mysqli和php創建一個不存在的表

Question

我的代碼有什麼問題？

它返回以下錯誤：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'EXIST test1 (id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, firstname VARCHAR(3' at line 1

<?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指令應該是：

$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 
)"; 

- > NOT EXIST 小號