If Statement with image

[daskemper]

I am still a novice, but I am working on an idea.

 

I have a table with a field: judges_ccpadv that contains only 3 possible characters = B, R, or N

 

My desire is to display a simple little icon for any of those 3 characters.

 

I tried:

if ($data['judges_ccpadv'] = "B") {
        echo "<img src='images/b.png'>\n";
}
if ($data['judges_ccpadv'] = "R") {
        echo "<img src='images/r.png'>\n";
}
if ($data['judges_ccpadv'] = "N") {
        echo "<img src='images/n.png'>\n";
}

 

The output is any image that is not in the if statement. So if "B" is in field, r.png and n.png are both displayed.

 

What am I doing wrong here? Assistance is greatly appreciated.

 

Thanks!
 

[requinix]

= is for assignment, == is for comparison.

[daskemper]

Dang it! simple stuff! Thanks a lot!

[kicken]

If the image names really are the same as the field letter you could simplify the code to just

$letter = strtolower($data['judges_ccpadv']);
echo "<img src='images/{$letter}.png'>\n";