How to launch two sql queries using result of first one in second?

Question

I have 2 queries and i want to use result of first query in second one.
Following does not work for me:

$id     = $_GET['uid'];
$app_id = $_GET['apid'];
$sql    = "insert into tbl_sc (client_id,status) values ($id,1)";
mysql_query($sql) or die ($sql);
$result = mysql_insert_id();
echo $result;

$sql    = "insert into tbl_ms(m_name, ng_ID, status)
           values ($app_id,$result ,1)";
$result = mysql_query($sql) or die ($sql);

Is there any other way to get same result?

Answer 1

You could have used MySQL LAST_INSERT_ID() function. This way all this mess with insert id will be gone.

$sql    = "insert into tbl_sc (client_id,status) values ($id,1)";
if(mysql_query($sql)){
    $sql    = "insert into tbl_ms(m_name, ng_ID, status)
               values ($app_id, LAST_INSERT_ID() ,1)";
    $result = mysql_query($sql);
    if($result){
        // Process your result
    }else{
        // second query failed!
        die (mysql_error());
    }
}else{
    // first query failed!
    die (mysql_error());
}

Answer 2

$result contains an SQL resource, not the id.

$insert_id = mysql_insert_id();
$sql = "INSERT INTO tbl_ms(m_name, ng_ID, status) 
        VALUES ($app_id, $insert_id, 1)";

Don't forget to sanitize user input to avoid injection attacks.

Answer 3

$result in your code will always contain a boolean, and if it was successful, when used in the next query, this will always be 1 . You echo d the value you need, but you didn't catch it in a variable so it could be used in the next query.

Try this:

$id = mysql_real_escape_string($_GET['uid']);
$sql = "INSERT INTO tbl_sc
          (client_id, status)
        VALUES
          ($id, 1)";
mysql_query($sql) or die ("MySQL error with query ( $sql ): ".mysql_error());

$app_id = mysql_real_escape_string($_GET['apid']);
$insertId = mysql_insert_id();
$sql = "INSERT INTO tbl_ms
          (m_name, ng_ID, status) 
        VALUES
          ($app_id, $insertId ,1)";
mysql_query($sql) or die ("MySQL error with query ( $sql ): ".mysql_error());

You MUST escape user input before using it in a query - you don't want a visit from Bobby Tables ...

Answer 4

Make a variable $insertedID = mysql_insert_id(); just before the second $sql variable ! And in the second $sql query replace the $result with $insertedID

It should solve your problem !

Answer 5

In the second query just use

insert into tbl_ms(m_name, ng_ID, status)
           values ($app_id,last_insert_id() ,1)

no need to play this via PHP!