PHPを使用してMYSQLを接続し、存在しない場合はデータベースを作成

Question

Imは最良の方法と練習でPHPを習得しようとしていますが、エラーがないのでコードごとに適切な出力を得ることはできません！ はイムが出力echo "Database $dbname created successfully\n";

<?php 
$user = "root"; 
$pwd = ""; 
$server = "localhost"; 
$dbname = "xyz"; 

//Connecting to MYSQL 
$db_conn = mysqli_connect($server,$user,$pwd); 
if (!$db_conn) { 
    die("Connection Error".mysqli_error()); 
} 
echo "Connected Successfully"; 

$db_select = mysqli_select_db($db_conn,$dbname); 
if (!$db_select) { 
    // If we couldn't, then it either doesn't exist, or we can't see it. 
    //Create Database Query 
    $db_create = "CREATE DATABASE $dbname"; 
    $db_selected = mysqli_query($db_conn,$db_create); 

    if ($db_selected) { 
     echo "Database $dbname created successfully\n"; 
     mysqli_select_db($db_conn,$dbname); 
    } else { 
     echo 'Error creating database: ' . mysql_error() . "\n"; 
    } 
} 

?> 

Answer 1

を得ていないので、私はそれのうちのソリューションを持って存在していない場合には、データベースを作成するには良い方法です！ 天気予報の良い方法とベストプラクティスをチェックしてデータベースを追加してください！

$db_conn = mysqli_connect($server,$user,$pwd); 
if (!$db_conn) { 
    die("Connection Error".mysqli_error()); 
} 
echo "Connected Successfully"; 
$db_select = mysqli_select_db($db_conn,$dbname); 
if (!$db_select) { 
    // If we couldn't, then it either doesn't exist, or we can't see it. 
    //Create Database Query 
    $db_create = "CREATE DATABASE IF NOT EXISTS $dbname"; 
    $db_selected = mysqli_query($db_conn,$db_create); 

    if ($db_selected) { 
     echo "Database $dbname created successfully\n"; 
    } else { 
     echo 'Error creating database: ' . mysqli_error() . "\n"; 
    } 
} 
else{ 
    echo "Database $dbname is connected"; 
}