我的代码有什么问题?
它返回以下错误:
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();
?>
最佳答案
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S
关于php - mysqli 和 php 创建一个不存在的表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36012548/