error in your SQL syntax...ear 'Resource id #3' ???

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  • tania
    New Member
    • Jul 2006
    • 4

    error in your SQL syntax...ear 'Resource id #3' ???

    i have this query:
    $query = mysql_query("SE LECT F_ID,F_TITLE,A_ FNAME,A_LNAME FROM (take_part INNER JOIN film ON F_ID = T_P_ID)INNER JOIN actor ON A_PID = ACTOR_ID WHERE A_PID=\"$a_id\" ");
    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 'Resource id #3' at line 1
  • tania
    New Member
    • Jul 2006
    • 4

    #2
    plese help me....
    thanks in advanced!

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