이 SQL 쿼리가 결과를 정확하게 전달하지 않는 이유는 무엇입니까?

Question

결과가 group by tblpersonalInfo.applicantID으로 정확하게 게시되기를 원하지만이 쿼리는 그룹과 함께 결과를 tblpersonalInfo.applicantID으로 표시하지만이 쿼리는 점진적으로 결과를 보여줍니다. 당신은 당신이 SELECT 절에 요약 수 있으며, 두 개의 하위 쿼리 하나에 UNION해야 할 수도 있습니다처럼 모든 쿼리에 대해 확실

SELECT DISTINCT 

    t1+t2, 

    tblPersonalInfo.ApplicantId, 
    tblPersonalInfo.Applicantname 
FROM (
    SELECT SUM(tblExpeRange.score)as t2 
    FROM tblexperience 
    LEFT OUTER JOIN tblExpeRange 
     ON tblexperience.ExpRange=tblExpeRange.expeRange 

    GROUP BY tblexperience.ApplicantId 
    ) tblexperience, 
    ( 
    SELECT SUM(tblGradPoint.score) AS t1 
    FROM tblAcademicInfo 
     LEFT OUTER JOIN tblGradPoint 
      ON tblAcademicInfo.cgpa=tblGradPoint.[cgpa/division] 
    GROUP BY tblAcademicInfo.ApplicantId 
    ) tblAcademicInfo, 
    tblPersonalInfo 

    INNER JOIN tblCircular 
     ON tblPersonalInfo.Cirname = tblCircular.Cirname 
WHERE tblCircular.Cirname=(tblPersonalInfo.Cirname) 

RETURN 

Answer 1

100 %하지만 ... 소리

SELECT sum(T) AS d, 
     Y.ApplicantId, 
     Y.Applicantname 
FROM (
     SELECT tblexperience.ApplicantId, 
       SUM(tblExpeRange.score)as T 
     FROM tblexperience 
       LEFT OUTER JOIN tblExpeRange 
        ON tblexperience.ExpRange=tblExpeRange.expeRange 
     GROUP BY tblexperience.ApplicantId 
     UNION 
     SELECT tblAcademicInfo.ApplicantId, 
       SUM(tblGradPoint.score) AS T 
     FROM tblAcademicInfo 
       LEFT OUTER JOIN tblGradPoint 
        ON tblAcademicInfo.cgpa=tblGradPoint.[cgpa/division] 
     GROUP BY tblAcademicInfo.ApplicantId 
     ) X 
     INNER JOIN tblPersonalInfo Y 
      ON X.ApplicantId = Y.ApplicantId 
     INNER JOIN tblCircular Z 
      ON Y.Cirname = Z.Cirname --(*) 
--WHERE Z.Cirname=(Y.Cirname) --.....maybe not required as the line above (*) deals with this 
GROUP BY Y.ApplicantId, 
     Y.Applicantname 

Answer 2

당신 하위 쿼리 tblAcademicInfo과 tblexperience 사이에 아무런 연결도하지 않으므로 CROSS JOIN을 만들고 여기서 첫 번째 값부터 두 번째 값까지의 모든 값을 합산합니다. 어쩌면 당신은 각각의 하위 쿼리 선택 목록에 ApplicantId을 추가 시도하고 그들 사이 tblPersonalInfo

SELECT DISTINCT 

    t1+t2, 

    tblPersonalInfo.ApplicantId, 
    tblPersonalInfo.Applicantname 
FROM  
    (
     SELECT 
      tblexperience.ApplicantId 
      ,SUM(tblExpeRange.score)as t2 
     FROM tblexperience 
     LEFT OUTER JOIN tblExpeRange 
      ON tblexperience.ExpRange=tblExpeRange.expeRange 
     GROUP BY tblexperience.ApplicantId 
    ) tblexperience, 
    INNER JOIN 
    ( 
     SELECT 
      tblAcademicInfo.ApplicantId 
      ,SUM(tblGradPoint.score) AS t1 
     FROM tblAcademicInfo 
     LEFT OUTER JOIN tblGradPoint 
      ON tblAcademicInfo.cgpa=tblGradPoint.[cgpa/division] 
     GROUP BY tblAcademicInfo.ApplicantId 
    ) tblAcademicInfo ON tblAcademicInfo.ApplicantId = tblexperience.ApplicantId 
    INNER JOIN tblPersonalInfo ON tblexperience.ApplicantId = tblPersonalInfo.ApplicantId 
    INNER JOIN tblCircular 
     ON tblPersonalInfo.Cirname = tblCircular.Cirname 

편집에 가입하는 데 사용해야합니다 또한, 당신은 당신이 공통 테이블 식을 사용하는 경우이 쿼리는 훨씬 더 읽기 보일 것 생각하지 않는다 및 별칭 :

WITH CTE_Experience AS 
(
    SELECT 
     e.ApplicantId 
     ,SUM(er.score) AS t2 
    FROM tblexperience e 
     LEFT JOIN tblExpeRange er ON e.ExpRange = er.expeRange 
    GROUP BY e.ApplicantId 
) 
,CTE_AcademicInfo AS 
(
    SELECT 
     ai.ApplicantId 
     ,SUM(gp.score) AS t1 
    FROM tblAcademicInfo ai 
     LEFT JOIN tblGradPoint gp ON ai.cgpa = gp.[cgpa/division] 
    GROUP BY ai.ApplicantId 
) 
SELECT 
    ai.t1 + ex.t2, 
    pin.ApplicantId, 
    pin.Applicantname 
FROM tblPersonalInfo pin 
    INNER JOIN CTE_Experience ex ON ex.ApplicantId = pin.ApplicantId 
    INNER JOIN CTE_AcademicInfo ai ON ai.ApplicantID = pin.ApplicantId 
    INNER JOIN tblCircular ci ON ci.Cirname = pin.Cirname 

Answer 3

이 시도 :

SELECT 
    tblPersonalInfo.ApplicantId, 
    tblPersonalInfo.ApplicantName, 
    (tblExperience.t2+tblAcademicInfo.t1) as d 
FROM tblPersonalInfo 
INNER JOIN (
    SELECT SUM(tblExpeRange.score) as t2, 
     tblExperience.ApplicantId 
    FROM tblExperience 
    LEFT OUTER JOIN tblExpeRange 
     ON tblExperience.ExpRange=tblExpeRange.ExpRange 
    GROUP BY tblExperience.ApplicantId 
)tblExperience ON tblPersonalInfo.ApplicantId=tblExperience.ApplicantId 
INNER JOIN (
    SELECT SUM(tblGradPoint.score) as t1, 
     tblAcademicInfo.ApplicantId 
    FROM tblAcademicInfo 
    LEFT OUTER JOIN tblGradPoint 
     ON tblAcademicInfo.cgpa=tblGradPoint.[cgpa/division] 
    GROUP BY tblAcademicInfo.ApplicantId 
)tblAcademicInfo ON tblPersonalInfo.ApplicantId=tblAcademicInfo.ApplicantId 
INNER JOIN tblCircular 
    ON tblPersonalInfo.Cirname=tblCircular.Cirname