使用php连接MYSQL并创建数据库（如果不存在）

Question

我试图用最好的方法和实践来学习php，但由于没有错误，我无法按照代码获得正确的输出！ 如果不存在该数据库是创建数据库的好方法，因为我没有得到输出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

我有一个解决方案！ 请检查天气是添加数据库的好方法和最佳实践！

$db_conn = mysqli_connect($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";
}