Question

首先，请原谅语言，我一直无法将我的问题解析为实际的英语，所以如果有人可以编辑以使其更清楚，那将有所帮助。

我一直在努力争取这个。我需要一个查询，根据每个组，从过去的N天开始，跳过最近的一个并检索下一个T天。这是经典“LIMIT with GROUP”问题的一个版本，事实上，我尝试过的其中一个查询无效，使用了该格式。

MRE如下：

CREATE TABLE `trying` (id INTEGER PRIMARY KEY AUTO_INCREMENT, types1 TEXT, stuffs INTEGER, dates DATE);
INSERT INTO `trying`(types1, stuffs, dates) VALUES("one",123,'2015-09-06');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("one",67,'2015-09-05');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("one",45,'2015-09-04');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("one",98,'2015-09-03');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("one",89,'2015-09-02');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("two",56,'2015-09-02');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("two",34,'2015-09-01');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("two",98,'2015-08-31');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("two",34,'2015-08-30');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("two",12,'2015-08-29');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("three",3,'2015-09-06');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("three",8,'2015-09-04');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("three",80,'2015-09-02');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("three",9,'2015-09-01');
INSERT INTO `trying`(types1, stuffs, dates) VALUES("three",6,'2015-08-31');

在表trying中，有三种types1'类型'：'一'，'二'和'三'，每组有5个观察值。请注意，各组之间的日期并不相似，它们之间甚至可能存在差距（so no dates BETWEEN, like in this question）。

在这个例子中，我希望得到一个包含每组三个中间值的表。因此，跳过第一个和最后一个值，预期输出将如下所示：

types1  stuffs  dates
one     67      2015-09-05
one     45      2015-09-04
one     98      2015-09-03
two     34      2015-09-01
two     98      2015-08-31
two     34      2015-08-30
three   8       2015-09-04
three   80      2015-09-02
three   9       2015-09-01

一些无效的查询：

SELECT types1, stuffs, dates FROM trying  GROUP BY types1 LIMIT 2,4;
/*this returned the following */
types1  stuffs  dates
two     56      2015-09-02


SELECT trying.* FROM (SELECT types1, stuffs, dates FROM trying)  GROUP BY trying.types1 OFFSET 2,4;
/*threw out an error: Every derived table must have its own alias */

Answer 1

可能有效的一种方法是使用用户变量对每个组中的行进行编号，然后将结果限制为行数在所需时间间隔内的行：

SELECT id, types1, stuffs, dates
FROM (
  SELECT 
    id, types1, stuffs, dates,
    (
      CASE types1
      WHEN @type 
      THEN @row := @row + 1 
      ELSE @row := 1 AND @type := types1 END
    ) + 1 AS row
  FROM trying p,
  (SELECT @row := 0, @type := '') r
  ORDER BY types1, dates asc  
) src
WHERE row BETWEEN 2 AND 4
ORDER BY id;

查询1的

Sample SQL Fiddle

或者，如果您始终要删除每个组中的第一行和最后一行，则可以使用左连接到派生表，该表返回每个组的最大和最小日期：

select t.* from trying t
left join (
    select types1, min(dates) min_dates, max(dates) max_dates 
    from trying group by types1
    ) minmax 
    on t.types1 = minmax.types1 
    and t.dates in (minmax.max_dates, minmax.min_dates)
where minmax.types1 is null;

查询2的

Sample SQL Fiddle

使用您的示例数据，两个查询都返回相同的结果：

| id | types1 | stuffs |                       dates |
|----|--------|--------|-----------------------------|
|  2 |    one |     67 | September, 05 2015 00:00:00 |
|  3 |    one |     45 | September, 04 2015 00:00:00 |
|  4 |    one |     98 | September, 03 2015 00:00:00 |
|  7 |    two |     34 | September, 01 2015 00:00:00 |
|  8 |    two |     98 |    August, 31 2015 00:00:00 |
|  9 |    two |     34 |    August, 30 2015 00:00:00 |
| 12 |  three |      8 | September, 04 2015 00:00:00 |
| 13 |  three |     80 | September, 02 2015 00:00:00 |
| 14 |  three |      9 | September, 01 2015 00:00:00 |

Answer 2

select types1,stuffs,dates from (
select @rank:=if(@prev_cat=types1,@rank+1,1) as rank,
  types1,stuffs,dates,@prev_cat:=types1

from trying,(select @rank:=0, @prev_cat:="")t
order by types1, dates desc
  ) temp

  where rank between 2 and  4

获得每组的最后[n，n + t]天

2 个答案: