我看到了很多相同的问题,但无法解决我的问题。
如果我运行此代码:
<?php
include_once($_SERVER['DOCUMENT_ROOT'].'/config.php');
$servername = HOST;
$username = USERNAME;
$password = PASSWORD;
$dbname = DB;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// sql to create table
$sql = "CREATE TABLE IF NOT EXISTS Articls (
id INT(10) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
name VARCHAR(254) COLLATE utf8_persian_ci NOT NULL
) DEFAULT COLLATE utf8_persian_ci";
if ($conn->query($sql) === TRUE) {
echo "Table Articls created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
/////////////////////////////////////////////////////////////////////////
$sql = "CREATE TABLE IF NOT EXISTS Tags (
id INT(10) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
id_articls INT(10) UNSIGNED NOT NULL,
name VARCHAR(256) COLLATE utf8_persian_ci NOT NULL,
FOREIGN KEY(Tags.id_articls) REFERENCES Articls(Articls.id)
) DEFAULT COLLATE utf8_persian_ci";
if ($conn->query($sql) === TRUE) {
echo "Table Tags created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
$conn->close();
?>
我收到此错误消息:(如果删除FOREIGN KEY,则可以正常工作)
成功创建了表Articls创建表时出错:无法创建
表'admin_wepar.Tags'(错误号:150)
编辑
如果将
into Articls.id
和Tags.id_articls
更改,则会出现此错误:成功创建了表关节创建表时出错:您有一个
您的SQL语法错误;检查与您的手册相对应的手册
MySQL服务器版本,可在'FOREIGN KEY附近使用正确的语法
(Tags.id_articls)参考Articls(Articls.id))DEFAULT COLLA'at
5号线
最佳答案
您需要声明Articls.id
和Tags.id_articls
签名或未签名
关于php - 无法在FOREIGN KEY上创建表(errno:150),我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/27679917/