RE: [PHP] Generate inline image

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:mfg@engsoftwarecenter.com] > Sent: Monday, May 13, 2002 2:42 PM > To: Martin Towell > Cc: gufrank@engsoftwarecenter.com; 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. > > >