Question

书面案例陈述以再次获得结果，我需要将结果列与Job_Level列进行比较，如果Job_Level包含结果中存在的任何一个值，则它应该在新列（ExpectedOutput）中返回true或false

select Email, Job_Levels,Answer,
case when answer = 'Assessor / Trainer' then 1
when answer = 'Administrator / Coordinator/ Assistant' then 2
 when answer = 'Adviser' then 3
  when answer = 'Apprentice' then 4
  when answer = 'Deputy / Vice Principal' then 5
  when answer = 'Chief Executive / Principal' then 6 
when answer = 'Technician' then 18
  when answer = 'Worker / Practitioner' then 19
 end as result  
 from TBL_CandidateInfo
 inner join tal_users
 on Userid = UserID_FK
  inner join [ABC_Migration].[dbo].[ABCSeeker]
 on email = susername
WHERE ProfileQuestion='What is your preferred job role / level?'

当前输出

Email         Job_Levels    Answer                result
a@gmail.com   2,1,16        Assessor / Trainer    1
b@gmail.com   2,1,16        Teaching / Lecturing  16
c@yahoo.com   12,16,18      Learner Suppor        12

预期输出：

Email         Job_Levels    Answer               result  ExpectedOutput
a@gmail.com   2,1,16        Assessor / Trainer   1        True
b@gmail.com   2,1,16        Teaching / Lecturing 16       True
c@yahoo.com   16,18         Learner Suppor       12       False

Answer 1

编辑：虽然此答案已被接受，但请注意此解决方案有一些警告。有关更简单的方法，请参阅@GuidoG的答案。

我认为您正在寻找这样的东西：

SELECT  Email
,       Job_Levels
,       Answer
,       result
,       CASE
            WHEN Job_Levels LIKE '%' + CONVERT(VARCHAR(2), result) + '%' THEN 'true'
            ELSE 'False'
        END AS ExpectedOutput
FROM    (
            SELECT Email, Job_Levels,Answer,
            case 
                    WHEN answer = 'Assessor / Trainer' then 1
                    WHEN answer = 'Administrator / Coordinator/ Assistant' then 2
                    WHEN answer = 'Adviser' then 3
                    when answer = 'Apprentice' then 4
                    WHEN answer = 'Deputy / Vice Principal' then 5
                    WHEN answer = 'Chief Executive / Principal' then 6 
                    WHEN answer = 'Technician' then 18
                    WHEN answer = 'Worker / Practitioner' then 19
             end as result  
             from TBL_CandidateInfo
             inner join tal_users
             on Userid = UserID_FK
              inner join [ABC_Migration].[dbo].[ABCSeeker]
             on email = susername
            WHERE ProfileQuestion='What is your preferred job role / level?'
        ) T

Answer 2

一种选择是使用CharIndex函数

declare @Job_Levels varchar(100) = '2,1,16'

select case when charindex('16' + ',', @Job_Levels + ',') > 0 then 'true' else 'false' end

请注意，我添加了一个额外的内容，以确保在查找1时不会得到TRUE，并且没有1而是16。
换句话说，需要Extra来确保寻找1不会对值16返回true，而只会对确切的值1返回

您的查询将如下所示

select t.Email,
       t.Job_Levels,
       t.Answer,
       t.result,
       case when charindex(t.result + ',', t.Job_Levels + ',') > 0 then 'true' else 'false' end as ExpectedOutput
from   ( select Email, Job_Levels,Answer,
                case when answer = 'Assessor / Trainer' then 1
                     when answer = 'Administrator / Coordinator/ Assistant' then 2
                     when answer = 'Adviser' then 3
                     when answer = 'Apprentice' then 4
                     when answer = 'Deputy / Vice Principal' then 5
                     when answer = 'Chief Executive / Principal' then 6 
                     when answer = 'Technician' then 18
                     when answer = 'Worker / Practitioner' then 19
                end as result
         from TBL_CandidateInfo
           inner join tal_users on Userid = UserID_FK
           inner join [ABC_Migration].[dbo].[ABCSeeker] on email = susername
         WHERE ProfileQuestion='What is your preferred job role / level?'
      ) t

Answer 3

使用此。请注意，如果我们在[co]之前和之后不使用'，'。[Job_Levels]结果可能不正确。例如，ID 11113在LIKE％11％语句中都显示为真实结果，但113不是正确结果。但是我的查询使用的是LIKE％，11，％，并且对于，113返回false，

SELECT      co.*
            ,CASE WHEN Convert(varchar(MAX),',' + [co].[Job_Levels] + ',') Like '%,' + Convert(varchar(MAX),[co].[result]) + ',%' THEN
                1 --true
            ELSE 
                0 --false
            END as ExpectedOutput
FROM        (
                select Email, Job_Levels,Answer,
                case when answer = 'Assessor / Trainer' then 1
                when answer = 'Administrator / Coordinator/ Assistant' then 2
                 when answer = 'Adviser' then 3
                  when answer = 'Apprentice' then 4
                  when answer = 'Deputy / Vice Principal' then 5
                  when answer = 'Chief Executive / Principal' then 6 
                when answer = 'Technician' then 18
                  when answer = 'Worker / Practitioner' then 19
                 end as result  
                 from TBL_CandidateInfo
                 inner join tal_users
                 on Userid = UserID_FK
                  inner join [ABC_Migration].[dbo].[ABCSeeker]
                 on email = susername
                WHERE ProfileQuestion='What is your preferred job role / level?'
            ) co

比较两列数据案例语句

3 个答案: