where the error with query[php]

lse123 · #1 (**permalink**) 11-11-10, 16:08

where the error with query[php]:

Code:

$queryAPPEAR2 = "select p.* s.* from $Products p join $Sides s on s.Product_ID = p.Product_ID AND p.Visibility=1 order by s.AA";  

$result2 = @mysql_query($queryAPPEAR2,$linkid);
$count2 = @mysql_num_rows($result2);
....................
line 214; mysql_free_result($result2);

// $Sides subset of $Products
this gives

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in /home/content/p/o/l/polisphotos/html/store/admin/sidesFeatured.php on line 214

r937 · #2 (**permalink**) 11-11-10, 16:42

Quote:

Originally Posted by lse123

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

that is simply a generic php message that says your mysql query crapped out

you want to run the query outside of php, so that you can get the actual mysql error message

(i know what the problem is, but i want to teach you how to find it)

lse123 · #3 (**permalink**) 11-11-10, 16:52

OK, I run in PHPMyAdmin and get error, but not identify yet where, is it the case?
I can NOT run CLI since shared hosting environment have for hosting...

lse123 · #4 (**permalink**) 11-11-10, 17:06

is it because there are common fields?

r937 · #5 (**permalink**) 11-11-10, 17:07

Quote:

Originally Posted by lse123

OK, I run in PHPMyAdmin and get error

and the error message was ... ?

lse123 · #6 (**permalink**) 11-11-10, 17:17

is it the comma? I modified to [remove join, is it required?]
select Products.* SidesFeaturedList.* from $Products, $Sides where SidesFeaturedList.Product_ID=Products.Product_ID AND Products.Visibility=1 order by SidesFeaturedList.AA
get:
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 '.* from Products, SidesFeaturedList where SidesFeaturedList.Product_ID=Products.' at line 1

lse123 · #7 (**permalink**) 11-11-10, 17:20

was the comma now functions thks

r937 · #8 (**permalink**) 11-11-10, 17:36

Quote:

Originally Posted by lse123

was the comma

yes it was

now please put back the JOIN

lse123 · #9 (**permalink**) 11-11-10, 18:09

is it join required...and you say this, I do not think so?
final now:

$queryAPPEAR2 = "select Products.*, SidesFeaturedList.* from $Products, $Sides where SidesFeaturedList.Product_ID=Products.Product_ID AND Products.Visibility=1 order by SidesFeaturedList.AA";

r937 · #10 (**permalink**) 11-11-10, 20:17

JOIN syntax is better

Code:

SELECT Products.*
     , SidesFeaturedList.* 
  FROM Products
INNER
  JOIN SidesFeaturedList 
    ON SidesFeaturedList.Product_ID = Products.Product_ID 
 WHERE Products.Visibility = 1 
ORDER 
    BY SidesFeaturedList.AA

the missing comma that caused the error was in the SELECT clause

there should be no commas in the FROM clause

guelphdad · #11 (**permalink**) 11-12-10, 09:22

Quote:

Originally Posted by lse123

is it join required...and you say this, I do not think so?
final now:

$queryAPPEAR2 = "select Products.*, SidesFeaturedList.* from $Products, $Sides where SidesFeaturedList.Product_ID=Products.Product_ID AND Products.Visibility=1 order by SidesFeaturedList.AA";

This is still a JOIN, your join is the SidesFeaturedList.Product_ID=Products.Product_ID in the WHERE clause.

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