On Monday 02 January 2006 7:54 pm, Ralph wrote:
I'm building a URL based on data extracted from mySQL and I want the
text space.gif used as my default value should the variable $picture
by NULL.
The results of the code below kick back a value for $picture when it
is set in the database, but ignores the space.gif if the database
result is empty.
If I flip the test from ! isset to isset then I get the space.gif for
each result.
I've had no luck searching out an answer.
Thanks in advance.
--------------The Code-----------------------------
<?php do { ?>
<img src="images/<?php
$picture=$row_rst_screenings['Picture'];
if(! isset($picture)) {
$picture='space.gif';
}else{
$picture=$row_rst_screenings['Picture'];
}
echo $picture;
?>"><br>
<p><span class="style2"><?php echo $row_rst_screenings['Title']; ?></
span><br />
<span class="style3"><?php echo $row_rst_screenings['Director']; ?></
span></p>
<p class="style4"><?php echo $row_rst_screenings['Blurb']; ?></p>
<p class="style3">Showtime: <?php echo $row_rst_screenings
['ShowTime']; ?> <?php echo $row_rst_screenings['ShowDate']; ?></p>
<p class="style3">Location: <?php echo $row_rst_screenings
['Location']; ?></p>
<?php } while ($row_rst_screenings = mysql_fetch_assoc
($rst_screenings)); ?>
---------------END-----------------
Seeing as you're querying, use mysql_num_rows() to check.