This:
<img src="$image" alt="$alt" />returns this:
<img src="<img src="/ss/ss/assets/" alt="" />" alt="Image Alt Text" />I want $image to return ONLY the image PATH and NOT the <img> tag.
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returns this:
hknight,
I think what you want do to is reference the field inside the $image: $image.url
What is the name of the field that the url gets stored in? Use that field name.
Terry
Use:
$image.Link