从php mysql中的另一个视图创建视图

Question

I am trying to create a view from another view from php and im not getting any error but it simple does not create a view. I can manually create view from another in the mysql console but not from php. Any idea where I am going wrong?

function createTransaction_file($db,$file_id){

      $sql = "CREATE VIEW transaction_file AS SELECT context,transaction_type,starttime,stoptime,stoptime - starttime AS runtime,correlator,parent_correlator,iteration FROM transactions WHERE file_id =" . $file_id . " ORDER BY starttime" ;

      //echo $sql;

      if($stmt = $db->prepare($sql)){

         /* execute query */
         $stmt->execute();
      }else{

       echo "File id is ". $file_id;

       echo "Error code ({$db->errno}): {$db->error}";

       die("Could not create transaction_iter view");
    }

    /* close statement */
    $stmt->close();

}

function createTransaction_iter($db,$iteration){

      $sql = "CREATE VIEW transaction_iter AS SELECT context,transaction_type,starttime,stoptime,stoptime - starttime AS runtime,correlator,parent_correlator,iteration FROM transaction_file WHERE iteration = ".$iteration." LIMIT 500;" ;

      //echo $sql;

      if($stmt = $db->prepare($sql)){

         /* execute query */
         $stmt->execute();


      }else{

       echo "File id is ". $file_id;

       echo "Error code ({$db->errno}): {$db->error}";

       die("Could not create transaction_iter view");
    }

    /* close statement */
    $stmt->close();

}

createTransaction_file($db,$file_id);
createTransaction_iter($db,$iteration);

Answer 1

I think you are simply not getting the details of the error back. Again, that if/else block is not what you meant to do I believe. Try something like this:

$stmt = $db->prepare($sql);

if (! $stmt->execute()) {
    $arr = $stmt->errorInfo();
    print_r($arr);
}  

UPDATE: So as expected that error message suggests a lack of required permissions for your user. I'm guessing your views were created with different rights. Here is a similar SO posting: mysql forgets who is logged in: command denied to user ''@'%' that discusses how you can get insight into the rights of those views/tables.