[Resolved] $key = "object_name"; $key->var Not Working

davidjam

Hello,

I am having problems again using a variable variable (or what I think that is) on an object. For instance, this works:

$name = new field("required");

Therefore,

$name->required == true;

Now I have a posted array, that holds all the names (as keys) for each of the objects.

For instance:

foreach($_POST['line'] as $key=>$value) {
     if($key->required) {
          .. validate ..
     }
}

For some reason $key is not translating to "name" (or any other object for that matter)

Now if, for debugging purposes, I put a die command in the foreach, like:

die("Key is: " . $key);

It shows up as:

Key is: name

Grateful in advance,

David.

Shrike

If $key is just the name of the array index, presumably you want

$value->required

davidjam

No, I really do want the name of the $key and not its $value, as the key name is the same as my object names. But thanks mr shrike.

OK, so I just tried something and it worked! I added an extra '$' in front of the $key:

foreach($_POST['line'] as $key=>$value) {
     if($$key->required) {
          .. validate ..
     }
}

So Variable Variables IS the answer after all. But it seems like I still don't really know how they work (yes I have read the manual page 50 times!). Can someone please give me their own way of explaining variable variables, I would be most obliged... 🙂

Shrike

Ah right, I thought you were storing the object in the value portion of the array index (pretty daft since your using $_POST). The reason $$key works is that $key just resolves to the index name (i.e. without the $ needed to treat it as a variable.)

davidjam

Spot on, I say! Of course, just like you said, the object name is just a name, but requires a '$' to be seen as a variable. Maybe this has nothing at all to do with variable variables and is just a PHP feature...

Thanks

Shrike

It is exactly variable variables 😉