我看到很多同样的问题,但我无法解决我的问题。
如果我运行这段代码:
<?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:(如果我删除 FOREIGN KEY 它会起作用)
表文章创建成功创建表时出错:无法创建
表“admin_wepar.Tags”(错误号:150)
Edit
如果有改变into Articls.id
and Tags.id_articls
我收到这个错误:
表文章创建成功创建表时出错:您有一个
SQL 语法错误;检查与您对应的手册
MySQL 服务器版本,用于在“FOREIGN KEY”附近使用正确的语法
(Tags.id_articls) 参考文章(Articls.id) ) 默认 COLLA' at
5号线
您需要声明两者Articls.id
and Tags.id_articls
签名或未签名
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