Detecting/creating MySQL database

Why this code does not recognize the existing database?

$result = mysql_query("SHOW DATABASES LIKE $database", $conn);
echo $result."       aaaaaaaaaaaaaaaaaaa<br>";
if ($result) 
{
    echo ("Datebase <font color='#FF0000'><b> $database </b></font> already exist.<br /><br />");
}
else
{
    mysql_query("create database IF NOT EXISTS $database CHARACTER SET utf8 COLLATE utf8_general_ci");
    echo ("Datebase <font color='#FF0000'><b> $database &开发者_Python百科lt;/b></font> is succesfully created.<br /><br />");
}

This code make my somekinda sick...

Here is a good code:

<?php

$query = "SHOW DATABASES LIKE ". $database;
$mysqlquery = mysql_query($query);
if($mysqlquery){
    echo '<b>Database already exist.<br>';
}
else
{
    $createquery = "CREATE DATABASE IF NOT EXISTS ". $database ." CHARACTER SET utf8 COLLATE utf8_general_ci";
    $mysqlcreatequery = mysql_query($createquery);
    if($mysqlcreatequery){
        echo 'Database '. $database .' created. ';
    }
    else
    {
        echo 'Database is not created!';
    }
}
?>

You forgot the quotes (always necessary when using strings in SQL) :

$result = mysql_query("SHOW DATABASES LIKE '$database'", $conn);

As the pattern is a string, you'll have to quote it:

"SHOW DATABASES LIKE '$database'"

The database is a string as stated by Dr.Molle. Also, watch out for wildcards. An _ is a single character wildcard, so when you look for the database order_2010 it will also match orders2010, which will lead to the assumption that order_2010 is indeed an existing database.

Small chance, but I thought I should mention it. :)