Sql 查询性能调整-搜索其他选项

我需要知道是否存在其他方法来执行此查询

表

考试

具有以下结构：

EXAM_DATE DATE
SUBJECT VARCHAR2(50);
GRADE NUMBER;

这个概念是从考试中获取统计数据

select EXAM_DATE,
        SUBJECT,        
        (SELECT COUNT(1) 
            from EXAMS
          where GRADE IN (9,10) 
            AND SUBJECT = EXA.SUBJECT 
            AND EXAM_DATE = EXA.EXAM_DATE) outstanding,
        (select count(1)
            from EXAMS 
          where GRADE IN (4,5,6,7,8) 
            AND SUBJECT = EXA.SUBJECT 
            AND EXAM_DATE = EXA.EXAM_DATE) approved,
        (select count(1)
            from EXAMS
          where GRADE IN (0,1,2,3) 
            AND SUBJECT = EXA.SUBJECT 
            AND EXAM_DATE = EXA.EXAM_DATE) disapproved,
FROM EXAMS EXA
GROUP BY EXAM_DATE,SUBJECT;

---

谢谢

只需使用条件聚合。不需要子查询：

select EXAM_DATE, SUBJECT,        
       SUM(GRADE IN (9,10) THEN 1 ELSE 0 END) as outstanding,
       SUM(GRADE IN (4,5,6,7,8) THEN 1 ELSE 0 END) as approved,
       SUM(GRADE IN (0,1,2,3) THEN 1 ELSE 0 END) as disapproved,
FROM EXAMS EXA
GROUP BY EXAM_DATE, SUBJECT;

---

只需使用条件聚合。不需要子查询：

select EXAM_DATE, SUBJECT,        
       SUM(GRADE IN (9,10) THEN 1 ELSE 0 END) as outstanding,
       SUM(GRADE IN (4,5,6,7,8) THEN 1 ELSE 0 END) as approved,
       SUM(GRADE IN (0,1,2,3) THEN 1 ELSE 0 END) as disapproved,
FROM EXAMS EXA
GROUP BY EXAM_DATE, SUBJECT;

---

只需使用条件聚合。不需要子查询：

select EXAM_DATE, SUBJECT,        
       SUM(GRADE IN (9,10) THEN 1 ELSE 0 END) as outstanding,
       SUM(GRADE IN (4,5,6,7,8) THEN 1 ELSE 0 END) as approved,
       SUM(GRADE IN (0,1,2,3) THEN 1 ELSE 0 END) as disapproved,
FROM EXAMS EXA
GROUP BY EXAM_DATE, SUBJECT;

---

只需使用条件聚合。不需要子查询：

select EXAM_DATE, SUBJECT,        
       SUM(GRADE IN (9,10) THEN 1 ELSE 0 END) as outstanding,
       SUM(GRADE IN (4,5,6,7,8) THEN 1 ELSE 0 END) as approved,
       SUM(GRADE IN (0,1,2,3) THEN 1 ELSE 0 END) as disapproved,
FROM EXAMS EXA
GROUP BY EXAM_DATE, SUBJECT;

---

我可能会使用一个轴来获得结果，因为您正在将列中的值（gradeNumber）转换为行中的值（gradeNumber的计数）。试试这个：

-- make and fill an example table
DECLARE @Exams TABLE
( Exam_Date     DATE
 ,[Subject]     VARCHAR(250)
 ,GradeNumber   INT)

INSERT @Exams VALUES ('2014-01-15','Escapology',1)
INSERT @Exams VALUES ('2014-01-15','Escapology',9)
INSERT @Exams VALUES ('2014-01-15','Escapology',5)
INSERT @Exams VALUES ('2014-01-15','Escapology',3)
INSERT @Exams VALUES ('2014-01-16','Art',8)
INSERT @Exams VALUES ('2014-01-16','Art',7)
INSERT @Exams VALUES ('2014-01-16','Chemistry',0)
INSERT @Exams VALUES ('2014-01-16','Chemistry',5)
INSERT @Exams VALUES ('2014-01-16','Chemistry',9)
INSERT @Exams VALUES ('2014-01-16','Chemistry',2)
INSERT @Exams VALUES ('2014-01-17','Maths',5)
INSERT @Exams VALUES ('2014-01-17','Maths',6)
INSERT @Exams VALUES ('2014-01-18','Geology',10)

-- set up the grade boundaries
DECLARE @A INT = 9
DECLARE @B INT = 4
DECLARE @C INT = 0

-- get the count of grades by exam and date
SELECT * FROM
(SELECT
     Exam_Date
     -- list the exam date twice as one will be used to make the count
    ,Exam_Date  AS ExamDate
    ,[Subject]
    ,CASE WHEN GradeNumber >= @A THEN 'Outstanding'
        WHEN GradeNumber >= @B THEN 'Approved'
        ELSE 'Disapproved' END AS Result
FROM @Exams) AS Up
PIVOT (COUNT(Exam_Date) FOR Result IN ([Outstanding],[Approved],[Disapproved])) AS PVT

---

我可能会使用一个轴来获得结果，因为您正在将列中的值（gradeNumber）转换为行中的值（gradeNumber的计数）。试试这个：

-- make and fill an example table
DECLARE @Exams TABLE
( Exam_Date     DATE
 ,[Subject]     VARCHAR(250)
 ,GradeNumber   INT)

INSERT @Exams VALUES ('2014-01-15','Escapology',1)
INSERT @Exams VALUES ('2014-01-15','Escapology',9)
INSERT @Exams VALUES ('2014-01-15','Escapology',5)
INSERT @Exams VALUES ('2014-01-15','Escapology',3)
INSERT @Exams VALUES ('2014-01-16','Art',8)
INSERT @Exams VALUES ('2014-01-16','Art',7)
INSERT @Exams VALUES ('2014-01-16','Chemistry',0)
INSERT @Exams VALUES ('2014-01-16','Chemistry',5)
INSERT @Exams VALUES ('2014-01-16','Chemistry',9)
INSERT @Exams VALUES ('2014-01-16','Chemistry',2)
INSERT @Exams VALUES ('2014-01-17','Maths',5)
INSERT @Exams VALUES ('2014-01-17','Maths',6)
INSERT @Exams VALUES ('2014-01-18','Geology',10)

-- set up the grade boundaries
DECLARE @A INT = 9
DECLARE @B INT = 4
DECLARE @C INT = 0

-- get the count of grades by exam and date
SELECT * FROM
(SELECT
     Exam_Date
     -- list the exam date twice as one will be used to make the count
    ,Exam_Date  AS ExamDate
    ,[Subject]
    ,CASE WHEN GradeNumber >= @A THEN 'Outstanding'
        WHEN GradeNumber >= @B THEN 'Approved'
        ELSE 'Disapproved' END AS Result
FROM @Exams) AS Up
PIVOT (COUNT(Exam_Date) FOR Result IN ([Outstanding],[Approved],[Disapproved])) AS PVT

---

我可能会使用一个轴来获得结果，因为您正在将列中的值（gradeNumber）转换为行中的值（gradeNumber的计数）。试试这个：

-- make and fill an example table
DECLARE @Exams TABLE
( Exam_Date     DATE
 ,[Subject]     VARCHAR(250)
 ,GradeNumber   INT)

INSERT @Exams VALUES ('2014-01-15','Escapology',1)
INSERT @Exams VALUES ('2014-01-15','Escapology',9)
INSERT @Exams VALUES ('2014-01-15','Escapology',5)
INSERT @Exams VALUES ('2014-01-15','Escapology',3)
INSERT @Exams VALUES ('2014-01-16','Art',8)
INSERT @Exams VALUES ('2014-01-16','Art',7)
INSERT @Exams VALUES ('2014-01-16','Chemistry',0)
INSERT @Exams VALUES ('2014-01-16','Chemistry',5)
INSERT @Exams VALUES ('2014-01-16','Chemistry',9)
INSERT @Exams VALUES ('2014-01-16','Chemistry',2)
INSERT @Exams VALUES ('2014-01-17','Maths',5)
INSERT @Exams VALUES ('2014-01-17','Maths',6)
INSERT @Exams VALUES ('2014-01-18','Geology',10)

-- set up the grade boundaries
DECLARE @A INT = 9
DECLARE @B INT = 4
DECLARE @C INT = 0

-- get the count of grades by exam and date
SELECT * FROM
(SELECT
     Exam_Date
     -- list the exam date twice as one will be used to make the count
    ,Exam_Date  AS ExamDate
    ,[Subject]
    ,CASE WHEN GradeNumber >= @A THEN 'Outstanding'
        WHEN GradeNumber >= @B THEN 'Approved'
        ELSE 'Disapproved' END AS Result
FROM @Exams) AS Up
PIVOT (COUNT(Exam_Date) FOR Result IN ([Outstanding],[Approved],[Disapproved])) AS PVT

---

我可能会使用一个轴来获得结果，因为您正在将列中的值（gradeNumber）转换为行中的值（gradeNumber的计数）。试试这个：

-- make and fill an example table
DECLARE @Exams TABLE
( Exam_Date     DATE
 ,[Subject]     VARCHAR(250)
 ,GradeNumber   INT)

INSERT @Exams VALUES ('2014-01-15','Escapology',1)
INSERT @Exams VALUES ('2014-01-15','Escapology',9)
INSERT @Exams VALUES ('2014-01-15','Escapology',5)
INSERT @Exams VALUES ('2014-01-15','Escapology',3)
INSERT @Exams VALUES ('2014-01-16','Art',8)
INSERT @Exams VALUES ('2014-01-16','Art',7)
INSERT @Exams VALUES ('2014-01-16','Chemistry',0)
INSERT @Exams VALUES ('2014-01-16','Chemistry',5)
INSERT @Exams VALUES ('2014-01-16','Chemistry',9)
INSERT @Exams VALUES ('2014-01-16','Chemistry',2)
INSERT @Exams VALUES ('2014-01-17','Maths',5)
INSERT @Exams VALUES ('2014-01-17','Maths',6)
INSERT @Exams VALUES ('2014-01-18','Geology',10)

-- set up the grade boundaries
DECLARE @A INT = 9
DECLARE @B INT = 4
DECLARE @C INT = 0

-- get the count of grades by exam and date
SELECT * FROM
(SELECT
     Exam_Date
     -- list the exam date twice as one will be used to make the count
    ,Exam_Date  AS ExamDate
    ,[Subject]
    ,CASE WHEN GradeNumber >= @A THEN 'Outstanding'
        WHEN GradeNumber >= @B THEN 'Approved'
        ELSE 'Disapproved' END AS Result
FROM @Exams) AS Up
PIVOT (COUNT(Exam_Date) FOR Result IN ([Outstanding],[Approved],[Disapproved])) AS PVT

---

你对表演有什么问题？你对表演有什么问题？你对表演有什么问题？你对表演有什么问题？我看到我的解决方案被标记为答案，但看过Gordon的，我会采用他的，比我简单得多的方法，虽然我肯定会参数化等级边界，所以使用。。。未付金额（等级号>=@A时为1，否则为0结束时为1）。。。等等。我看到我的解决方案被标记为答案，但看过Gordon的，我会采用他的，比我简单得多的方法，虽然我肯定会参数化等级边界，所以使用。。。未付金额（等级号>=@A时为1，否则为0结束时为1）。。。等等。我看到我的解决方案被标记为答案，但看过Gordon的，我会采用他的，比我简单得多的方法，虽然我肯定会参数化等级边界，所以使用。。。未付金额（等级号>=@A时为1，否则为0结束时为1）。。。等等。我看到我的解决方案被标记为答案，但看过Gordon的，我会采用他的，比我简单得多的方法，虽然我肯定会参数化等级边界，所以使用。。。未付金额（等级号>=@A时为1，否则为0结束时为1）。。。等等