Posted by d on 01/25/06 14:23

"Ubantu Rococo" <paul.x.lee@baesystems.com> wrote in message
news:43d767e6$1_1@glkas0286.greenlnk.net...
> >
>> I don't think imagejpeg works as you think it does. It outputs an
>> image's actual binary data, so you're essentially trying to output a jpeg
>> image with HTML tags before and after it - which won't work. What you
>> need to do, is put the PHP code only into a file, called say image.php.
>> Then, in your HTML document (the rest of the code above), just put an
>> image tag in: <img src="image.php">. That will display the image in the
>> HTML page. You can even have javascript to work out where you click on
>> the image, and then update its url like this:
>>
>> <img src="image.php?x=10&y=22">, and the image.php can then add something
>> at those co-ordinates to demonstrate it has been selected.
>
>
> Thanks, I understand now, I think....
>
> Could I just put the imagecopy routine in an external file (called
> getWorldMap.php, with the getWorldMap() function contained
> within) and then do:
>
> <?php include("getWorldMap.php");
>
> $wm = getWorldMap(); // $wm is the resource,copied from WorldMap
>
> imagejpg($wm);
>
> ?>

You could indeed. That script would output the jpeg, so it has to be on its
own, and with the correct header sent. I think if you're not going to be
using the code anywhere else, just keep it in one flat file. That would
make it a lot easier to comprehend ;)

Also, remember to call imagedestroy() on your image handles to free up the
memory they use. Otherwise you might find yourself running out of resources
on your webserver, if its not tidying up after itself.

dave

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