How about the obvious way?
<img src="generateimage.php?id=3">
-Rasmus
On Tue, 28 May 2002, Engineering Software Center wrote:
> 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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