Question

我有两个表：一个，主表显示星期几，当天显示汤。桌上的汤，有一份可能的汤清单。

+-----+------------+--------+
| id  | day        |id_soup |
+-----+------------+--------+
| 1   | Friday     | 1      |
| 2   | Saturday   | 1      |
| 3   | Sunday     | 2      |
+-----+------------+--------+

+-----+------------+
| id  | soup       | 
+-----+------------+
| 1   | potatoes   | 
| 2   | carrots    | 
| 3   | peas       |
+-----+------------+

要获得星期五汤（字符串不是id），我可以使用这样的查询：

SELECT ma.id, ma.day, ma.id_soup, so.soup
FROM main ma, soup so
WHERE ma.id_soup = so.id
AND ma.id = 1

我得到这样的东西（输出）

+-----+------------+---------+----------+
| id  | day        | id_soup | soup     |
+-----+------------+---------+----------+
| 1   | Friday     | 1       | potatoes |
+-----+------------+---------+----------+

如果星期五只提供一汤，这种情况很有效。

问题是周五将有两个汤（土豆和豌豆 - 一系列汤）

我们需要另一张桌子吗？

EDITED

+-----+---------+---------+
| id  | id_day  | id_soup |     
+-----+---------+---------+
| 1   | 1       | 1       |
| 2   | 1       | 2       |
| 3   | 2       | 2       |
| 4   | 2       | 3       |
| 5   | 3       | 1       |
| 6   | 3       | 3       |
+-----+---------+---------+

除了主表（id，day，id_soup）的其他字段，还可以在单个查询中获取汤数组（土豆，豌豆）？

构建此查询的最佳方法是什么？

编辑：

期望的输出（仅用于说明）：

+-----+------------+---------+------------------+
| id  | day        | id_soup | soup             |
+-----+------------+---------+------------------+
| 1   | Friday     | 1       | potatoes, peas   |
+-----+------------+---------+------------------+

EDITED

问题：如果需要的输出如下：

+-----+------------+-------------------+
| id  | day        | soup              |
+-----+------------+---------+---------+
| 1   | Friday     | potatoes, carrots |
+-----+------------+---------+---------+
| 2   | Saturday   | carrots, peas     |
+-----+------------+---------+---------+
| 3   | Sunday     | potatoes, peas    |
+-----+------------+---------+---------+

Answer 1

是的，如果每天有多个汤，您将需要一个中间表。您可以使用GROUP_CONCAT声明中的SELECT为所有汤写一天。

http://dev.mysql.com/doc/refman/5.7/en/group-by-functions.html#function_group-concat

SELECT ma.id, ma.day, GROUP_CONCAT(DISTINCT so.soup) as soup
FROM main ma
LEFT JOIN soupmain sm ON(ma.id = sm.id_main)
LEFT JOIN soup so ON(so.id = sm.id_soup)
WHERE ma.id = 1
GROUP BY ma.id

Answer 2

您可以使用group_concat

SELECT ma.day,  groupconcat(so.soup)
FROM main ma, soup so
WHERE ma.id_soup = so.id
AND ma.id = 1
group by ma.day

加入两个表，从一个表中获取一组值

2 个答案: