Array isn't being seen by function.. storing in variable a no-no?

Question

I've googled around and didn't see the answer. I have an array I'm storing in a variable that I'm trying to pass to a function..

$myArr = 'array('item1', 'item2')';

require('script.php'); //where actual function is

makeCode($myArr);

When I use makeCode(array('item1', 'item2')); it works fine.. I've even tried to add global $myArr to makeCode, but that didn't work either.

I'm thinking it's a scope problem, but maybe I'm misusing the string. print_r($myArr) prints properly, it just isn't passing or something.

The function basically just compares $myArr values and if it matches what's in the function's array, it outputs the correct HTML, so I didn't list it. It works, just not the variable.. Thanks!

--makeCode()--
 function makeCode($listArr){
 /* global $myArr; //Tried this */
   $output = '';
   $items = array(
      'item1'        =>  "Code for item1",
      'item2'         =>  "Code for item2"
       )
   /* $myArr = $listArr; //tried this too */
        foreach ($listArr as $val) {
            if(array_key_exists($val, $items)){
                if(strlen($output)>0) $output .="|"; //Add Sytnax
                $output .="$items[$val]";
            }
        }
  }

That's pretty much it.

Jim · Accepted Answer · 2013-02-21 01:49:44Z

4

$myArr = 'array('item1', 'item2')';

$myArr is a string here. I'm not sure that's what you meant. Try:

$myArr = array('item1', 'item2');

If this is what you meant then this will behave differently than calling

makeCode(array('item1', 'item2'));

Because this is calling using an actual array.

answered Feb 21, 2013 at 1:49

Jim

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12 Comments

Casey Dwayne Over a year ago

Ah, I see. By doing that, it only loads the first item from the $myArr. What I need $myArr to do is output array('item1', 'item2') inside, just like it is on the last line in your post.

Casey Dwayne Over a year ago

It does print Array ( [0] => item1 [1] => item2 ) however. Would I have to format this array? They're on the same page I just wanted to keep it up top with the rest of the overrides.

Jim Over a year ago

@kcdwayne I'm not sure I follow, maybe you could add the code for makeCode to your question along with what output you expect?

Casey Dwayne Over a year ago

Sure. I've updated with the function. I'm just trying to get $myArr to work in makeCode() as makeCode($myArr) being the same as if I used makeCode(array('item1', 'item2')); . Right now the output completely disappears. (or just prints value 0 and skips the rest)

Jim Over a year ago

@kcdwayne There's a small bug in your method: foreach ($listArr as $val) is using $listArr but the start of the method uses $listArray

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