Question

表1

Mail              Special Quote (Y/N)

g@hotmail.com     Blank 
s@ho.com          Blank
sss@j.com         Blank

表2

Email Address     Dept

g@hotmail.com     Config 
s@ho.com          Finance
sss@j.com         Marketing

现在我希望表1更新为Y，表2中匹配的电子邮件地址来自Finance或Marketing。

UPDATE TABLE 1 
Set Special Quote to be Y 
where in TABLE 2 the Dept is Finance or Marketing for the matching email address.

可能只是大脑冻结，需要一些咖啡。

想想加入电子邮件地址然后为Y和N位投入一些案例陈述？

Answer 1

不需要案例陈述。

update table1 set special_quote = 'Y'
where mail in ( select email_address from table2 where dept in ('Finance','Marketing' ))

Answer 2

尝试以下方法：

UPDATE A
SET A.SpecialQuote = 'Y'
FROM Table1 A
INNER JOIN Table2 B
ON A.Mail = B.EmailAddress
WHERE B.Dept IN ('Finance','Marketing')

Answer 3

我更像是一个MySQL用户，所以这可能是错的，但尝试这样的事情：

UPDATE table1 SET specialquote = 'Y'
WHERE Mail IN (SELECT EmailAddress FROM table2 WHERE Dept IN ('Finance','Marketing'))

Answer 4

您非常接近，您需要加入邮件字段中的两个表格，然后检查部门是否在财务/营销部门

UPDATE t1
SET t1.[Special Quote] = 'Y'
FROM Table1 t1
INNER JOIN Table2 t2
    ON t1.mail = t2.mail
WHERE t2.Dept in ('Finance','Marketing')

这将仅更新包含的项目，如果您要更新Y / N，则可以执行以下操作（请参阅SQL Fiddle with Demo）

UPDATE t1
SET t1.sq = CASE WHEN t2.dept IS NOT NULL THEN 'Y' ELSE 'N' END
FROM  table1 t1
LEFT JOIN table2  t2
    ON t1.mail = t2.mail
    AND t2.Dept in ('Finance','Marketing')

Answer 5

你走在正确的轨道上。这是语法：

update table1
    set SpecialQuote = 'Y'
    where mail in (select emailaddress from table2 where dept in ('Finance', 'Marketing')

Answer 6

更新表1 设置[特殊报价] ='Y' 来自Table1 t1 加入Table2 t2 在t1。[电子邮件地址] = t2。[电子邮件地址] 其中t2.dept（'Finance'，'Marketing'）

未经测试的样本，所以要明白这一点，但相信它足以说明这一点。祝你好运！

编辑：根据OP的“Y / N更新”评论，更改如下......尽管还有其他几种微妙的不同方式可以完成同样的事情。希望这对你有所帮助。

Update Table1
   Set [Special Quote]=(case when t2.dept='Finance' 
                             then 'Y'
                             when t2.dept='Marketing'
                             then 'Y'
                             else 'N' 
                        end) 
  from Table1 t1
  join Table2 t2
    on t1.[Email Address]=t2.[Email Address]

Answer 7

create table #tbl1 (mail varchar(50),quote varchar(1))
create table #tbl2 (mail varchar(50),dept varchar(10))

insert into #tbl1 values ('a@a.com','')
insert into #tbl1 values ('b@b.com','')
insert into #tbl1 values ('c@c.com','')
insert into #tbl1 values ('d@d.com','')
insert into #tbl1 values ('e@e.com','')

insert into #tbl2 values ('a@a.com','config')
insert into #tbl2 values ('b@b.com','finance')
insert into #tbl2 values ('c@c.com','marketing')
insert into #tbl2 values ('d@d.com','other')
insert into #tbl2 values ('e@e.com','skivers')


update #tbl1
set quote = 'Y'
where mail in (SELECT mail FROM #tbl2 WHERE dept IN ('finance','marketing'))

select * from #tbl1

drop table #tbl1
drop table #tbl2

Answer 8

我喜欢使用CTEs来帮助澄清查询的不同部分，因此这是一种方法。如果你不喜欢它还有其他的。

with getval as (
select mail,dept from test2
where dept in ('Marketing','Finance')
)


update test1
set quote = case when dept is NOT null then 'Y' else 'N' end 
from test1 t1
left outer join getval t2 on t1.mail = t2.mail

无论如何，我使用左连接来获取所有记录，然后使用案例来排序部门出现的位置（意味着记录在CTE中找到）。

... OR

update table1
set quote = case when dept is NOT null then 'Y' else 'N' end 
from table1 t1
left outer join table2 t2 on t1.email = t2.email and t2.dept in ('Marketing','Finance')

那也应该有用。

需要案例陈述欺骗

8 个答案: