Many thanks for your thoughts, and appologies for not being more specific.
I attempted the printf method suggested but the same results appear.
Basically, the constant ALBUM_IMG_DIR is a directory on the server and therefore when executed an image should be displayed based on this dir. I have used this constant elsewhere on different pages in the following way:
<img src="<?php echo ALBUM_IMG_DIR.$row['al_image']; ?>" border="0">
And this works perfectly where I avoid the html concatenation process.
With this in mind I have also re-written the the code without the use of echoing html and I still get the same result - the image is simply not showing!