SQL如何使用具有不同连接条件的相同字段？答案

【问题标题】：SQL how to use the same field with different join condition?SQL如何使用具有不同连接条件的相同字段？
【发布时间】：2017-03-02 22:11:35
【问题描述】：

好的，我遇到了这个问题。我有以下表格：

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       |

我对这个查询很好：

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

结果：

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         |

但现在我需要像这样添加商店栏：

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 |

这就是我到目前为止的方式，但它总是在 shop_name 字段上返回 null ：

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

我认为问题在于：“GROUP_CONCAT(CONCAT(CASE WHEN m.id = s.member_id THEN m.full_name END) SEPARATOR ', ') shop_name” shop_name 和 vendor 来自 member.role_id 上的同一字段

有人可以帮助我吗？我将不胜感激:)

【问题讨论】：

A 在0 之后应该是什么意思？这对我来说似乎是一个语法错误。
m.id = s.member_id 将始终为真，因为它是连接中的 ON 条件。
@Barmar 对不起我的错，当我写下我已经解决的问题时，我觉得我的手指滑了

标签： mysql sql concatenation concat group-concat

【解决方案1】：

您必须加入member 两次。一次根据fruits.vendor_id获取厂商名，另外根据shop_base获取节目名。

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

【讨论】：