Oracle子查询建议

时间:2015-02-27 18:59:09

标签: oracle subquery

我有以下2个查询。我很难将第二个查询纳入第一个查询;这两个查询都可以通过jobid加入。基本上我要做的是通过加入jobid,即total_opens和unique_open_cnt,将第二个查询中的2个总列合并到第一个查询中。我希望这很清楚。任何想法都非常感谢。谢谢。

SELECT u.fname, u.lname, j.job_id, j.title, ja.SENT_DATE, 
count(distinct ja.email_id) as total_sent,
COUNT(je.EMAIL_ID) as total_clicks
            FROM jobs_table j , user_table u ,
             job_alerts ja
            left join job_eml_track je
            on ja.EMAIL_ID = je.EMAIL_ID
            WHERE j.user_id = u.user_id 
            and ja.JOB_ID = j.JOB_ID
            and ja.JOB_ID = 116 
            group by j.job_id, j.title,ja.SENT_DATE,j.EMPLOYER_ID,u.fname, u.lname    

SELECT COUNT(*) AS total_opens , 
   COUNT(DISTINCT userd) AS unique_open_cnt,
   REGEXP_REPLACE(v.id, '^testnew_(\d+)', '\1', 1, 1, 'i') as jobid
 FROM  basetable v
WHERE v.id LIKE 'testnew%'
and REGEXP_REPLACE(v.id, '^testnew_(\d+)', '\1', 1, 1, 'i') = 116
group by REGEXP_REPLACE(v.id, '^testnew_(\d+)', '\1', 1, 1, 'i')

2 个答案:

答案 0 :(得分:1)

这可能类似于

select * from 
(<your first query>) q1
inner join
(<your second query>) q2
  on q1.job_id = q2.jobid

答案 1 :(得分:1)

您应该可以使用WITH clause。我没有检查语法,但像

with a AS
(SELECT u.fname, u.lname, j.job_id, j.title, ja.SENT_DATE, 
count(distinct ja.email_id) as total_sent,
COUNT(je.EMAIL_ID) as total_clicks
            FROM jobs_table j , user_table u ,
             job_alerts ja
            left join job_eml_track je
            on ja.EMAIL_ID = je.EMAIL_ID
            WHERE j.user_id = u.user_id 
            and ja.JOB_ID = j.JOB_ID
            and ja.JOB_ID = 116 
            group by j.job_id, j.title,ja.SENT_DATE,j.EMPLOYER_ID,u.fname, u.lname  ),
b as
(
SELECT COUNT(*) AS total_opens , 
COUNT(DISTINCT userd) AS unique_open_cnt,REGEXP_REPLACE(v.id, '^testnew_(\d+)', '\1', 1, 1, 'i') as jobid
 FROM  basetable v
WHERE v.id LIKE 'testnew%'
and REGEXP_REPLACE(v.id, '^testnew_(\d+)', '\1', 1, 1, 'i') = 116
group by REGEXP_REPLACE(v.id, '^testnew_(\d+)', '\1', 1, 1, 'i'))
select a.fname, a.lname -- and so on....
from a, b
where a.jobid = b.jobid
order by a.jobid