using variable variables to populate array results in error

Question

I have the following code:

function filterUsers(array $setOfAllUsers) {
   if (empty($setOfAllUsers)) {
      return array(array(), array());
   }

   $activeUsers   = array();
   $inactiveUsers = array();
   foreach($setOfAllUsers as $userRow) {
       $var = ($userRow['IsActive'] ? '' : 'in') . 'activeUsers';

       $$var[$userRow['CID']]['Label'] = $userRow['UserLabel'];
       // Error happens here ---^

       $$var[$userRow['CID']]['UserList'][$userRow['UID']] = array(
           'FirstName' => $userRow['FName'],
           'LastName'  => $userRow['LName'],
           ... More data
       );
   }

   return array($activeUsers, $inactiveUsers);
}

I get the following error: Warning: Illegal string offset 'Label' in ...

How can I fix this? I tried defining Label part first like this: $$var[$userRow['CID']] = array(); $$var[$userRow['CID']]['Label'] = ''; but did not work.

To make things clear what I am trying to achieve is this:

if ($userRow['IsActive']) {
   $activeUsers[$userRow['CID']]['Label'] = $userRow['UserLabel'];

   $activeUsers[$userRow['CID']]['UserList'][$userRow['UID']] = array(
           'FirstName' => $userRow['FName'],
           'LastName'  => $userRow['LName'],
           ... More data
   );
} else {
   $inactiveUsers[$userRow['CID']]['Label'] = $userRow['UserLabel'];

   $inactiveUsers[$userRow['CID']]['UserList'][$userRow['UID']] = array(
           'FirstName' => $userRow['FName'],
           'LastName'  => $userRow['LName'],
           ... More data
   );
}

Instead of repeating above in if/else I wanted to achieve it using $$

How do you create $setOfAllUsers and what does that array look like. Example only please. — RiggsFolly
– RiggsFolly, Commented Aug 14, 2015 at 14:23
@RiggsFolly its a function so its being passed in, in some cases its retrieved from database in some cases its built on the fly — GGio
– GGio, Commented Aug 14, 2015 at 14:25
This error is often when you don't have array. In this case it's $setOfAllUsers. Please check it with var_dump — Kristiyan
– Kristiyan, Commented Aug 14, 2015 at 14:25
@marian0 thats intentional since I want to access "variable of a variable" — GGio
– GGio, Commented Aug 14, 2015 at 14:25

arbogastes · Accepted Answer · 2015-08-14 14:40:14Z

4

Try using ${$var} instead of $$var.

EDIT

From PHP Manual (http://php.net/manual/en/language.variables.variable.php):

In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.

edited Aug 14, 2015 at 14:40

answered Aug 14, 2015 at 14:28

arbogastes

1,3489 silver badges10 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

arbogastes Over a year ago

Sure, I don't understand why someone give this answer "not useful".

Mihai Matei · Accepted Answer · 2015-08-14 14:34:32Z

3

function filterUsers(array $setOfAllUsers) {
   if (empty($setOfAllUsers)) {
      return array(array(), array());
   }

   $users = array(
     'inactiveUsers' => array(), 
     'activeUsers'   => array()
   );
   foreach($setOfAllUsers as $userRow) {
       $status = ($userRow['IsActive'] ? '' : 'in') . 'activeUsers';
       $users[$status][$userRow['CID']] = array();
       $users[$status][$userRow['CID']]['Label'] = $userRow['UserLabel'];

       $users[$status][$userRow['CID']]['UserList'] = array(
           $userRow['UID'] => array(
             'FirstName' => $userRow['FName'],
             'LastName'  => $userRow['LName'],
           )
       );
   }

   return $users;
}

answered Aug 14, 2015 at 14:34

Mihai Matei

24.3k5 gold badges34 silver badges50 bronze badges

1 Comment

GGio Over a year ago

Thank you, this seems like a cleaner version of what I am trying to do and wont confuse anyone :). I guess I forgot to make things simpler and not over complicate it.

John Bupit · Accepted Answer · 2015-08-14 14:29:08Z

0

Try the following:

$myUsers = $$var;
$myUsers[...][...] = ...;

The problem with using $$var[...][...] is that first $var[...][...] is evaluated and then it tries to find the variable named $$var[...][...].

answered Aug 14, 2015 at 14:29

John Bupit

10.7k10 gold badges45 silver badges79 bronze badges

2 Comments

GGio Over a year ago

I tried that, it returns empty since it does not modify the $activeUsers and $inactiveUsers arrays

John Bupit Over a year ago

Although this doesn't answer the original question using $$, but try that using $myUsers = ($userRow['IsActive'] ? $activeUsers : $inactiveUsers);.

Marc B · Accepted Answer · 2015-08-14 14:34:42Z

Don't use var-vars like this. You run into a PHP syntax glitch. And even if there wasn't a syntax glitch, you shouldn't be using var-vars in the first place. They make for near-impossible-to-debug code.

$x = 'foo';
$foo = array();
$$x[0] = 1;
var_dump($x);  // string(3) "foo"
var_dump($foo); // array(0) { }
$$x[1][2] = 3;
PHP Notice:  Uninitialized string offset: 2 in php shell code on line 1

Note how the array didn't get modified by the $$x[0] assignment, and how the 2+ dimensional assigment causes "undefined string offset". Your var-var isn't being treated as an array - it's being treated as a STRING, and failing, because it's a syntax glitch. You can treat strings as arrays, but not using var vars:

$bar = 'baz';
$bar[1] = 'q';
echo $bar; // output: bqz

There appears to be NO way to use a var-var as an array, especially a multi-dimensional array.

Collectives™ on Stack Overflow

Using variable variables to populate array results in error

4 Answers 4

1 Comment

1 Comment

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

1 Comment

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related