Issue with variable in function array_key_exists

[son]

I use the following data and function successfully on another page that has lots of other stuff going on (and a test page with only this bit):

while (list($catID, $catNameSelect, $catParentSelect) = mysqli_fetch_array($catResult, MYSQLI_BOTH))  {  $categoryNR[(string)$catID] = $catID;  $categoryNames[(string)$catID] = $catNameSelect;	  $catParent = (string)$catParentSelect;	   if(!array_key_exists($catParent, $childrenTree))  $childrenTree[$catParent] = array();  $childrenTree[$catParent][] = (string)$catID;  }function renderTree($parent = 0, $subCounter = 0){global $cid;global $categoryNR;    global $categoryNames;    global $childrenTree;	 if($parent != 0)  {  echo "<option value=\"webster.php?cid=" . $categoryNR[$parent] .  "\"";  echo " style=\"padding-left:" . $subCounter . "px;\">", $categoryNames[$parent], "</option>\n";  $subCounter = $subCounter + 10;  }    $children = $childrenTree[$parent];	 if(count($children) > 0)  { //If node has children	    foreach($children as $child)	    renderTree($child, $subCounter);	 }  }renderTree();  //This renders the hierarchical tree

However, on the relevant page it shows drop down ok, but when you look in code you can see that there is an issue...Undefined variable: childrenTreearray_key_exists() [<a href=function.array-key-exists'>function.array-key-exists</a>]: The second argument should be either an array or an object I do not get this. It is working fine on other page/s with same query and the lot. Why is it saying that childrenTree is not defined in this case? Where could I look? Thanks,Son

[justsomeguy]

Why is it saying that childrenTree is not defined in this case?

That much should be obvious (it's because it's not defined), maybe you need to initialize it before you start the loop.

[son]

In my desperation I have tried already above this bit of code: $childrenTree = 0; $childrenTree = ''; $childrenTree = FALSE;The first about undefined variable indeed gone, but still 'The second argument should be either an array or an object'. In addition, I did not do anything else on other pages where it is working... Could it be that other variable infringe with the code? I do not have another $childrenTree, but maybe another one is causing this? Son

[justsomeguy]

If it's an array, then start it off as an array: $childrenTree = array();

[son]

But this is where the my problem is to understand. I only want to initialise when there are children as

if(!array_key_exists($catParent, $childrenTree))  $childrenTree[$catParent] = array();  $childrenTree[$catParent][] = (string)$catID;

However, going through all the thrown error codes after 'Undefined variable: childrenTree' it also repeatetly complains 'Undefined index'. Checking all those indexes in detail I can clearly see that those are all the ones without children. The one category with children does not come up in error code. Also, that it works on other page exactly like it is totally confuses me... Any ideas where else I could check? Son

Edited April 27, 2012 by son

[justsomeguy]

If you're getting an error that $childrenTree is undefined in the while loop when you're building the arrays, the solution is to initialize that variable to an empty array before the loop like I showed. Inside the function that prints everything you also want to check if the $parent index is set before trying to access it.

[son]

I did set

$childrenTree = array();

just before loop as suggested and had then also added a check if $parent is set in function as

	 if(isset($parent) && $parent != 0)  {

I still get the same 'Undefined index: ' for all the entries that are not parents themselves... The one that is a parent to two entries is fine... Son

[justsomeguy]

Which line is the error on? You need to check if that index exists before you try to access it.