If Else Statment

[liquidd]

Hey all,

I am fairly new to PHP development coming off of ASP. I am working an if, else statement that had to do with photos.
I am using the theory:
If photo is available
add photo
else
add blank filler

The problem i am having is here:

<?php

$Pic2 = $row_details['photo2'];
if ($Pic2 <> "")
echo '<img src="http://url/media/$row_details['photo2']" width="125" height="54">';
Else
echo '<img src="images/defaultwhite.jpg" width="125" height="54">' ;

?>

Of course i am getting a Parse error because it is looking for '.' or ';'. I am not sure where it would go for the info to be pulled from the database.

Is this thought even close? Thank for your help..

[laserlight]

You need to use a little concatenation:

<?php

$Pic2 = $row_details['photo2'];
if ($Pic2 <> "")
    echo '<img src="http://url/media/' . $row_details['photo2'] . '" width="125" height="54">';
else
    echo '<img src="images/defaultwhite.jpg" width="125" height="54">' ;

?>

[liquidd]

Ugh, I feel dumb. I had the first half of '. at one time, but didnt close it out. Thanks.