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I have set up a column in msql database for a link to my images direction of my website. I set this up as a text field with no character length. In that field I have put only the picture file name, example: TestPic.jpg.
in my code I have put the following:
| <img src="images/"<?php echo $row_DisplayDetails['PicLink']; ?> width="530" height="310" alt=""/> |
I've tried many altered versions of this statement, but get only a picture frame, square with a tiny generic picture icon. If I reference the filename directly, the picture shows up.
Can someone PLEASE help me out.
Thanks, not only were my quotes not in the right place, I made it simple by including in my database field the "image/picname.jpg" and excluding it in my code.
Thanks for responding!!!
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Thanks, not only were my quotes not in the right place, I made it simple by including in my database field the "image/picname.jpg" and excluding it in my code.
Thanks for responding!!!
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