Enhancement Request

From: Date: Wed, 31 May 2000 06:02:57 +0000
Subject: Enhancement Request
Groups: php.version4 
Request: Send a blank email to php-version4+get-15959@lists.php.net to get a copy of this message
Application: Pulling .jpg files out of a database into the in-memory file and then exported directly to the html. Workarounds: 1) Not putting image files in the database 2) Export to a temp file and use <img src=x.jpg> 3) Separate .php file using header(image/jpg) and use <img src=xxx.php?id=y> Just wishing for a "clean solution." Best bet so far is just calling an external php file with ? params. Wish I could just call one of my classes member methods to load/dump it instead of having to hack my way with external files. 1 4line snippet of code for a whole file gets kinda annoying. I can put it in the DB, get it out, but it's becoming a PITA to display it properly. Maybe someone can suggest a workaround? -Szii An another note: print "<img src=\"$ad->OutputPhoto($cat,$valObject->AdID)\">"; echo $ad->OutputPhoto($cat,$valObject->AdID); print "$ad->OutputPhoto($cat,$valObject->AdID)"; The first "print" call says UndefinedProperty: OutputPhoto The "echo" call works as expected. The second "print" call works as expected. Can anyone explain why being inside the <img src=> tag would cause it to become an undefined property?

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