1

这是我的 SQL 小提琴：http ://sqlfiddle.com/#!2/90d45/3

表方案：

CREATE TABLE domain_rankings (domain_id int, rank int, create_date datetime);
INSERT INTO domain_rankings VALUES (1, 0, "2012-01-01");
INSERT INTO domain_rankings VALUES (1, 2, "2012-01-02");
INSERT INTO domain_rankings VALUES (1, 1, "2012-01-03");
INSERT INTO domain_rankings VALUES (2, 0, "2012-01-01");
INSERT INTO domain_rankings VALUES (2, 1, "2012-01-02");
INSERT INTO domain_rankings VALUES (2, 2, "2012-01-03");
INSERT INTO domain_rankings VALUES (3, 1, "2012-01-01");
INSERT INTO domain_rankings VALUES (4, 3, "2012-01-01");
INSERT INTO domain_rankings VALUES (4, 2, "2012-01-02");
INSERT INTO domain_rankings VALUES (4, 1, "2012-01-03");

我想获得比较第一个条目和最后一个条目（按日期）排名上升的 domain_id 的计数。

所以在这种情况下，所有排名上升的计数应该是 2 (domain_id: 1, 2)。

domain_id 3 不应包含在计数中，因为它只有一个条目。因此，可能需要 HAVING COUNT(*) > 1 的子查询。

domain_id 4 也不应该包含在计数中，除非我反转查询并希望排名下降。

我应该如何解决这个问题？我知道我需要子查询，但我不知道从哪里开始。JSFiddle 应该让您了解我被困在哪里。

尝试 - 这将返回有效的行，但不正确地返回行的 COUNT： 

SELECT domain_id, COUNT(DR.domain_id)
FROM domain_rankings DR
WHERE
  (SELECT rank
   FROM domain_rankings
   WHERE domain_rankings.domain_id = DR.domain_id
   ORDER BY create_date ASC LIMIT 1
  ) > (
    SELECT rank
    FROM domain_rankings
    WHERE domain_rankings.domain_id = DR.domain_id
    ORDER BY create_date DESC LIMIT 1
  )
GROUP BY DR.domain_id
HAVING count(*) > 1

最终答案（两个选项中较慢的一个）：

SELECT COUNT(a.domain_id) cnt
FROM (
  SELECT domain_id,MIN(create_date) mind, MAX(create_date) maxd
  FROM domain_rankings
  GROUP BY domain_id
  HAVING COUNT(*) > 1
) master
JOIN  domain_rankings a 
  ON a.domain_id = master.domain_id 
 AND a.create_date = master.mind
JOIN  domain_rankings b
  ON b.domain_id = master.domain_id 
 AND b.create_date = master.maxd
WHERE a.rank < b.rank
4

2 回答 2

1

Maybe not the simplest approach, but you can for example use a subquery to get the min and max date per domain_id, and use regular joins to get the corresponding lines. Then you can just compare the ranks at the end;

SELECT COUNT(a.domain_id) cnt
FROM ( SELECT domain_id,MIN(create_date) mind, MAX(create_date) maxd
       FROM domain_rankings GROUP BY domain_id) master
JOIN  domain_rankings a 
  ON a.domain_id = master.domain_id AND a.create_date = master.mind
JOIN  domain_rankings b
  ON b.domain_id = master.domain_id AND b.create_date = master.maxd
WHERE a.rank < b.rank

Another version is to just use LEFT JOINS to filter out the first and last row, and compare the ranks;

SELECT COUNT(a.domain_id) cnt
FROM domain_rankings a JOIN domain_rankings b
  ON a.domain_id = b.domain_id AND a.rank < b.rank
LEFT JOIN domain_rankings c
  ON a.domain_id = c.domain_id AND a.create_date > c.create_date
LEFT JOIN domain_rankings d
  ON b.domain_id = d.domain_id AND b.create_date < d.create_date
WHERE c.domain_id IS NULL and d.domain_id IS NULL;

An SQLfiddle with both.

于 2013-10-10T19:46:56.943 回答
0 
SELECT count(*)
FROM TABLE AS a
INNER JOIN TABLE AS b ON a.domain_id=b.domain_id
AND a.date=b.date-interval 2 DAY
AND a.rank<b.rank
于 2013-10-10T19:47:12.553 回答