"Dynamic" variables?

[Inkybro]

Hello, I'm writing a script in which I need to obtain different variable names, this is what I have (doesn't work):

		for($k=1;$k<5;$k++){
			for($l=1;$l<7;$l++){
				if($k==1){
					$people[$l] = ${"people_".$l.""};
				}elseif($k==2){
				}elseif($k==3){
				}elseif($k==4){
				}		
			}
		}
		printf($people);

It returns a blank array. I have confirmed that the variables exist so it SHOULD work. Thanks in advance.

[Rodney_H_]

How about a little more information about what you are trying to do? Because it does look like you are creating variables, people_1, people_2, people_3, people_4, people_5, people_6, but you are not assigning any value to them.

ALso, elseif $k == 2, 3, or 4 ? What do you want to do there??

[thorpe]

as far as i can see (and its pretty hard to see as your code is so poorly formatted), you never actually put anything inside the $people_1, people_2 etc etc variables. can you tell us what the point of this code is? what are you trying to do?

and please.... for the sake of readability, put some spaces in it!

[Inkybro]

The variables are assigned in a page before it and sent as a "POST". I've manually gone in and changed the dynamicness to, say, $people_1, which worked. So it's something to do with the dynamic part of it. If you need the full source, that isn't a problem. The if($k) is to determine which "set" of things it's checking (people, situation, setting or theme).

[Weedpacket]

Hmm... then you might have better success looking at $POST['people_1'], $POST['people_2'], etc. for the information.

[Inkybro]

Well that's what I was saying. I've tried that and get the same result, so I tried just manually setting it as $people_1 and it worked. I also tried $_POST['people_1'] manually. Worked as well. I might have another way to do this, though ideas are still accepted.

[Weedpacket]

Well, if it worked then what's the problem?