RE: [PHP] ImageColorAllocate() Problem

From: Martin Towell (martin.towell_at_world.net)
Date: 10/03/03 

To: "'Jed R. Brubaker'" <jed.brubaker@psych.utah.edu>,  php-general@lists.php.net
Date: Fri, 3 Oct 2003 16:28:40 +1000

I think it might be because you're passing a string to the function instead
of a hex value...

try changing it to this and see if it works

$color = ImageColorAllocate($im, hexdec($hex1), hexdec($hex2),
hexdec($hex3));

HTH
Martin

-----Original Message-----
From: Jed R. Brubaker [mailto:jed.brubaker@psych.utah.edu]
Sent: Friday, 3 October 2003 4:23 PM
To: php-general@lists.php.net
Subject: [PHP] ImageColorAllocate() Problem

I am having a bizzare problem with the imagecolorallocate() function. Maybe
all of you can help me!

Here is the code:

$hex1 = "0x".$color{0}.$color{1};
$hex2 = "0x".$color{2}.$color{3};
$hex3 = "0x".$color{4}.$color{5};

$color = ImageColorAllocate($im, $hex1, $hex2, $hex3);

Simple enough! $color starts off as a hex color code minus the #, and then
is pulled into each of the 3 alpha values by way for the hex variables. It
is later outputted in text:

ImageTTFText ($im, $cursive_size, 0, $text1_x, $text1_y, $color, $font1,
$cursive_text);

Problem: The text is always black.
Something strange: If I just type in a randomly selected color, it works!

Is there some reason that the values in my variables wouldn't be passing?

Thanks in advance,
Jed R. Brubaker 

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