CakePHP 3错误：SQLSTATE [42000]：语法错误或访问冲突：1064

Question

我收到此错误：

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'，但是我不确定。 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.* 。 如您所知，这会创建不正确的SQL别名，例如Colleges__* 。 而是选择一个表的所有列，您需要传递一个表对象。

所以您可能想要做类似的事情：-

->select($college_admins->Colleges)

假设Colleges与您的CollegeAdmins表相关联。

Answer 2

您不能为colleges。*加上别名，因为这是指colleges表中的所有列，而别名是指单个列（或表或子查询）。 您需要列出colleges表中的所有字段，并为每个字段提供一个别名，例如

select colleges.ig as colleges_id, colleges.field1 as colleges_field1, ...

sql中没有语法提供这种方式的别名。 您可能想做的是访问mysql在php中返回的元数据，以检索每个字段的表名。