UNSET doesnt remove/unset array values permanently

Question

I have the following script for a Google maps application. The information about a specific place is sent via ajax to the $array array which is then added to the $SESSION['infomarker']. This part works fine. But the user should also be able to delete a place (If they regret the input).

This is handled by the removeElementWithValue() function, which takes the places lat and lng as arguments ($val1 and $val2 in this sample script). The array which contains the 'val1' and 'val2' values then disappears, like they should. But when I change the $val1 and $val2 variables to 'val3' and 'val4', the array which contained 'val1' and 'val2' returns while the array which contains 'val3' and 'val4' is unset.

I thought that unset() 'unset' the array in question for good, or do I need to do something else?

<?php
session_start();


if(isset($_POST['lat']) && !empty($_POST['lat']) ) {
$array = array(
                'titel' => $_POST['titel'], 
                'comment' => $_POST['comment'], 
                'lat' => $_POST['lat'], 
                'lng' => $_POST['lng'] 
                );


$_SESSION['infomarker'][] = $array;


   }    

  $val1 = 'val1';

  $val2 = 'val2';


  $newarray = removeElementWithValue($_SESSION['infomarker'], "titel", "comment", $val1, $val2);

 function removeElementWithValue($array, $key1, $key2, $value1, $value2){
 foreach($array as $subKey => $subArray){
      if($subArray[$key1] == $value1) {
      if($subArray[$key2] == $value2) 

         unset($array[$subKey]);

      }

   }
return $array;

}

print_r($newarray);

Colin M · Accepted Answer · 2013-01-08T15:53:56.770

You are modifying the $array that is being passed to your removeElementWithValue method. You are passing the $_SESSION array into the method.

unset is doing what it's supposed to, by removing the value from $array. But, your method is actually receiving a copy of the $_SESSION variable since PHP will pass arrays by-value not by-reference.

You can prefix the $array variable in the function declaration with an amperstand (&) character, which will inform PHP to pass that variable to the function by-reference, and will cause any modifications made to the array inside of the function to change the source array ($_SESSION).

Clarification:

Calling a function creates a scope in the "stack" for that method. PHP, by default, passes everything by-value (for those who want to argue that objects are passed by reference - they're not. The reference to the object is passed by-value).

What does that mean? Well, it's easier than it seems. Take the code below:

function myfunc($value) {
  $value = $value + 1;
}

$testing = 1;
myfunc($testing);
var_dump($testing);

This will output int(1). Notice how, even though we increment the value inside of the function - the result doesn't change the $testing variable. This is because the $testing variable was declared in a different scope inside of the stack. As far as PHP saw, when we called the method, we actually called this myfunc(1) (notice the number instead of a variable). The variable's value was grabbed from memory, and that value was passed to the method.

This is where passing by-reference comes in handy. You can instruct PHP to pass by-reference by prefixing any argument in the function definition with an amperstand, so you get this:

function myfunc(&$value) {
  $value = $value + 1;
}

$testing = 1;
myfunc($testing);
var_dump($testing);

The addition of the simple ampersand will now make this code print int(2).

So what's the difference? Well, you'll remember that I said the first time we are basically calling myfunc(1). The second time, we are effectively calling myfunc(0xff940ac4) where 0xff940ac4 is the memory address to the $testing variable. Now, any changes we make to $value inside of the method will change the data at memory address 0xff940ac4, instead of the value that was "copied" into the scope for this method.

Ok, not quite sure whta you mean by this, but I added the ampersand and it did the trick. I will have to read up on this! Thank's a lot though! — user1009453, Jan 08 '13 at 15:43
Ok, not quite sure what this means, but I added the ampersand and that did the trick. I will have to read up on this! Thanks a lot! — user1009453, Jan 08 '13 at 15:48
Very nice explanation, even I umnderstand the difference now :) Thank's for taking the time to explain this! — user1009453, Jan 09 '13 at 16:11

score 1 · Answer 2 · answered Jan 08 '13 at 15:36

That is, because in PHP, by default arrays aren't passed by-reference, but by-value. So, the function that you are using, will remove the values from the $array inside the function. But that won't modify the function. If you want to have that behaviour, you should have PHP pass the value by-reference.

You can do this by changing the function definition like this:

function removeElementWithValue(&$array, $key1, $key2, $value1, $value2)
{ ... }

UNSET doesnt remove/unset array values permanently

2 Answers2