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. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
