If Esle Clause

[Rossi]

Hi, I am trying to create an if else clause that based on whether the database has information in a particular column a hyperlink will or will not be displayed.

I have currently done the following

<?php $pdf = $row_getProducts['part_pdf']; if ($pdf == "NULL" ) { echo "testing"; } else echo "" ?>

I have reversed the display so I can check it is working ie If NULL it will display testing! however when working I would swap this with the empty echo “”

I think it might be the first line that is the problem but dont know why. If anyone can help me it would be much appreciated.

Regards
Ross.

[sarthakpatel]

semicolon is missing after else echo

Otherwise there’s no problem in your code and i think it should work. If it doesn’t, then let me know.

[Rossi]

thanks for that, I did as you said:

<?php $pdf = $row_getProducts['part_pdf']; if ($pdf == "NULL") { echo "test"; } else echo ""; ?>

still gives nothing, it is as if it is not checking the part_pdf from the database, if I remove this and replace with a basic “xxxx” comparison it works.

I have no idea why it wont work.

[sarthakpatel]

while comparison with null we don’t put quotes " ". If you do so it will consider as a string instead of null

So, try this:

[php]<?php
$pdf = $row_getProducts[‘part_pdf’];
if ($pdf == NULL) {
echo “test”;
}
else echo “”;
?>[/php]

[Rossi]

I think that has done the trick! thank you very very much! i knew it had to be somthing simple but couldn’t put my finger on it.