我有一个查询,它返回以下结构:
start_date | pid | uid | type | total
2019-11-10 | 2006933595591018006 | 1812803885757105697 | recommended | 9
2019-11-10 | 2006933595591018006 | 1812803885757105697 | actual | 3现在我想得到一个总数,ratio,在type = actual/type = recommended之间的比率。对于给定的集合,应该只有两个结果,即每个type中的一个。
期望产出:
start_date | pid | uid | ratio
2019-11-10 | 2006933595591018006 | 1812803885757105697 | 0.3333 我假设这可以通过一个group by子句和一个case select之类的方法来完成?到目前为止,我一直无法做到这一点。
发布于 2019-11-14 21:23:25
您可以不使用任何CASE或GROUP BY来完成此操作:
edb=# create table my_table (start_date date, pid text, uid text, type text, total int);
CREATE TABLE
edb=# insert into my_table values ('2019-11-10', '2006933595591018006','1812803885757105697','recommended',9);
INSERT 0 1
edb=# insert into my_table values ('2019-11-10', '2006933595591018006','1812803885757105697','actual',3);
INSERT 0 1
edb=# SELECT a.start_date, a.pid, a.uid, a.total/r.total as ratio
FROM my_table a
JOIN my_table r ON (a.start_date=r.start_date AND a.pid=r.pid AND a.uid=r.uid)
WHERE a.type = 'actual'
AND r.type = 'recommended';
start_date | pid | uid | ratio
---------------------+---------------------+---------------------+------------------------
2019-11-10 00:00:00 | 2006933595591018006 | 1812803885757105697 | 0.33333333333333333333
(1 row)https://dba.stackexchange.com/questions/253337
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