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Variable Variables Pointing to Arrays or Nested Objects

Is it possible to create a variable variable pointing to an array or to nested objects? The php docs specifically say you cannot point to SuperGlobals but its unclear (to me at least) if this applies to arrays in general.

Here is my try at the array var var.

     // Array Example
     $arrayTest = array('value0', 'value1');
     ${arrayVarTest} = 'arrayTest[1]';
     // This returns the correct 'value1'
     echo $arrayTest[1];
     // This returns null
     echo ${$arrayVarTest};   

Here is some simple code to show what I mean by object var var.

     ${OBJVarVar} = 'classObj->obj'; 
     // This should return the values of $classObj->obj but it will return null  
     var_dump(${$OBJVarVar});    

Am I missing something obvious here?

like image 809
K-2052 Avatar asked Mar 02 '26 08:03

K-2052


2 Answers

Array element approach:

  • Extract array name from the string and store it in $arrayName.
  • Extract array index from the string and store it in $arrayIndex.
  • Parse them correctly instead of as a whole.

The code:

$arrayTest  = array('value0', 'value1');
$variableArrayElement = 'arrayTest[1]';
$arrayName  = substr($variableArrayElement,0,strpos($variableArrayElement,'['));
$arrayIndex = preg_replace('/[^\d\s]/', '',$variableArrayElement);

// This returns the correct 'value1'
echo ${$arrayName}[$arrayIndex];

Object properties approach:

  • Explode the string containing the class and property you want to access by its delimiter (->).
  • Assign those two variables to $class and $property.
  • Parse them separately instead of as a whole on var_dump()

The code:

$variableObjectProperty = "classObj->obj";
list($class,$property)  = explode("->",$variableObjectProperty);

// This now return the values of $classObj->obj
var_dump(${$class}->{$property});    

It works!

like image 195
johnnyArt Avatar answered Mar 03 '26 21:03

johnnyArt


Use = & to assign by reference:

 $arrayTest = array('value0', 'value1');
 $arrayVarTest = &$arrayTest[1];

 $arrayTest[1] = 'newvalue1'; // to test if it's really passed by reference

 print $arrayVarTest;
like image 41
Lukman Avatar answered Mar 03 '26 23:03

Lukman



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