CODEWITHSUNDEEP
phpmysql
Adriana
Adriana

Reputation: 8604

How do I select mysql database in php?

I have this code:

               if(!mysql_connect($host,$user,$passwd)){
                    die("Hoops, error! ".mysql_error());
                }

...no error from here.

                if(!mysql_select_db($db,$connect)){
                    $create_db = "CREATE DATABASE {$db}";
                    mysql_query($create_db,$connect);
                    mysql_query("USE DATABASE {$db}",$connect);
                }

..."no database selected" error from here. I would like to select database if it exists and if doesn't then create it and select it.

Why is my code not right?

Thank you in advance

Upvotes: 0

Views: 1592

Answers (7)

Adriana
Adriana

Reputation: 8604

I would like to thank to all of you, however I found fault on my side. This script was in class and one of variables were not defined inside this class. So I'm really sorry. I don't know how to consider the right answer, but I noticed my mistake after reading Clayton's answer about not properly set parameters, so I guess he is the winner ;)

Upvotes: 0

roa3
roa3

Reputation: 901

you could try w3schools website. They have a very simple and easy to learn tutorial for selecting database. The link is : http://www.w3schools.com/php/php_mysql_select.asp Hope this help :)

Upvotes: 0

PartialOrder
PartialOrder

Reputation: 2960

Where are you saving the value returned by mysql_connect()? Don't see it here. I assume $host, $user, $password and $db are properly set ahead of time. But you're passing a param to mysql_select_db that may not be properly set.

$connect = mysql_connect($host,$user,$passwd);
if (!$connect) {
    die('Could not connect: ' . mysql_error());
}
if(!mysql_select_db($db,$connect)) ...

Start by checking to see if you can select without the CREATE query first. Try a simple SELECT query to start. If you can connect, select the db, and execute a SELECT query, that's one step. Then try the CREATE query. If that doesn't work, it's almost certainly a matter of permissions.

Upvotes: 3

Noah
Noah

Reputation: 15320

Why not simply use the CREATE DATABASE IF NOT EXISTS syntax instead?

Something like this ...

  $con = mysql_connect('localhost');
  $sql = 'CREATE DATABASE IF NOT EXISTS {$db}';
  if (mysql_query($sql, $con)) {
    print("success.\n");
  } else {
    print("Database {$db} creation failed.\n");
  }
  if(!mysql_select_db($db,$connect)){
    print("Database selection failed.\n");
  }

Upvotes: 1

Zak
Zak

Reputation: 25205

You might need database create permissions for the user attempting to create the database. Then you need to operate on a valid connection resource. $connect never looks to be assigned to the connection resource.

Upvotes: 1

Ali
Ali

Reputation: 267059

Change the line

mysql_query($create_db,$connect);
mysql_query("USE DATABASE {$db}",$connect);

To

mysql_query($create_db,$connect);
mysql_select_db($db);*

and it should work.

Upvotes: 0

Tom Haigh
Tom Haigh

Reputation: 57815

You should check the return value of mysql_query() - currently if any of those calls fail you won't know about it:

if(!mysql_select_db($db,$connect)){
    if (!mysql_query("CREATE DATABASE $db", $connect)) {
        die(mysql_error());
    }

    if (!mysql_select_db($db, $connect)) {
        die(mysql_error());
    }
}

Upvotes: 0

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