php - CakePHP 3 错误 : SQLSTATE[42000]: Syntax error or access violation: 1064

Question

我收到这个错误:

Error: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'AS `Colleges__*` FROM college_admins CollegeAdmins LEFT JOIN colleges Colleges O' at line 1

这是给出此错误的 SQL 查询:

SELECT Colleges.* AS `Colleges__*` FROM college_admins CollegeAdmins LEFT JOIN colleges Colleges ON Colleges.id = (CollegeAdmins.college_id) WHERE CollegeAdmins.user_id = :c0 LIMIT 20 OFFSET 0

我启用了 quoteIdentifiers config\app，但它导致了这个新错误:

Error: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'AS `Colleges__*` FROM `college_admins` `CollegeAdmins` LEFT JOIN `colleges` `Col' at line 1

查询变为:

SELECT `Colleges`.* AS `Colleges__*` FROM `college_admins` `CollegeAdmins` LEFT JOIN `colleges` `Colleges` ON `Colleges`.`id` = (`CollegeAdmins`.`college_id`) WHERE `CollegeAdmins`.`user_id` = :c0 LIMIT 20 OFFSET 0

我认为它将“Col from Colleges”作为关键字“COL”，但我不确定。如何解决这个问题？

这是生成 MySQL 查询的 CakePHP 代码:

return $college_admins->find()
    ->select(['Colleges.*'])
    ->leftJoinWith('Colleges')
    ->where(['CollegeAdmins.user_id' => $userId]);

Answer 1

您不能在 CakePHP ORM 查询 (CakePHP 3.x) 中使用 Colleges.*。正如您所发现的那样，这会创建不正确的 SQL 别名，例如 Colleges__*。要选择表格的所有列，您需要传递一个表格对象。

所以你可能想要做这样的事情:-

->select($college_admins->Colleges)

假设 Colleges 与您的 CollegeAdmins 表关联。