mysqli 和 php 创建一个不存在的表

Question

我的代码有什么问题吗？

它返回以下错误：

您的 SQL 语法有错误；检查与您的 MySQL 服务器版本相对应的手册，了解在第 1 行的 'EXIST test1 ( id INT(6) UNSIGNED AUTO_INCRMENT PRIMARY KEY,firstname VARCHAR(3') 附近使用的正确语法

<?php
$servername = "localhost";
$username = "root";
$password = "passtest";
$database = "daily";
$table = "test1";


$conn = new mysqli($servername, $username, $password, $database);


if (mysqli_connect_error()) {
    die("Database connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";



$sql = "CREATE TABLE IF NOT EXIST $table (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL
)";

if ($conn->query($sql) === TRUE) {
    echo "Table MyGuests created successfully";
} else {
    echo "Error creating table: " . $conn->error;
}

$conn->close();
?> 

Answer 1

你的 SQL 指令应该是：

$sql = "CREATE TABLE IF NOT EXISTS $table (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL
)";

--> 不存在