Here’s my code. In case it isn’t obvious what I’m trying to do, I want to only show the image if there is info in the field. And the link around the image uses the info in the field. The mailto: is coming up blank. So I’m guessing/hoping I’m doing something wrong with the way I’m printing/calling it. I would greatly appreciate any help.
< ?php print $email; ?> knocks it out of the HTML it’s in. So instead of an image wrapped in a URL, I get email@example.com"> IMAGE
pmac627, your solution doesn’t work for me either. I still get a blank value for my string. And I’ve tested other fields, to make sure it isn’t just this field. I will look into isset though. Thanks for the recommendation.