bad sql statement

[stew]

SELECT * FROM gallery WHERE img_cat = '$cat' ORDER BY uploaded_on DESC LIMIT $limit, 12

hey there guys... the above SQL statement doesn't work... does anyone know why?!?! oh and i know that the field names are correct!

stew

[njm]

What happens when you run it? What does mysql_error() tell you?

hth, njm

[keith73]

Originally posted by stew

SELECT * FROM gallery WHERE img_cat = '$cat' ORDER BY uploaded_on DESC LIMIT $limit, 12

hey there guys... the above SQL statement doesn't work... does anyone know why?!?! oh and i know that the field names are correct!

stew [/B]

where are $cat and $limit set.

It may be the case that $limit is not set so you are getting an error because LIMIT , 12 is not correct.

put that statement into a variable and output the variable. Also output mysql_error(). These are good debugging methods that will save you the trouble of waiting for an answer on the forum.

• keith

[stew]

mysql_fetch_array(): supplied argument is not a valid MySQL result resource

and with mysql_error() it says:

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 ' 12' at line 1

ive checked the manual and its a bit hard to understand, also ive run in thru phpmyadmin and it works.

[keith73]

Originally posted by stew
mysql_fetch_array(): supplied argument is not a valid MySQL result resource

and with mysql_error() it says:

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 ' 12' at line 1

ive checked the manual and its a bit hard to understand, also ive run in thru phpmyadmin and it works.

output the query to the screen.
the problem is obviously with the limit statement.

again, where are $limit and $cat being set in the code?

• keith

[stew]

shit right... sorry... i fixed it.

i added the mysql_error() thing and realised what the problem was ,

cheers