php - 创建表时出错 : Key column doesn't exist

Question

这个问题已经有答案了:

Can't create table (errno: 150) on FOREIGN KEY (2 个回答)

已关闭 8 年前。

我尝试创建这些表:

$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";



$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"

;

第一个表创建成功，但第二个表出现此错误:

Error creating table: Key column 'Tags.id_articls' doesn't exist in table

如果我删除 Tags.在Tags.id_articls和Tags.id_articls我得到 errno: 150

如果我尝试一下

   $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"

我收到此错误:

Error creating table: 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 '.id_articls) REFERENCES Articls(Articls.id) ) DEFAULT COLLATE utf8_persian_ci' at line 5

Answer 1

Tags 表中的外键应该是 id_articls 而不是 Tags.id_articls