PHP if statement to display gif

[limessl]

I have a table in which each line is dependant on an SQL lookup to determine whether the contents of each row should be displayed or not.

 

 

Here's the code I'm using :-

 

<?php if ($row["3"] != "0.00" ) echo "<img src="images/stories/button01.gif" />";?>

 

Obviously it doen't work, else I'd not be posting on here. I am an absolute novice in coding, having picked up everything so far on a mix of google and pot luck. Sadly, neither of those methods has helped!

 

FWIW, the cell before this one

 

has the same syntax except says  - echo "3 Years" and it displays fine.

 

If I omit the PHP the image file displays so it isn't a problem with the URL.

[trq]

You need to escape double quotes within a double quoted string.

 

<?php if ($row["3"] != "0.00" ) echo "<img src=\"images/stories/button01.gif\" />";?>

[Goldeneye]

I have a table in which each line is dependant on an SQL lookup to determine whether the contents of each row should be displayed or not.

 

 

Here's the code I'm using :-

 

<?php if ($row["3"] != "0.00" ) echo "<img src="images/stories/button01.gif" />";?>

 

Obviously it doen't work, else I'd not be posting on here. I am an absolute novice in coding, having picked up everything so far on a mix of google and pot luck. Sadly, neither of those methods has helped!

 

FWIW, the cell before this one

 

has the same syntax except says  - echo "3 Years" and it displays fine.

 

If I omit the PHP the image file displays so it isn't a problem with the URL.

 

Either that, or use single-quotes for your strings so you don't have to escape double-quotes:

<?php if ($row["3"] != "0.00" ) echo '<img src="images/stories/button01.gif" />';?>

[abazoskib]

Or do not echo HTML. Break up your PHP blocks. You'll thank yourself later, and the overhead of going in and out of PHP is much less than echoing tons of HTML.