Undefined function

windows · Jan 9, 2008

PHP is outputtin that the imagecreatefromgif() and png is undefined. I have PHP 4.3.0 for Windows... here's the line in error: 
 
$ThePic = imagecreatefrompng($PicFile); 
 
But, I don't understand why it should be undefined... I have the latest version of PHP and the gd.dll (Or something close to that... at school so I can't check.)

This is really pissing me off... Thanks In Advance for any and all help.it's missing the gd library i'm afraid - you'll need to check php.ini to make sure it's enabled. and for php help i'd suggest heading over to the php builder board (<a class="postlink" href="http://www.phpbuilder.com/board">http://www.phpbuilder.com/board</a>)Thanks for your reply. I have corrected the php.ini file, and no longer get errors. However, execution stops after calling imagecreatefromgif(); 
 
$PlayerPic = imagecreatefromgif($PicFile); 
echo "TEST"; 
 
$PicFile = A picture that is in the same dir as the script. Such as mypic.gif. (This has been checked, the var has the right value.) 
 
However, TEST is never echoedno no no no no that's not how it works. You cant just call a function like that and it will display an image - think about it in terms of the PHP creating html - you need an <img src=http://www.webdeveloper.com/forum/archive/index.php/...> tag to get an image, and that's not what that function creates. 
 
What you need to do is create a makeimage.php file (i've posted the one i use for jpegs just for you to look at) and then call it from your main php file like so: 
 
echo "<img src=http://www.webdeveloper.com/forum/archive/index.php/\"makeimage.php?directory=1&file=this.jpg\"> 
 
here's my makeimage.php file: 
 
 
<? 
 
// if (!$max_width) 
$max_width = 100; 
// if (!$max_height) 
$max_height = 100; 
 
$file = $directory."/".$file; 
 
$size = GetImageSize($file); 
$width = $size[0]; 
$height = $size[1]; 
 
$x_ratio = $max_width / $width; 
$y_ratio = $max_height / $height; 
 
if ( ($width <= $max_width) && ($height <= $max_height) ) { 
$tn_width = $width; 
$tn_height = $height; 
} 
else if (($x_ratio * $height) < $max_height) { 
$tn_height = ceil($x_ratio * $height); 
$tn_width = $max_width; 
} 
else { 
$tn_width = ceil($y_ratio * $width); 
$tn_height = $max_height; 
} 
 
$src = <a class="postlink" href="http://www.webdeveloper.com/forum/archive/index.php/ImageCreateFromJpeg($file">http://www.webdeveloper.com/forum/archi ... Jpeg($file</a>); 
$dst = ImageCreate($tn_width,$tn_height); 
ImageCopyResized($dst, $src, 0, 0, 0, 0, 
$tn_width,$tn_height,$width,$height); 
header("Content-type: image/jpeg"); 
ImageJpeg ($dst, '', 90); 
ImageDestroy($src); 
ImageDestroy($dst); 
 
?> 
 
given as your code is generating an image, putting things like echo into it dont do anything or will confuse it - they'll have to go in your main script.Thanks so much. I now understand this a whole lot better. I put the image function in it's own file. Now it creates a 32x32 picture, and works almost perfectly. One problem: 
 
echo "<img src=http://www.webdeveloper.com/forum/archive/index.php/\"findPlayerpic.php?PicNum=".$ThePicNum."\">"; 
The actual file: 
<HTML> 
<?php 
echo $PicNum; 
// Just testing that var for now... 
?> 
$PicNum is null. Why? How can I retrieve it from the URL?you dont (correctly) have globals on - use $_GET['variable name'] to get to the URL variables: 
 
echo "<img src=http://www.webdeveloper.com/forum/archive/index.php/\"findPlayerpic.php?PicNum=$ThePicNum\">"; 
 
then: 
 
echo $_GET['PicNum']

Undefined function

windows

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