Question

虽然我之前曾问过类似的问题，但我试图对它应用相同的技术，它只是不能正常工作，错误和各种各样的观点。

我为此创建了一个sqlfiddle; http://sqlfiddle.com/#!2/b1a29

我正在尝试创建一个select函数，它将返回animal_id，animal_name，animal_type_name，shelter_name，animal_type_id和location_name。

我试图用以下代码擅长它，但很明显我错过了一些东西;

$query = $this->db->query('SELECT animal_id, animal_name, animal_type_name, shelter_name, shop_id, location_name
                                FROM animals a
                                INNER JOIN shelter s ON s.shop_id = a.shop_id
                                INNER JOIN location l ON l.location_id = s.location_id
                                INNER JOIN animal_types at ON at.animal_type_id = a.animal_type_id');

Answer 1

问题是您正在选择不明确或存在于多个表中的列名。在这种情况下，它是shop_id。为了更好地执行查询，您需要指定列shop_id的来源，

SELECT animal_id, animal_name, animal_type_name, 
       shelter_name, s.shop_id, location_name
FROM animals a
INNER JOIN shelter s ON s.shop_id = a.shop_id
INNER JOIN location l ON l.location_id = s.location_id
INNER JOIN animal_types at ON at.animal_type_id = a.animal_type_id

SQLFiddle Demo

Answer 2

运行代码时，会产生错误：

专栏＆＃39; shop_id＆＃39;在字段列表中是不明确的：SELECT animal_id， animal_name，animal_type_name，shelter_name，shop_id，location_name 来自动物的INNER JOIN避难所s on s.shop_id = a.shop_id INNER JOIN location l ON l.location_id = s.location_id INNER JOIN ON at.animal_type_id = a.animal_type_id
中的animal_types

从错误：您的shop_id列属于更多的一个表，因此您必须使用表名/表别名对其进行限定。

将您的查询更改为：

SELECT animal_id, animal_name, animal_type_name, shelter_name, s.shop_id, location_name
                                FROM animals a
                                INNER JOIN shelter s ON s.shop_id = a.shop_id
                                INNER JOIN location l ON l.location_id = s.location_id
                                INNER JOIN animal_types at ON at.animal_type_id = a.animal_type_id

mysql选择问题，多表

2 个答案: