Jeez, this is simple. The browser is caching the image. There's no real
way to fix that. There are some headers that may encourage the browser to
not cache, but there is no sure way to control whether the users will have
this setting overridden.
imagedestroy($pic); is only freeing the memory used on the server while the
script is executing. It's nice if you're doing a whole bunch of processing
after generating the image but otherwise the memory is automatically freed
when the script exits. It will do nothing to affect the copy that now sits
in the browser's cache.
In other words, you'll just have to hit reload.
You can also search for specifics of the headers that are meant to stop a
browser from caching.
"@ Darwin" <[EMAIL PROTECTED]> wrote in message
> The reason "imagedestroy()" may have not worked for you is that maybe
> imagedestroy() is misspelled? If you misspelled it in the code then it
> work. Just a suggestion.
> - Darwin
> > -----Original Message-----
> > From: Robbert van Andel [mailto:robbert@;vafam.com]
> > Sent: Thursday, November 07, 2002 9:30 PM
> > To: [EMAIL PROTECTED]
> > Subject: [PHP] Graphics question
> > I have created a webpage that builds a table and graph using the GD
> > functions. When I switch to another page or select a different month to
> > build the table for, the graph.jpg does not rebuild. I have to press
> > reload to make the graph render using the new data. I have used
> > imagedistroy($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?
> > --
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