
Originally Posted by
netwiseit
Hello all,
Newbie to posting here but I have a question:
I've used the code above and when i view source the image location is "images\testpic.jpg" which is correct as that's the image location entered into the record that has been input.
When I wish to display a dynamic image on a webpage i.e. viewcostume.php the image displays correctly but when I use the script it doesn't display the image because the file location is "images\testpic.jpg" instead of "testpic.jpg" which is what is required for the script in order to work correctly.
Has their been an update to resolve this issue / is their a work around that will mead that I have to alter the database structure in order to compensate for the "images\"part of the database record.
I'm also looking for a simple image upload script that will allow the image to be uploaded after the record has been added to the database.
Any help would be greatly appreciated.
netwiseit.
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