RE: [PHP] Generate inline image

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: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. > > > > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >