Unset not working in multiple foreach statements (PHP)

Question

Can someone tell me what is wrong with this code :

foreach ($data as $region ):

  foreach ($region as $type):

   foreach ($type as $type2):

    foreach ($type2 as $key=>$val):

    if ($val=='background-color: FFFFFF;' || $val=='')  unset($type2[$key]);

    endforeach;

   endforeach;

  endforeach;

endforeach;

After print_r($data) seems that the data array is the same and unset is not working

Read this very carefully: http://php.net/manual/en/control-structures.foreach.php — deceze, Aug 08 '13 at 10:49

score 2 · Answer 1 · answered Aug 08 '13 at 10:49

2

Your loop is operating on copies of the original elements; changes to $type2 will not be visible in $data because $type2 is a copy.

You can solve this by iterating over all arrays by key, then indexing into $data with those keys to remove the value:

foreach ($data as $k1 => $region ):

  foreach ($region as $k2 => $type):

   foreach ($type as $k3 => $type2):

    foreach ($type2 as $k4 =>$val):

    if ($val=='background-color: FFFFFF;' || $val=='') {
        unset($data[$k1][$k2][$k3][$k4]);
    }

    endforeach;

   endforeach;

  endforeach;

endforeach;

Of course this is ugly, but that's four nested loops will do. There is also the option if iterating by reference instead of grabbing keys, but personally I dislike that because of the nice opportunity to write bugs by reusing the abandoned references after the loop has ended. Especially in this case I dislike it to the fourth power.

answered Aug 08 '13 at 10:49

Jon

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Especially in this case i dislike your method, because you generate alot of copies (depending on array sizes of course), which takes MUCH more cpu time (and some more RAM) than passing a reference. and the only advantage that you noted for that is that bad code after that will not be as bad (as far as i understood that) – x4rf41 Aug 08 '13 at 11:12
@x4rf41: PHP uses copy-on-write, so we are talking about very lightweight copies (only extra ZVALs are created) -- it's exactly the same as when iterating by reference, both as regards CPU time and RAM. Where do you base your assertions? – Jon Aug 08 '13 at 11:41
@x4rf41: Actually, what happens is exactly the reverse. Compare [this](http://ideone.com/q5BQHr) to [this](http://ideone.com/a6ZM2R). References in PHP should be used *very* sparingly and only if you know exactly what you are doing. I was surprised myself to see how bad references are in this case. – Jon Aug 08 '13 at 11:52
Tks Jon your code is working just fine ! x4rf41, i know that this code is not "to optimal" or elegant but im looping trough a small size array so performance shouldnt be affected. – user2029029 Aug 08 '13 at 17:34

score -1 · Answer 2 · answered Aug 08 '13 at 10:51

-1

use this, it should work:

foreach ($data as &$region ):

  foreach ($region as &$type):

   foreach ($type as &$type2):

    foreach ($type2 as $key=>$val):

    if ($val=='background-color: FFFFFF;' || $val=='')  unset($type2[$key]);

    endforeach;

   endforeach;

  endforeach;

endforeach;

the array values are passed as reference because of the & i put before the value variables. The unset will work this way

answered Aug 08 '13 at 10:51

x4rf41

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Don't forget to unset the references! – AmazingDreams Aug 08 '13 at 10:52
@AmazingDreams what do you mean by that? The code works exactly as wanted – x4rf41 Aug 08 '13 at 10:58
Try setting `$region = "Some value";` and see what happens to `$data`. – AmazingDreams Aug 08 '13 at 11:47
@AmazingDreams obviously the according array value of $data will be set – x4rf41 Aug 08 '13 at 11:48
Exactly. Now try it in a production environment, in another part of the application. – AmazingDreams Aug 08 '13 at 11:48
i assume its a local variable within a function, so its no problem at all unless you reuse the reference again inside the function, but i see why inexperienced programmers would make a mistake there – x4rf41 Aug 08 '13 at 11:50
@x4rf41: Please see the examples I have provided under the comments to my answer, and next time think twice about the condescending attitude towards "inexperienced programmers". – Jon Aug 08 '13 at 11:56
I think it is better to assume that it is not if you don't see it wrapped in a function. – AmazingDreams Aug 08 '13 at 11:57

Unset not working in multiple foreach statements (PHP)

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