Question

    $tran = "START TRANSACTION;";  

    $tran_res = mysql_query($tran);  

    $qry_1 = "INSERT INTO docList (doc_ip , doc_country , doc_ref) VALUES ('$ip' , '$country' , '$http_ref');";  
    $res_1 = mysql_query($qry_1);  
    if(!$res_1)
        die ("qry1 fail " . mysql_error() );  

    $ins_id = mysql_insert_id();  
    if(!$ins_id) 
        die ("ins id fail " . mysql_error() );  
    echo "<b>$ins_id</b>";  

    $qry_2 = "INSERT INTO docContent (doc_id , cont_date , cont_title , cont_aim , cont_obj , cont_theory , cont_sw , cont_code) VALUES ('$ins_id' , '$dt' , '$title' , '$aim' , '$obj' , '$th' , '$sw' , '$code');"; 

    $res_2 = mysql_query($qry_2);  
    if(!$res_2)
        die("qry2 fail " . mysql_error() );  `

The execution of above is returning the following error:

2 qry fail 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 'login'); if($query->num_rows()>0) return $query->result_array(); } ' at line 1

In effect the execution of $qry_2 is failing, but I am perplexed by the error it is showing (there is no such code on line 1 as mentioned in the error note). Moreover, the query ($qry_2) executes properly in the MySql console.

Answer 1

You haven’t posted the real query as received by the MySQL server, but I’d dare say you haven’t used mysql_real_escape_string() to inject your data into your SQL.

(Are you trying to insert PHP code in the database?)