我用一个函数外键将两个表链接在一起。
我犯了一个错误:“errno:150”外键约束的格式不正确。
我找不出问题出在哪里。请,我将感谢任何建议。
谢谢你的帮助。
这是我用php编写的脚本(见下文)。
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db = "myDB";
$conn = new mysqli($servername, $username, $password, $db);
if ($conn->connect_error) {
die ( "Connection failed: " . $conn->connection_error);
}
echo "Connected successfully";
$sql = "CREATE DATABASE IF NOT EXISTS myDB";
if ($conn->query($sql) === TRUE) {
echo "Database created successfully";
} else {
echo "Error creating database: " . $conn->error;
}
$sql = "CREATE TABLE IF NOT EXISTS us_tAddress (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
country VARCHAR(100),
city VARCHAR(100),
zip VARCHAR(100),
street VARCHAR(100),
created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
updated_at DATETIME DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP
) ";
if ($conn->query($sql) === TRUE) {
echo "Table us_tAddress created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
$sql = "CREATE TABLE IF NOT EXISTS us_tUser (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
address_id INT (6),
first_name VARCHAR(100),
last_name VARCHAR(100),
position VARCHAR(100),
created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
updated_at DATETIME DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
FOREIGN KEY (address_id) REFERENCES us_tAddress (id)
) ";
if ($conn->query($sql) === TRUE) {
echo "Table us_tUser created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
$conn->close ();
?>
1条答案
按热度按时间dddzy1tm1#
您的字段具有不同的数据类型。
引用的列是类型为id int(6)的无符号自动增量主键的us\u taddress中的id。
引用列是来自类型为int(6)的us\u tuser的address\u id。
必须将表us\u taddress中id列的类型更改为signed,或将表us\u tuser中字段address\u id的数据类型更改为unsigned。