[SOLVED] How to display alternative image if database record is empty?

[hemangi]

HI everyone,

 

I am beginner in PHP and I need help with little code.

I am making a web site in Dreamweaver using MySql database and PHP.

I have image address in row 10 of the database table and I display them well but they only exist for some of the records, not all. But when I place the recordset on the page and test it in browser-it displays the image when the recordset contains it's address, but it displayes an empty image (like broken link one) for all the rest of the records.

What I need is simple if else statement in php to tell it to display alternative image or nothing when the row record is empty.

I have this code, but is obviously wrong because it gives me an error:

 

<?php if ($row_rsList['WL']==""){
echo "images/emptyGraphic.gif"}?>
              <?php else

        echo $row_rsList['WL'] />

 

"rsList" is my recordset in dreamweaver which contains all the table rows and "WL" is row number 10.

I place instances of all the row in the page, like phone:... email:... etc and I also would like to know how to make them not show if the record is empty. Instead I have now empty line where this record was supposed to be. I would like it to ignore the line and bring the next record directly under it.

 

Can you please help me !!!

 

Regards,

Emiliya

[Coreye]

You could do this:

<?php
if(!$row_rsList['WL']) {
	echo "images/emptyGraphic.gif"
} else {
	echo $row_rsList['WL'];
}
?>

[JD*]

Or a slimmer way:

 

<?php 
    echo (!$row_rsList['WL'] ? "images/emptyGraphic.gif" : $row_rsLiost['WL']);
?>

 

This does a comparison of two cases.

[keeB]

Slimmer != better. This guy is writing in dreamweaver. That tells you he's not ready for short hand if statements.

[hemangi]

Thanks very much guys, it all works fine now! I like it slim 

and I am a girl