mysql select issue，multi tables

Question

雖然我之前提出過類似的問題，但我試圖將相同的技術應用於它，但它並不像它應該的那樣工作，並且出現錯誤和各種各樣的問題。

我爲此創建了一個sqlfiddle; http://sqlfiddle.com/#!2/b1a29

我試圖創建一個select函數，它將返回animal_id，animal_name，animal_type_name，shelter_name，animal_type_id和location_name。

我試圖用下面的代碼有一個很好，但很明顯，我失去了一些東西;

$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

問題是，您正在選擇不明確的列名或存在於多個表上的列名。在這種情況下，它是shop_id。爲了執行查詢好了，你需要指定列shop_id應該來自，

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

當我運行你的代碼，它產生錯誤：

Column 'shop_id' in field list is ambiguous: 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

出錯：您的shop_id列屬於更多的表，因此您必須使用表名/表別名進行限定。

查詢改成這樣：

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