Question

我有一个filter <- unique(unlist(group[sapply(group, function(x) length(x) == 1)]));表：

road_insp

我可以选择最近的检查，每条道路：

create table road_insp
(
    insp_id integer,
    road_id integer,
    insp_date date,
    condition number,
    insp_length number
);

--Run each insert statement, one at a time.
INSERT INTO road_insp (insp_id, road_id, insp_date, condition, insp_length)  VALUES (1, 100, #1/1/2017#, 5.0, 20);
INSERT INTO road_insp (insp_id, road_id, insp_date, condition, insp_length)  VALUES (2, 101, #2/1/2017#, 5.5, 40);
INSERT INTO road_insp (insp_id, road_id, insp_date, condition, insp_length)  VALUES (3, 101, #3/1/2017#, 6.0, 60);
INSERT INTO road_insp (insp_id, road_id, insp_date, condition, insp_length)  VALUES (4, 102, #4/1/2018#, 6.5, 80);
INSERT INTO road_insp (insp_id, road_id, insp_date, condition, insp_length)  VALUES (5, 102, #5/1/2018#, 7.0, 100);
INSERT INTO road_insp (insp_id, road_id, insp_date, condition, insp_length)  VALUES (6, 102, #5/1/2018#, 7.5, 120);

+---------+---------+-----------+-----------+-------------+
| insp_id | road_id | insp_date | condition | insp_length |
+---------+---------+-----------+-----------+-------------+
|       1 |     100 | 1/1/2017  |         5 |          20 |
|       2 |     101 | 2/1/2017  |       5.5 |          40 |
|       3 |     101 | 3/1/2017  |         6 |          60 |
|       4 |     102 | 4/1/2018  |       6.5 |          80 |
|       5 |     102 | 5/1/2018  |         7 |         100 |
|       6 |     102 | 5/1/2018  |       7.5 |         120 |
+---------+---------+-----------+-----------+-------------+

但是，正如您所看到的，每个日期可以对每个日期进行多次检查（检查SELECT b.insp_id, b.road_id, b.insp_date, b.condition, b.insp_length FROM road_insp b WHERE b.insp_date=( select max(insp_date) from road_insp a where a.road_id = b.road_id ); +---------+---------+-----------+-----------+-------------+ | insp_id | road_id | insp_date | condition | insp_length | +---------+---------+-----------+-----------+-------------+ | 1 | 100 | 1/1/2017 | 5 | 20 | | 3 | 101 | 3/1/2017 | 6 | 60 | | 5 | 102 | 5/1/2018 | 7 | 100 | | 6 | 102 | 5/1/2018 | 7.5 | 120 | +---------+---------+-----------+-----------+-------------+和#5）。结果是每条道路返回多条记录。

相反，如果每个道路有多次检查，每个日期，我想通过选择最长的检查来打破平局。

#6

如何在MS Access查询中执行此操作？

Answer 1

您可以改为使用order by和top：

SELECT ri.*
FROM road_insp as ri
WHERE ri.insp_id = (select top 1 ri2.insp_id      --removed brackets on top(1)
                    from road_insp as ri2 
                    where ri2.road_id = ri.road_id 
                    order by ri2.insp_date desc, ri2.insp_length desc,
                             ri2.insp_id
                   );

选择最大值，使用决胜局

1 个答案: