INSERT INTO resource (interviewId) VALUES (SELECT resinterviewId FROM resinterview WHERE email IN (SELECT email FROM resinterview GROUP BY email HAVING Count(email) >= 2))
--resource表 select t.[email] from ( select [email],count(*) cnt from [resource] group by [email] ) t where t.[cnt]>1 ---------------------- --resinterview表 select t.[email] from ( select [email],count(*) cnt from [resinterview] group by [email] ) t where t.[cnt]>1
--两个表之间相同email select a.[email] from [resource] a inner join [resinterview] b on a.[email]=b.[email]
二楼的不对,insert into 是直接给resource表新增了数据,我需要的将email相同的resinterview表中的resinterviewId赋给resource中的InterviewId。
(interviewId)
VALUES (SELECT resinterviewId
FROM resinterview
WHERE email IN (SELECT email
FROM resinterview
GROUP BY email
HAVING Count(email) >= 2))
select t.[email] from
(
select [email],count(*) cnt from [resource]
group by [email]
) t
where t.[cnt]>1
----------------------
--resinterview表
select t.[email] from
(
select [email],count(*) cnt from [resinterview]
group by [email]
) t
where t.[cnt]>1
select a.[email] from [resource] a
inner join [resinterview] b
on a.[email]=b.[email]