Question

我有一张这样的表

id - price - date
1  - 10    - 01/01/2017
2  - 10    - 02/01/2017
3  - 10    - 03/01/2017
4  - 10    - 04/01/2017
5  - 10    - 05/01/2017
6  - 20    - 06/01/2017
7  - 20    - 07/01/2017
8  - 20    - 08/01/2017
9  - 20    - 09/01/2017
10 - 10    - 10/01/2017
11 - 10    - 11/01/2017
12 - 10    - 12/01/2017
13 - 10    - 13/01/2017

我希望以这种方式显示结果

10 - 01/01/2017 - 05/01/2017
20 - 06/01/2017 - 09/01/2017
10 - 10/01/2017 - 13/01/2017

我的查询是

SELECT price, min(date), max(date) FROM mytable GROUP BY price

但结果是

20 - 06/01/2017 - 09/01/2017
10 - 10/01/2017 - 13/01/2017

我可以使用mysql查询获得正确的结果，还是必须找到php解决方案？

Answer 1

这是一个空白和岛屿问题。您可以使用变量来检测具有相同price值的连续记录的孤岛：

SELECT price, MIN(`date`) AS start_date, MAX(`date`) AS end_date
FROM (
   SELECT id, price, `date`,
          @rn := @rn + 1 AS rn,
          -- If price remains the same then increase island population,
          -- otherwise reset it
          @seq := IF(@p = price, @seq + 1,
                      IF(@p := price, 1, 1)) AS seq
   FROM mytable
   CROSS JOIN (SELECT @p := 0, @rn := 0, @seq := 0) AS vars
   ORDER BY `date`) AS t
GROUP BY price, rn - seq

Answer 2

SELECT price, min(date), max(date) FROM mytable GROUP BY price order by price

MySQL：按列分组并获取日期范围

2 个答案: