[英]How do I efficiently use SQL join
表
工作
JobId, JobName
的JobDetail
JobDetailId, JobId, AcctNum, PerformDate
一个Job
包含由JobId
链接的许多JobDetails
。
对于某些JobNames
,(JobeName1,JobName2),我想要一个仅包含一个JobDetail
的列表,其中Performdate
在过去6个月内。
对于上述标准,我应该获得两个记录。 什么是正确的SQL语句?
我已经做到了这一点,而不仅仅是获得正确的陈述。
SELECT
COUNT(J.JobId)
FROM
Job J
WHERE
J.JobName IN ('Doc1', 'Doc2')
GROUP BY
J.JobName
[编辑]
SQL Server 2005
[EDIT2]
SELECT
J.JobName, JD.AcctNum
FROM
Job J
JOIN JobDetail JD ON J.JobId = JD.JobId
WHERE
J.JobName IN ('DOC1', 'DOC2')
GROUP BY
J.JobName
该语句产生:
J.JobId在选择列表中无效...未包含在聚合函数中。
[EDIT3]
http://sqlfiddle.com/#!3/3ef21/1/0
我想要由PerformDate为Doc1和DOC2提供的最新AcctNum。
因此,从SQLFIDDLE中的数据中,我应该为每个JobName返回2个ROWS。
[EDIT4]
我的预期数据是JobDetail表中的*行:
JobDetailId 1
jobDetailId 3
干得好:
SELECT COUNT(J.JobId) FROM Job J
INNER JOIN JobDetails D ON J.JobId = D.JobId
WHERE J.JobName IN ('Jobname1', 'Jobname2')
AND D.PerformDate >= DateAdd(mm, -6, getdate())
GROUP BY J.JobName
我已经在TSQL(SQL Server)中编写了此代码。 对于其他DBMS,您将必须对DateAdd(mm, -6, getdate())
使用相关的函数。
更新:要选择最新AcctNum
的DOC1
和DOC2
。
Select D.JobID, Acctnum from JobDetail D Inner Join ( SELECT J.JOBID, MAX(PerformDate) as LatestPerformDate FROM Job J INNER JOIN JobDetail D ON J.JobId = D.JobId WHERE J.JobName IN ('DOC1', 'DOC2') Group by J.JobID ) X on D.JobID = X.JobID And D.PerformDate = X.LatestPerformDate
更新2:当DOC2
可以在Job
表中以不同的JobID
出现两次时。
Select D.JobID, Acctnum from
JobDetail D Inner Join Job J on J.JobID = D.JobID
Inner join
(
SELECT J.JOBname, MAX(PerformDate) as LatestPerformDate FROM Job J
INNER JOIN JobDetail D ON J.JobId = D.JobId
WHERE J.JobName IN ('DOC1', 'DOC2')
group by j.jobname
) X on J.JobName = X.JobName
And D.PerformDate = X.LatestPerformDate
问题中sqlfiddle中发布的输入数据的输出:
JOBID ACCTNUM
--------------------
1 1235
4 1238
SELECT j.*, d.*
FROM Job j
JOIN (SELECT JobId, MAX(PerformDate) maxDate
FROM JobDetail
WHERE PerformDate > DATEADD(mm, -6, GETDATE())
GROUP BY JobId) md
ON j.JobId = md.JobId
JOIN JobDetail d
ON d.JobId = md.JobId AND d.PerformDate = md.maxDate
WHERE j.JobName IN ('DOC1', 'DOC2');
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