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In this function create I am trying to create a database. However, it fails to create and displays the error "Error creating database: Access denied." SO this means there is a connection and makes it passed the first check. Then when the mysql_query is called, it doesnt return true. Can anyone help? 

function create($dbName, $tbName, $fields, $types_sizes, $PK)
        {
            $sql = null;
            $con = mysql_connect("borg.cs.up.ac.za","username", "password");

            if (!$con)
            {
                die('Could not connect: ' . mysql_error());
            }
            else 
                {
                  if(mysql_query("CREATE DATABASE". $dbName, $con))
                    {
                        echo "Database created";
                        $dbExists = 1;
                    }
                   else
                    {
                        echo "Error creating database: " . mysql_error();
                        $dbExists = 0;
                    }
                   $i = 0;
                    if(($con)&&($bdExists = 1))
                    {
                        $sql . "CREATE TABLE " . $tbName . "(";
                        while($i < count($fields))
                        {
                            $sql . $fields[$i] . " " + $types_sizes[$i] . ",";
                            $i++;
                        }
                    if($PK != null)
                    {
                        $sql . "PRIMARY KEY (" . $PK .")";
                    }
                    }
                mysql_query($sql,$con);
                mysql_close($con);
            }
         }
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2  
You probably do not have the necessary privileges to create databases on the account you connect to the server. – Jeroen Bolle May 14 '12 at 13:11
Also you may set prefix ($dbName_.'tablename') in name for separate tables – Sergey May 14 '12 at 13:12
Create the database using PhpMyAdmin or MySQL Workbench then just create the tables via code. It seems you do not have the required permissions to create a database. – Omtara May 14 '12 at 13:13
I do have the full amount of privileges so thats not a problem.I have an idea its got something to do with the connection, and not the query. – Chris May 14 '12 at 13:16
@Chris the error message exactly telling you that you don't have the privileges. You wouldn't have got that message had it been connection issue. and mysql_* is deprecated officailly. You should learn mysqli or better PDO. – itachi May 14 '12 at 13:17

2 Answers

up vote -1 down vote

you have not given space in the line

if(mysql_query("CREATE DATABASE". $dbName, $con)), please provide space after "Create Database" -> "create database ".$dbname, because otherwise it will be database+name combined.

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this query didn't run in the 1st place because access was denied. – itachi May 14 '12 at 13:21
let the user run it once and then decide whether it is working or not. – Gaurav Bansal May 14 '12 at 13:22
OK so @Gaurav Bansal. Thanks. There was a space missing. But now its giving me this error "Error creating database: Access denied for user 's11028786'@'%' to database 'books' ", any ideas? i have to have access, because i successfully connect in my previous function. – Chris May 14 '12 at 13:26
2  
If you get an error such as Access denied for user 'monty'@'localhost' to database 'menagerie' when attempting to create a database, this means that your user account does not have the necessary privileges to do so. ----- mysql dev guide. – itachi May 14 '12 at 13:27
up vote 0 down vote

Try this. This is untested.Changed some lines in your code.

function create($dbName, $tbName, $fields, $types_sizes, $PK)
    {
        $sql = null;
        $con = mysql_connect("borg.cs.up.ac.za","username", "password");

        if (!$con)
        {
            die('Could not connect: ' . mysql_error());
        }
        else 
            {
    if(mysql_select_db($dbName))
                {
                    echo "Database selected";
                    $dbExists = 1;
                }
               else
                {
                    echo "Error creating database: " . mysql_error();
                    $dbExists = 0;
                }
               $i = 0;
                if(($con)&&($dbExists == 1))
                {
                    $sql = "CREATE TABLE " . $tbName . "(";
                    while($i < count($fields))
                    {
                        $sql .= $fields[$i] . " " + $types_sizes[$i] . ",";
                        $i++;
                    }
                if($PK != null)
                {
                    $sql .= "PRIMARY KEY (" . $PK .")";
                }
                }
            mysql_query($sql,$con);
            mysql_close($con);
        }
     }
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