Weirdest PHP error of all time?

[ppowell]

		print_r("2hasResult = $hasResult<P>");

	/*---------------------------------------------------------------------------------------------------------------------------------------------------------
		Block to determine metadata.  If you are coming from DownloadView or RemoveView you will have to have
		a database instance to generate proper metadata for display, however, coming from CreateView or ResizeView you
		will not have any database instance, so metadata will be limited and derived from the image source itself.
	-----------------------------------------------------------------------------------------------------------------------------------------------------------*/
	$imageURL = ($album) ? "$imageURLPath/$album" : $imageURLPath;

	print_r("3hasResult = $hasResult<P>");

At no time are any of the variables even remotely related to one another, but watch this:

2hasResult =

3hasResult = /catalog/images/awards_2004

For some wacko reason $hasResult = $imageURL! I have never even used reference pointers for any of these variables, how is that possible?

Thanx
Phil

[tsinka]

Hi,

is this the complete code ? Does that happen if you execute exactly just the code you posted ?

Thomas

[ppowell]

It's not the complete code, but I used print_r() as placeholders to trace exactly when $hasResult literally becomes $imageURL and that is exactly where it occurs. At NO point anywhere in the method does $hasResult get any other value except in this way:

$hasResult = (is_array($result) && @sizeof($result) > 0) ? true : false;

That's it!

Phil

[tsinka]

Hi,

does it work if you change the code you posted in your first post like this:

$hasResult = "";
print_r("2hasResult = $hasResult<P>"); 

    /*--------------------------------------------------------------------------------------------------------------------------------------------------------- 
        Block to determine metadata.  If you are coming from DownloadView or RemoveView you will have to have 
        a database instance to generate proper metadata for display, however, coming from CreateView or ResizeView you 
        will not have any database instance, so metadata will be limited and derived from the image source itself. 
    -----------------------------------------------------------------------------------------------------------------------------------------------------------*/ 
    $imageURL = ($album) ? "$imageURLPath/$album" : $imageURLPath; 

    print_r("3hasResult = $hasResult<P>");

One more question ... which php version do you use ?

Thomas

[ppowell]

I can't change $hasResult to null in this case because it already has to have a value of true or false depending on a returned resultset from a db call:

$hasResult = (is_array($result) && @sizeof($result) > 0) ? true : false;

print_r("2hasResult = $hasResult<P>");

// PRODUCES 1 or NOTHING

$imageURL = ($album) ? "$imageURLPath/$album" : $imageURLPath;  

print_r("3hasResult = $hasResult<P>");

// PRODUCES /catalog/images/blah.jpg

I'm using PHP version 4.3.6

Phil

[tsinka]

Hi,

interesting ... I couldn't find anything about such a bug on the bug list.

Can you post just the lines of your code that contain $hasResult in any way ?

I tested the code in several ways with PHP 4.3.6 but wasn't able to reproduce that problem.

Thomas

[ppowell]

		$hasResult = (is_array($result) && @sizeof($result) > 0) ? true : false;


	// BLOCK TO DETERMINE IF THERE HAVE BEEN CHANGES TO IMAGE DIMENSIONS NOT UPDATED IN DB METADATA
	if ($hasResult && strcmp(strtolower($action), 'resize') == 0) {
	 list($image_width, $image_height) = @getimagesize("$locationPath/" . $this->id);
	 if (is_numeric($image_width) && is_numeric($image_height) && 
	     ((int)$image_width !== (int)$result[0]->image_width || (int)$image_height !== (int)$result[0]->image_height)
	     ) {
	   // PERFORM AN UPDATE ON THE RECORD IN THE DB
	   $actionPerformer = new ActionPerformer();
	   $actionPerformer->connect();
	   $sql = "UPDATE image SET image_width = $image_width, image_height = $image_height WHERE id = " . $result[0]->id;
	   $query =& new mySqlQuery($sql, $actionPerformer->getDBConn());
	   $query->runQuery();
	   $query = null;
	   $actionPerformer->disconnect();
	   $actionPerformer = null;
	   $result[0]->image_width = $image_width;
	   $result[0]->image_height = $image_height;
	 }
	}
	//---END OF IMAGE DIMENSION METADATA UPDATE BLOCK--------------------------------------------------------------------------



	/*---------------------------------------------------------------------------------------------------------------------------------------------------------
		Block to determine metadata.  If you are coming from DownloadView or RemoveView you will have to have
		a database instance to generate proper metadata for display, however, coming from CreateView or ResizeView you
		will not have any database instance, so metadata will be limited and derived from the image source itself.
	-----------------------------------------------------------------------------------------------------------------------------------------------------------*/
	$imageURL = ($album) ? "$imageURLPath/$album" : $imageURLPath;

I've tested inside the conditional thoroughly and it never enters the conditional at all, in fact, it retains its value after the conditional, it loses its value IMMEDIATELY after the $imageURL call!

Phil

[tsinka]

Hi,

would you mind to post the complete script as attachment (renamed to .txt) ?

It might be a buffer overflow or something like that in the worst case or just something with the php code but from what I saw so far ... 😕

Thomas

[ppowell]

Sorry that I am not allowed to do by law. The entire .php script is property of the United States Federal Government and I am forbidden by law to do anything with it (I can post reworded snippets of it, but that's all I can do legally).

All I can do is show you the code snippet exactly where the error occurs:

	print_r("3hasResult = $hasResult<P>");
	$imageURL = ($album) ? "$imageURLPath/$album" : $imageURLPath;
	print_r("4hasResult = $hasResult<P>");

And the resulting display:

3hasResult =

4hasResult = /catalog/images/blah.jpg

Phil

[tsinka]

Hi,

just to make sure that this isn't a print_r problem ... did you try to use echo instead of print_r ?

Thomas

[ppowell]

Yes, exact same results. The moment I do my Ternary Operator assignment to $imageURL $hasResult obtains the wrong value, literally, $hasResult = $imageURL.

Phil

[tsinka]

So ... with

echo "3hasResult = ".$hasResult."<br>"; 
if (isset($album) && strlen($album)>0) {
  $imageURL = $imageURLPath."/".$album;
} else {
  $imageURL = $imageURLPath; 
}
echo "4hasResult = ".$hasResult."<br>"; 

you still have the same problem ?

If yes, I guess that something is wrong with the PHP module.
You might want to try that script on another server if possible.

Thomas

[ppowell]

Yep, exact same problem even with that script!

Furthermore, this is also producing erroneous results:

$val = 'image|Image Maintainence';
list($mySection, $myDisplay) = explode('|', $val);
echo "val = $val<P>mySection=$mySection<p>myDisplay=$myDisplay<P>";

Produces:

val = image|Image Maintainence

mySection = image

myDisplay = image

WT Freak???

Phil

[tsinka]

Hi,

do you have a chance to test that code on another server with another PHP version ?

It might be a problem with the PHP module ... maybe a buffer problem.

Thomas

[ppowell]

Yes, I tested on the development server using PHP 4.3.2 with register_globals off and everything is just fine!

Phil

PS: Further testing on the other environment produced this wacko result:

8val = and myDisplay =

isset val?

Array ( [section] => person [action] => edit_person [PHPSESSID] => 6f37032f8b893689c3e1b89ef8e1215a ) inside getCount:

9val = 0 and myDisplay = 0

Here is the code that generates that stuff:

    if ($authBool && $action && class_exists(ucfirst($action) . 'View')) {
    print_r("7val = $val and myDisplay = $myDisplay<P>");
    print_r(isset($val));
     $junk = eval('$' . $action . 'View =& new ' . ucfirst($action) . 'View($id);');
    print_r("8val = $val and myDisplay = $myDisplay<P>");
    print_r("isset val? "); print_r(isset($val)); print_r("<P>");
     $displayHTML = ${$action . 'View'}->displayHTML();
    unset($val); unset($myDisplay);
    print_r("9val = $val and myDisplay = $myDisplay<P>");
    print_r("isset val? "); print_r(isset($val)); print_r("<P>");
     ${$action . 'View'} = null;
    } 

Between 8 and 9 I do a class method that returns HTML stuff and then physically unset both $val and $myDisplay, and yet they STILL retain values of 0 that it could have never received!!!

Phil

PS: Further testingt:

echo "val = $val<P>";
print_r(isset($val)); print_r("<P>");
print_r(strlen($val)); print_r("<P>");
print_r(is_numeric($val)); print_r("<P>");

Produces:

val = 0

According to this, $val does not exist and has a value of 0. Is that possible in any version of PHP for a value to not exist (not null, mind you, unset($val)! ) and yet have a declared value of 0 even though it is nonnexistent, much less non-numeric?

[ppowell]

The "weirdest PHP error of all time" was fixed by changing only ONE line!!!

$result = $query->getResult();

Instead of

$result =& $query->getResult();

It was a mistake on my part because such an assignment, if $query->getResult() were null, would point $result to the same value as $query->getResult and return it.. which then would mean that $result would point to a nonexistent entity in memory allocation stack and cause a memory buffer overflow!

Phil

[Weedpacket]

Maybe you should file a bug report on this one (assuming this is in 4.3.8): although it was a bug in your code that exposed it, it seems to me that "assigning references to null" either shouldn't be allowed, or some other measure should be taken on the part of the PHP engine, to avoid weirdness like this. It certainly shouldn't go tramping around the symbol table at random. For security reasons, if nothing else.

[ppowell]

Actually it's 4.3.6 (and this is ok in 4.3.2 which is really freaky). I agree, assigning a reference to null is usually a bad idea and that was a mistake on my part to do so (coding error).

Phil