GrahamMercer
-
Posts
4 -
Joined
-
Last visited
Content Type
Profiles
Forums
Events
Posts posted by GrahamMercer
-
-
I am getting the following error message
SQL error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'order by RaceSetupID' at line 1 SQL errno: 1064 SQL: select * from hesketh_racesetups where IndividualRaceID = order by RaceSetupID
The offending line of code is
$sql = "select * from xxxxxxx_racesetups where IndividualRaceID = $mrid order by RaceSetupID";
($mrid is previously set by the line $mrid = $row->IndividualRaceID;)
Thanks for any help that can be offered
Graham
-
You guys are awesome!Thanks so much for the code (and correction), your help is greatly appreciated.Talk about helpful, less than an hour from when I posted my question and the problem is solved!Cheers,Graham
-
Hi and first of all thanks to everyone here that takes the time to help out newbies such as myself. I have a conditional IF statement that allows me to add a link to a piece of text if that record has a website, but I would like to also be able to display the text without the link if there is no website associated with the record.The code that I am currently using is
<?php if($row_RSsales['website']!='') { ?><a href="<?php echo $row_RSsales['website']; ?>" target="_blank"><?php echo $row_RSsales['name']; ?></a><?php } ?>
The trouble with this is that if a particular record does not have a website then the code does not echo the business name. I would like to include an ELSE statement with the IF statement so that when a record does not have a website entry the business name is still echoed, but not hyperlinked.I have been fiddling with code but cannot figure out how to do it.
<?php if($row_RSsales['website']!='') { ?><a href="<?php echo $row_RSsales['website']; ?>" target="_blank"><?php echo $row_RSsales['name']; ?></a>;else <?php echo $row_RSsales['name']; ?><?php } ?>
doesn't work and I am not sure how to solve it.Any suggestions would be greatly appreciated!thanks,Graham
code debug help please?
in PHP
Posted
Ok thanks, I will look into that.