That would cause an error like "call to undefined function." But thanks for the suggestion.
On Thu, 2002-11-07 at 20:04, @ Darwin wrote: > The reason "imagedestroy()" may have not worked for you is that maybe > imagedestroy() is misspelled? If you misspelled it in the code then it won't > work. Just a suggestion. > > - Darwin > > > -----Original Message----- > > From: Robbert van Andel [mailto:robbert@;vafam.com] > > Sent: Thursday, November 07, 2002 9:30 PM > > To: [EMAIL PROTECTED] > > Subject: [PHP] Graphics question > > > > > > I have created a webpage that builds a table and graph using the GD > > functions. When I switch to another page or select a different month to > > build the table for, the graph.jpg does not rebuild. I have to press > > reload to make the graph render using the new data. I have used > > imagedistroy($pic) in the hopes that this will work, but it didn't. > > > > To elaborate on the situation, I first call imagecreate, then run the > > code to create the bar graph. I then use > > > > imagejpg($pic,$graphname) (or something ot that affect. I don't have > > access to the code at home). > > > > I then show the picture using > > > > echo "<img src=\"$graph\">"; > > > > Finally I call imagedestroy($pic). Do you have any suggestions? > > > > > > > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
