Re: Output Errors 2

In article <200210221805.MAA08355@www2.ebmstpa.com>, biorn@ebmstpa.com says... > Unless it is a typo, the line that reads: > > if (image == 'NULL') { echo '<p></p></td>'; } > > should read: > > if ($image == 'NULL') { echo '<p></p></td>'; } > ^ > > You forgot the $ in front of image. > > HTH > > MB > Bzdpltd@aol.com said: > > > Hi sorry back again. > > > > Thanks for helping me get the code right. Sorry being stupid again, I am > having problems with an if / else statement. > > > > I am trying the following but it does not seem to work, what the problem > is, that I am saying if the image name field in the database is empty to > print nothing, and if it contains data to print that field name, as the a > href is in the code which pulls the image, if the field is empty it should > show nothing, but what happens is the image is shown in the browser and its > not blank. > > > > Here is the revised code: > > > > echo ("<tr> \n"); > > echo ("<td width=\"72%\" > > valign=\"top\"><h3>$title</h3> \n"); > > echo ("<p>$description</p></td> \n"); > > echo ("<td width=\"28%\" valign=\"top\"> > > \n"); > > if (image == 'NULL') { echo > > '<p></p></td>'; } > > else { echo '<p><img src="' . $image . '" > > width="201" > height="160"></p></td>'; } > > echo ("</tr> \n"); > > }} > > > > > > If someone can see the problem it would be most helpful. In addition to Maureen's point, you are currently checking whether the variable $image contains the string NULL; you can test for NULL (without the quotes) or better, use the function empty() to test whether $image has a value. -- David Robley Temporary Kiwi! Quod subigo farinam