異なるジョイン条件で同じフィールドを使用する方法は？

Question

私はこの問題を抱えています。私は、次の表を持っている：

member 
id | username  | role_id | full_name  | 
1 | abc@email.com | 1  | administrator | 
2 | bcd@email.com | 2  | Sunkist   | 
3 | cde@email.com | 2  | BlueJam   | 
4 | def@email.com | 3  | Fresh Shop  | 
5 | efg@email.com | 3  | Other Shop  | 

role 
id | role   | 
1 | superadmin | 
2 | vendor  | 
3 | shop   | 

fruits 
id | fruit_name | barcode | vendor_id | 
1 | banana  | 12345 | 2  | 
2 | melon   | 23456 | 2  | 
3 | apel   | 34567 | 3  | 
4 | orange  | 45678 | 3  | 
5 | papaya  | 56789 | 2  | 

shop_base 
id | fruit_id | member_id | 
1 | 1  | 4  | 
2 | 1  | 5  | 
3 | 2  | 4  | 
4 | 2  | 5  | 
5 | 3  | 5  | 
6 | 4  | 5  | 
7 | 5  | 5  | 

私はこのクエリで元気です：

SELECT f.barcode, f.fruit_name, m.full_name AS vendor, f.id AS fruit_id 
FROM fruits AS f 
LEFT JOIN member AS m ON f.vendor_id = m.id 
WHERE f.vendor_id > 0 
GROUP BY f.barcode 
ORDER BY f.barcode DESC 

結果：

barcode | fruit_name | vendor | fruit_id | 
56789 | papaya  | Sunkist | 5   | 
45678 | orange  | BlueJam | 4   | 
34567 | apel   | BlueJam | 3   | 
23456 | melon   | Sunkist | 2   | 
12345 | banana  | Sunkist | 1   | 

が、今私はこのようなお店の列を追加する必要があります。

barcode | fruit_name | vendor | fruit_id | shop_name    | 
56789 | papaya  | Sunkist | 5   | Other Shop    | 
45678 | orange  | BlueJam | 4   | Other Shop    | 
34567 | apel   | BlueJam | 3   | Other Shop    | 
23456 | melon   | Sunkist | 2   | Fresh Shop, Other Shop | 
12345 | banana  | Sunkist | 1   | Fresh Shop, Other Shop | 

これは私がこれまでに得た方法ですが、方法はshop_nameフィールドにnullを返します。

SELECT f.barcode, f.fruit_name, m.full_name AS vendor, f.id AS fruit_id, GROUP_CONCAT(CONCAT(CASE WHEN m.id = s.member_id THEN m.full_name END) SEPARATOR ', ') shop_name 
FROM fruits AS f 
LEFT JOIN member AS m ON f.vendor_id = m.id 
LEFT JOIN shop_base AS s ON m.id = s.member_id 
WHERE f.vendor_id > 0 
GROUP BY f.barcode 
ORDER BY f.barcode DESC 

私はprobleがこの上だと思う： "GROUP_CONCAT（CONCAT（CASE m.id = s.member_id THEN m.full_name END）SEPARATOR ''）shop_name" shop_nameとvendorはmember.role_idの同じフィールドにあります

誰でも教えてください。私は非常に感謝します:)

Answer 1

あなたはmemberと二度参加する必要があります。一度fruits.vendor_idに基づいてベンダー名を取得し、shop_baseに基づいてショー名を取得するには別途取得してください。

SELECT f.barcode, f.fruit_name, m.full_name AS vendor, f.id AS fruit_id, GROUP_CONCAT(m1.full_name SEPARATOR ', ') shop_name 
FROM fruits AS f 
LEFT JOIN member AS m ON f.vendor_id = m.id 
LEFT JOIN shop_base AS s ON f.id = s.fruit_id 
LEFT JOIN member AS m1 ON s.member_id = m1.id 
WHERE f.vendor_id > 0 
GROUP BY f.barcode 
ORDER BY f.barcode DESC 

DEMO