error in mysql (mysql_fetch_array)

Mellow_Heart

hi all

befor two days ,, i transfeer my website and forums to another server and evreything done ,, Except the database of script (this script library for sound and video)
the Graphical interface for this script work Without any problem ,, but sections Links Does not work
this error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/alawda11/public_html/son/cat.php on line 30

and the Download Links

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/alawda11/public_html/son/details.php on line 24

and the control panel when i need to open ant sections
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 '' at line 1

The error number returned by the database: 1064

plz help me
i Tired of trying to solve this problem
thnx all

dagon

need to see the code that generated the errors

Mellow_Heart

this code for cat.php;

$getcatname = mysql_query("SELECT * FROM `sound_cat` WHERE catid = $catid");
while ( $get = mysql_fetch_array($getcatname)) {

and this code for details.php

$results =  mysql_query("SELECT * FROM sound_link where linkid=$linkid");
    while($row=mysql_fetch_array($results)){

almojahed

Ooooops
i have the same Problem ,, with the same Script .

dagon

add some error checking to the query, with mysql_error()

Mellow_Heart

There seems to be an error in the database
The error is: The implementation of the SQL statement is wrong
The database returns the following information about the 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''at line 1
The error number returned by the database: 1064

bradgrafelman

Where are the variables $catid and $linkid ever defined?

Also, when you're debugging SQL queries, it's often helpful to output the SQL query string itself when an error occurs. It is for this reason that you'll often see the queries stored in a variable instead of placed directly inside the mysql_query() call.

almojahed

would you like to say it again with simple Words pls ... now what i have to do ?
step by step pls.

Mellow_Heart

Could not run query (sound_cut): 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 'sound_cut' at line 1

bradgrafelman

Which query is that for?

If $catid or $linkid contain strings, then you must surround those values with single quotes in the SQL query.

almojahed

its not resolved , thanx anyway guys
may b can Mr Mellow Heart find any Solution
Good Luck

dagon

if you would answer bradgrafelman's questions we can get closer to the issue, but as we are not psychic, we cant help with out information.