Hi: all:
With helps from Martin, now I can create images on the fly.
I used code like <img src=\"generateimage.php\"> to create my
images. However, with this option, how can I pass a variable to
generateimage.php? The php scrip is supposed to take a variable: say
ImageID and looks up the MySQL database for all values and then draw the
graph. Anyone can give me some help?
Thank you,
Frank
On Mon, 13 May 2002, Martin Towell wrote:
> yep, then in the file generateimage.php you'd have
> <?
> $im = imagecreate(...);
> // ... image creation code here
> header("Content-Type: image/png"); // I do this anyway, I've found some
> browsers complain when you don't
> imagepng();
> imagedestroy($im);
> ?>
>
> obviously, the above "code" is for png, but you can use any that your gd
> library/broswer supports.
>
> -----Original Message-----
> From: Engineering Software Center [mailto:[EMAIL PROTECTED]]
> Sent: Monday, May 13, 2002 2:42 PM
> To: Martin Towell
> Cc: [EMAIL PROTECTED]; php
> Subject: RE: [PHP] Generate inline image
>
>
>
> Thanks for the reply. But how do I do that exactly?
> <img src=\"generateimage.php\"> ?
> is this correct?
>
>
> > HTML docs only contain text. If you want images "inline" then you use the
> > <img> tag.
> >
>
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