Hello,
I'm getting this 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 'IF cat_order = '3' THEN SET cat_order = '2' [ELSEI' at line 3
From this query:
$updateOrder = "
UPDATE
products_cat
IF
cat_order = '".mysql_real_escape_string($neworder)."'
THEN
SET
cat_order = '".mysql_real_escape_string($row_rs_sortvalues['cat_order'])."'
[ELSEIF
cat_order = '".mysql_real_escape_string($row_rs_sortvalues['cat_order'])."'
THEN
SET
cat_order = '".mysql_real_escape_string($neworder)."'
]
END IF
WHERE
cat_group_order = '".mysql_real_escape_string($row_rs_sortvalues['cat_group_order'])."'
AND cat_level = '".mysql_real_escape_string($row_rs_sortvalues['cat_level'])."'
";
mysql_select_db($database_conn, $conn);
$ResultOrder = mysql_query($updateOrder, $conn) or die(mysql_error());
When I echo the completed query this is what it looks like:
UPDATE
products_cat
IF
cat_order = '3'
THEN
SET
cat_order = '2'
[ELSEIF
cat_order = '2'
THEN
SET
cat_order = '3'
]
END IF
WHERE
cat_group_order = '2'
AND cat_level = '2'
Can anyone spot an error in that? This is my first time using an IF statement, am I using it correctly?
Thanks for any and all help!
Peter