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 Message Boards » » PHP problem: nested functions and variable scope Page [1]  
Ernie
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45943 Posts
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My PHP knowledge is limited, so I very well could be doing this ass-backwards

I have the following code in which doodoo() accepts a variable and depending on the value of the variable sets $crap[] to a particular set of values. Another function, peepee() is then created which (for the sake of simplicity) echos the values of $crap[].


function doodoo ($poop) {
if ($poop == "a") {
$crap = array(...);
}
elseif ($poop == "b") {
$crap = array(...);
}

function peepee () {
foreach ($crap as $var) {
echo $var;
}
}
}


This gives me an error that an invalid argument is supplied for the foreach loop. Passing $crap to peepee() gives me an error message that I am missing one argument for peepee(). I would think that defining $crap as a global variable would at least be a step in the right direction, but that does nothing for me.



[Edited on March 13, 2008 at 10:23 PM. Reason : ]

3/13/2008 10:17:24 PM

mienutzich
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4300 Posts
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$crap=array(..);
function doodoo ($poop) {
global $crap;
if ($poop == "a") {
$crap = array(...);
}
elseif ($poop == "b") {
$crap = array(...);
}

function peepee () {
global $crap;
foreach ($crap as $var) {
echo $var;
}
}
}

donno if this is how you were doing it before...

if crap is not set outside of the parent function and included as a global variable and then included again as a global in the child then it will not work

3/13/2008 10:54:26 PM

Ernie
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45943 Posts
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Well, I got it working. But $crap doesn't have to be defined outside of the initial function (that would negate the entire use of the function anyway), it only needs to be defined as a global variable in each of the functions. I was only defining it in the first. Thanks for the help.

3/13/2008 11:01:06 PM

DirtyMonkey
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4269 Posts
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i see you've got it working, but...

peepee() doesn't really need to be a function, and i'm not even sure if you can define a function inside of a function anyway. but, aside from that, $crap is ONLY defined if $poop == "a" OR $poop == "b". so if $poop = "c", $crap is never set. you should do something like this to make sure it's set.


function doodoo($poop) {
$crap = array();
if($poop == "a") {
$crap = array(...);
} else if ($poop == "b") {
$crap = array(...);
}

foreach($crap as $var) {
echo $var;
}
}


that way you know that $crap is always set, even if it's an empty array. if you ever decide to set $crap from a parameter of doodoo() - which would eliminate the need for that if/else block - you might want to wrap the foreach loop with if(is_array($crap)) to make sure the foreach won't fail.

p.s. you have dirty variable naming conventions

3/14/2008 2:18:06 AM

Ernie
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45943 Posts
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You can definitely nest functions in PHP and the inner function is necessary There's a lot more going on in the actual code; I just posted the bare bones for the sake of simplicity.

3/14/2008 7:46:15 AM

Stein
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19842 Posts
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The issue is that nested function or not, you still need to pass it the variable, which you weren't doing at first.

3/14/2008 8:27:15 AM

Ernie
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45943 Posts
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That was never really the problem. I knew the variable had to be passed for the foreach loop to run, I just didn't know how to make the foreach loop know the variable existed.

3/14/2008 8:34:36 AM

Talage
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5091 Posts
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I would love to hear more details on why you're using a nested function. Just out of curiosity.

3/14/2008 4:15:50 PM

Ernie
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45943 Posts
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And I would love to post the other 2000 lines of code that would clarify the matter

3/14/2008 5:58:10 PM

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