Question

问题

如果在表雇主中找到userID，我怎样才能得到一个简单的表userType = employee或userType = employer。

试图做

通过以下代码，我可以获得一个回报，即当他是一个雇主时，这个用户是雇主，当他不是我什么也得不到时。

SELECT (CASE WHEN (userID IS NOT NULL) THEN 'employer' ELSE 'employee' END) AS 'userType' FROM tblEmployer WHERE userID = 401

还尝试使用if语句

IF EXSISTS (SELECT userID FROM tblEmployer WHERE userID = 401) 
IF ((SELECT COUNT(*) FROM tblEmployer WHERE userID = 401) > 0) ELSE ...

但是这会给Assignment type not recognized. (near "IF" at position 0)

Answer 1

如果要准确返回一行，请在select子句中使用聚合或子查询：

select (case when exists (select 1
                          from tblEmployer e
                          where e.userId = 401
                         )
             then 'employer' else 'employee'
        end) as userType

可替换地：

select (case when count(*) > 0 then 'employer' else 'employee' end) as userType
from tblEmployer e
where e.userId = 401;

找到SQL IF值，ELSE做其他事情

1 个答案: