Question

我知道如何在PDO中创建数据库，但是我在进行“检查”时遇到了问题。如果你没有'this'数据库（数据库可能是mydb），我需要阅读，然后创建它，然后回显出“mydb已成功创建。”

如果已经存在具有该名称的数据库，则回显“mydb found”。就像我说我在这里使用PDO。我将展示到目前为止我所拥有的一切。我还将包括如何制作支票以制作我需要的表格。

提前感谢您的帮助！

<?php
//Database creation
require_once 'core/init.php';

$servername = Config::get('mysql/host');
$username = Config::get('mysql/username');
$password = Config::get('mysql/password');
$dbname = Config::get('mysql/db');
$createtable = false;


try {
    $conn = new PDO("mysql:host=$servername", $username, $password);
    // set the PDO error mode to exception
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $sql = "CREATE DATABASE IF NOT EXISTS $dbname";
    // use exec() because no results are returned
    $conn->exec($sql);
    echo "Database created successfully<br>";
     $createtable = true;
    }
catch(PDOException $e)
    {
    echo $sql . "<br>" . $e->getMessage();
    }

$conn = null;
$createtable = true;
} else {
  echo Config::get('mysql/db'). '- found!';
}

//Table1 creation
if($createtable){
try {
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);

    // sql to create table
    $table = 'surveys';
    $sql = "CREATE TABLE IF NOT EXISTS $table (
    id INT(11) AUTO_INCREMENT PRIMARY KEY,
    username VARCHAR(20) NOT NULL,
    password VARCHAR(64) NOT NULL,
    last_name VARCHAR(20) NOT NULL,
    first_name VARCHAR(20) NOT NULL,
    middle_name VARCHAR(20),
    address_1 VARCHAR(32) NOT NULL,
    address_2 VARCHAR(32),
    city VARCHAR(32) NOT NULL,
    state VARCHAR(12) NOT NULL,
    zip INT(12) NOT NULL,
    salt VARCHAR(32) NOT NULL
    )";

    // use exec() because no results are returned
    $conn->exec($sql);
    echo "Table {$table} created successfully";

//Table2 creation
    $table = 'users_session';
    $sql = "CREATE TABLE IF NOT EXISTS $table (
    id INT(11) AUTO_INCREMENT PRIMARY KEY,
    user_id INT(11) NOT NULL,
    hash VARCHAR(64) NOT NULL,
    )";

    $conn->exec($sql);
    echo "Table {$table} created successfully";
  }

catch(PDOException $e)
    {
    echo $sql . "<br>" . $e->getMessage();
    }

$conn = null;
} else {
  echo "Tables are already made and ready to go!";
}
?>

Answer 1

您可以使用SHOW TABLES：

$sql = "SHOW TABLES LIKE 'tableName'";
$tableExists = $conn->query($sql)->rowCount() > 0;
if( !$tableExists ){
    //create table
}

创建一个check（if语句）以使用PDO在PHP中创建数据库

1 个答案: