Re: [phplib] gd1.3 in php From: Bart-Jan Vrielink (bartjan <email protected>)
Date: 09/28/00

On 28 Sep 2000, Sanlong Irvan wrote:

> seem like i can't get the gd library work with php
>
> Here is the step i did
>
> 1. download it at http://www.boutell.com/gd/#getgd
> 2. tar -zxvf gd-1.8.3.tar.gz
> 3. cd gd-1.8.3
> 4. vi Makefile (edit according to your system).
> 5. make all
> 6. cp libgd.a /usr/local/gd1.3/ make sure you create this directory gd1.3

> Then i try the code
> <?
> #include /usr/local/gd1.3/libgd.a
> echo "
> <html>
> <body>
> <center>";
> $imagefile = "melb1.jpg";
> $src_img = imagecreatefromjpeg($imagefile);
> $new_w = imagesx($src_img)/4;
> $new_h = imagesy($src_img)/4;
> $dst_img = imagecreate($new_w,$new_h);
> imagecopyresized($dst_img,$src_img,0,0,0,0,$new_w,$new_h,imagesx($src_img),imagesy($src_img));
> imagejpeg($dst_img, "scratch/$imagefile");
> echo "<img src=scratch/$imagefile>\n";
> echo "</center></body></html>";
> ?>
>
> but it gives me this error
> Warning: ImageCreateFromJpeg: No JPEG support in this PHP build in /home/hosting/www.asiandotcom.com/www/tirvan/imagetest.php on line 8
>
> Any idea guys?
>
> Did i missed soemthing? do i have to recompile it again?

Did you (re)compile php with gd support ?? try <php? phpinfo() ?> to see
what's compiled into php.
The #include statement is something you would use when you would use libgd
in C, it's not needed in PHP. 

-- 
Tot ziens,

Bart-Jan

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