Jump to content


Photo

UPDATE not working... why?


  • Please log in to reply
3 replies to this topic

#1 solisis

solisis
  • New Members
  • Pip
  • Newbie
  • 2 posts

Posted 29 April 2006 - 06:00 AM

Obviously I'm new to this so please bear with me...

My UPDATE query works fine when updating a database column with an int value to it but when I use the same exact code to update something with, say, varchar or text it doesn't work.
I can think of no other issue besides the possibility that I have to prepare the text for insertion, though this doesn't make much sense because if I use INSERT it will do its job fine. Perhaps you can help me understand why UPDATE isn't accepting the query.

Here's the basics of the code...
$query = "UPDATE ".$table." SET description = ".$val." WHERE id = ".$g."";
mysql_db_query($database, $query) or die("Failed Query");

The necessary connection code and variable values are in tact because if I swap "$query =" with "print" it prints out the correct code as it should be.

If you have any questions or comments they would be appreciated. Thanks

#2 .josh

.josh
  • Staff Alumni
  • .josh
  • 14,871 posts

Posted 29 April 2006 - 06:52 AM

so what type of error msg does it give?
Did I help you? Feeling generous? Buy me lunch! 
Please, take the time and do some research and find out how much it would have cost you to get your help from a decent paid-for source. A "roll-of-the-dice" freelancer will charge you $5-$15/hr. A decent entry level freelancer will charge you around $15-30/hr. A professional will charge you anywhere from $50-$100/hr. An agency will charge anywhere from $100-$250/hr. Think about all this when soliciting for help here. Think about how much money you are making from the work you are asking for help on. No, we do not expect you to pay for the help given here, but donating a few bucks is a fraction of the cost of what you would have paid, shows your appreciation, helps motivate people to keep offering help without the pricetag, and helps make this a higher quality free-help community :)

#3 kenrbnsn

kenrbnsn
  • Staff Alumni
  • Advanced Member
  • 8,235 posts
  • LocationHillsborough, NJ, USA

Posted 29 April 2006 - 11:13 AM

Strings need to be delimited by single quotes in the query:
<?php
 $query = "UPDATE ".$table." SET description = '".$val."' WHERE id = ".$g."";
mysql_db_query($database, $query) or die("Failed Query: $query<br>".mysql_error());
?.

Ken


#4 solisis

solisis
  • New Members
  • Pip
  • Newbie
  • 2 posts

Posted 30 April 2006 - 04:23 AM

Thank you very much kenrbnsn. I wasn't aware that I'd have to do that.
It works perfectly now.

Crayon Violent- I wasn't getting an error because all i told it to do was print failed query if it failed.






0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users