Php mysql create database if not exists

Question

I want to create a database. Why is not the db created with this code?

$dbname = 'regulations_db';
    $con = mysql_connect("localhost","root","pass");
    if (!$con)
    {
        die('Could not connect: ' . mysql_error());
    }
if (mysql_num_rows(mysql_query("SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = '". $dbname ."'"))) {
        echo "Database $dbname already exists.";
    }
    else {
        mysql_query("CREATE DATABASE '". $dbname ."'",$con);
        echo "Database $dbname created.";
    }

This is working, but I think the first one is the best practice:

if (mysql_query("CREATE DATABASE IF NOT EXISTS regulations_db",$con))
    {
        echo "Database created";
    }
    else
    {
        echo "Error creating database: " . mysql_error();
    }

Community · Accepted Answer · 2017-05-23 10:31:22Z

26

Just do a simple mysql_select_db() and if the result is false then proceed with the creation.

As an example, check out the first answer here by another very smart StackOverflower.

<?php
// Connect to MySQL
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}

// Make my_db the current database
$db_selected = mysql_select_db('my_db', $link);

if (!$db_selected) {
  // If we couldn't, then it either doesn't exist, or we can't see it.
  $sql = 'CREATE DATABASE my_db';

  if (mysql_query($sql, $link)) {
      echo "Database my_db created successfully\n";
  } else {
      echo 'Error creating database: ' . mysql_error() . "\n";
  }
}

mysql_close($link);
?>

edited May 23, 2017 at 10:31

CommunityBot

11 silver badge

answered Jan 30, 2012 at 18:28

buley

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2 Comments

Julix Over a year ago

Update: now should use mysqli_select_db, right? -- also would it make sense to repeat "$db_selected = mysqli_select_db('my_db', $link);" in the success condition (i.e. now it should work) -- or did you only use it to check for existence of the database and have therefor no further need for it anyway?

Mona Jalal Over a year ago

mysql_select_db --> is not recognized for me.

Anyone · Accepted Answer · 2017-04-30 06:01:07Z

5

Three steps to fix this:

Don’t specify the database name when connecting.
Your SQL statement should be CREATE DATABASE IF NOT EXISTS php1.
Call mysqli_select_db($link, 'php1') to make that the default database for your connection.

answered Apr 30, 2017 at 6:01

Anyone

511 silver badge1 bronze badge

1 Comment

Julix Over a year ago

Isn't that what the second code that he describes effectively does?

Community · Accepted Answer · 2020-06-20 09:12:55Z

If you're using MySQLi Object-oriented method, you can use following code, this code is similar to previous answer and only the method is different, I just put this because if anyone using MySQLi Object-oriented method, you can use this code directly.

$servername = "localhost";
$username = "mysql_user";
$password = "user_password";
$dbName = "databaseName";

// Connect to MySQL
$conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

// If database is not exist create one
if (!mysqli_select_db($conn,$dbName)){
    $sql = "CREATE DATABASE ".$dbName;
    if ($conn->query($sql) === TRUE) {
        echo "Database created successfully";
    }else {
        echo "Error creating database: " . $conn->error;
    }
}

Furthermore you can refer W3school site here.

Good Luck! :D

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