php - 在 PHP 中为特定变量创建一个表 (MySQLi)

Question

此代码应检查用户是否在表中，如果不在，则为用户创建表。现在它只是将表命名为 $MyServer。我该如何使其名称成为变量 $MyServer.

<?php
    require "conn.php";
    echo "debug 1";
    $stmt = $conn->prepare("SELECT * FROM UserData WHERE username = ?");
    $stmt->bind_param('s',$username);
    $username = $_POST["username"];
    $stmt->execute();
    $stmt->store_result();

    echo "debug 2";
    if ($stmt->num_rows == 0){ // username not taken
        echo "debug 3";
        $stmt2 = $conn->prepare("INSERT INTO UserData (username, password) VALUES (?, ?)");
        $password =($_POST["password"]);
        $username =($_POST["username"]);
        $stmt2->bind_param('ss', $username, $password);
        $stmt2->execute();
        $MyServer =($_POST["username"]);
        $sql = ('CREATE TABLE $MyServer (
            id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
            username VARCHAR(30) NOT NULL
            )');
        if($conn->query($sql) === TRUE){
            echo "Table created successfully";
        } else {
            echo "Table is not created successfully ";
        }

        if ($stmt2->affected_rows == 1){
            echo 'Insert was successful.';

        }else{ echo 'Insert failed.';
            var_dump($stmt2);
        }
    }else{ echo 'That username exists already.';}

    ?>

Answer 1

为每个用户创建表不是一个好主意。您需要创建一个表来存储所有用户记录

但如果你真的想这样做。您需要将 ' 更改为 " 您的变量是否可以取值

$sql = "CREATE TABLE $MyServer (
            id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
            username VARCHAR(30) NOT NULL
            )";

看看区别:

http://docs.php.net/manual/en/language.types.string.php