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  • sql教程之GROUP BY中的WITH CUBE、WITH ROLLUP原理测试及GROUPING应用

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 前几天,看到一个群友用WITH ROLLUP运算符。由于自个儿没用过,看到概念及结果都云里雾里的,所以突然来了兴趣对生成结果测了一番。

    一、概念:

    WITH CUBE:生成的结果集显示了所选列中值的所有组合的聚合。

    WITH ROLLUP:生成的结果集显示了所选列中值的某一层次结构的聚合。

    GROUPING:当行由 WITH CUBE或WITH ROLLUP运算符添加时,该函数将导致附加列的输出值为 1;当行不由 CUBE 或 ROLLUP 运算符添加时,该函数将导致附加列的输出值为 0。仅在与包含 CUBE 或 ROLLUP 运算符的 GROUP BY 子句相关联的选择列表中才允许分组。

    二、测试:

    1、建立临时表

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CREATE TABLE #T0
(
    [GRADE] [VARCHAR](50) NULL,     --年级
    [CLASS] [VARCHAR](50) NULL,     --班级
    [NAME] [VARCHAR](50) NULL,      --姓名
    [COURSE] [VARCHAR](50) NULL,    --学科
    [RESULT] [NUMERIC](8,2) NULL    --成绩
)

CREATE TABLE #T1
(
    [ID] [INT] IDENTITY(1,1) NOT NULL,    --序号
    [GRADE] [VARCHAR](50) NULL,           --年级
    [CLASS] [VARCHAR](50) NULL,           --班级
    [NAME] [VARCHAR](50) NULL,            --姓名
    [COURSE] [VARCHAR](50) NULL,          --学科
    [RESULT] [NUMERIC](8,2) NULL          --成绩
)

CREATE TABLE #T2
(
    [ID] [INT] IDENTITY(1,1) NOT NULL,    --序号
    [GRADE] [VARCHAR](50) NULL,           --年级
    [CLASS] [VARCHAR](50) NULL,           --班级
    [NAME] [VARCHAR](50) NULL,            --姓名
    [COURSE] [VARCHAR](50) NULL,          --学科
    [RESULT] [NUMERIC](8,2) NULL          --成绩
)
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     2、插入测试数据

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INSERT INTO #T0 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT '2019','CLASS1','9A01','C#',100
UNION
SELECT '2019','CLASS1','9A02','C#',100
UNION
SELECT '2019','CLASS2','9B01','C#',100
UNION
SELECT '2019','CLASS2','9B02','C#',100
UNION
SELECT '2018','CLASS1','8A01','JAVA',100
UNION
SELECT '2018','CLASS1','8A02','JAVA',100
UNION
SELECT '2018','CLASS2','8B01','JAVA',100
UNION
SELECT '2018','CLASS2','8B02','JAVA',100
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    查询T0表结果:

    3、GROUP BY

    抛砖引玉,看看常用的GROUP BY排序:默认以SELECT字段顺序(GRADE->CLASS->NAME->COURSE)进行排序,以下两种查询结果是一样的。

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SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY GRADE,CLASS,NAME,COURSE

SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY GRADE,CLASS,NAME,COURSE
ORDER BY GRADE,CLASS,NAME,COURSE
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    4、WITH CUBE

    原理1:以GROUP BY字段依次赋以NULL值进行分组聚合。

    原理2:第1个字段(即GRADE字段)生成结果:除原始数据外,以第1个字段固定赋以NULL值,然后其它字段依次赋以NULL值进行分组聚合,结果由右往左进行排序

    下面开始测第1个字段的结果是怎么来的:

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INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT 
FROM #T0 
GROUP BY GRADE,CLASS,NAME,COURSE

INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT 'ZZ' GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT 
FROM #T0 
GROUP BY CLASS,NAME,COURSE

INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT 'ZZ' GRADE,'ZZ' CLASS,NAME,COURSE,SUM(RESULT) RESULT 
FROM #T0 
GROUP BY NAME,COURSE

INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT 'ZZ' GRADE,'ZZ' CLASS,'ZZ' NAME,COURSE,SUM(RESULT) RESULT 
FROM #T0 
GROUP BY COURSE

INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT 'ZZ' GRADE,'ZZ' CLASS,'ZZ' NAME,'ZZ' COURSE,SUM(RESULT) RESULT 
FROM #T0

--第1个字段结果排序由右往左
INSERT INTO #T2 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT GRADE,CLASS,NAME,COURSE,RESULT FROM #T1 WHERE ID BETWEEN 1 AND 27 ORDER BY COURSE,NAME,CLASS,GRADE

UPDATE #T2 SET GRADE=NULL WHERE GRADE='ZZ'
UPDATE #T2 SET CLASS=NULL WHERE CLASS='ZZ'
UPDATE #T2 SET NAME=NULL WHERE NAME='ZZ'
UPDATE #T2 SET COURSE=NULL WHERE COURSE='ZZ'
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    WITH CUBE的结果:

SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY GRADE,CLASS,NAME,COURSE
WITH CUBE

    自已测试的结果:

SELECT * FROM #T2

    结果与上面一致。

    其它字段优先跟哪个字段组合、最终怎样排序?呃,测过,没搞清楚……

    5、WITH ROLLUP

    原理1:除原始数据外,以GROUP BY最后1个字段(即COURSE字段)固定赋以NULL值,然后其它字段依次赋以NULL值进行分组聚合,结果由左往右进行排序

    这个跟WITH CUBE的第1个字段非常相象:一个是第1个字段,一个是最后1个字段;一个结果是由右往左排序,一个结果是由左往右排序。

    下面开始测结果是怎么来的:

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TRUNCATE TABLE #T1
TRUNCATE TABLE #T2

INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT 
FROM #T0 
GROUP BY GRADE,CLASS,NAME,COURSE

INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT GRADE,CLASS,NAME,'ZZ' COURSE,SUM(RESULT) RESULT 
FROM #T0 
WHERE NOT EXISTS (SELECT 1 FROM #T1 WHERE GRADE=#T0.GRADE AND CLASS=#T0.GRADE AND NAME=#T0.NAME AND COURSE='ZZ')
GROUP BY GRADE,CLASS,NAME

INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT GRADE,CLASS,'ZZ' NAME,'ZZ' COURSE,SUM(RESULT) RESULT 
FROM #T0 
WHERE NOT EXISTS (SELECT 1 FROM #T1 WHERE GRADE=#T0.GRADE AND CLASS=#T0.CLASS AND NAME='ZZ' AND COURSE='ZZ')
GROUP BY GRADE,CLASS

INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT GRADE,'ZZ' CLASS,'ZZ' NAME,'ZZ' COURSE,SUM(RESULT) RESULT 
FROM #T0 
WHERE NOT EXISTS (SELECT 1 FROM #T1 WHERE GRADE=#T0.GRADE AND CLASS='ZZ' AND NAME='ZZ' AND COURSE='ZZ')
GROUP BY GRADE

INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT 'ZZ' GRADE,'ZZ' CLASS,'ZZ' NAME,'ZZ' COURSE,SUM(RESULT) RESULT 
FROM #T0 

--结果排序由左往右
INSERT INTO #T2 (GRADE,CLASS,NAME,COURSE,RESULT)
SELECT GRADE,CLASS,NAME,COURSE,RESULT FROM #T1 ORDER BY GRADE,CLASS,NAME,COURSE

UPDATE #T2 SET GRADE=NULL WHERE GRADE='ZZ'
UPDATE #T2 SET CLASS=NULL WHERE CLASS='ZZ'
UPDATE #T2 SET NAME=NULL WHERE NAME='ZZ'
UPDATE #T2 SET COURSE=NULL WHERE COURSE='ZZ'
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    WITH ROLLUP的结果:

SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY GRADE,CLASS,NAME,COURSE
WITH ROLLUP

    自己测试的结果:

SELECT * FROM #T2

    结果与上面一致。

    6、GROUPING

    这个就比较容易理解了,WITH CUBE与WITH ROLLUP用法一样,先看结果:

SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT,GROUPING(COURSE) [GROUPING]
FROM #T0
GROUP BY GRADE,CLASS,NAME,COURSE
WITH ROLLUP

    上面GROUPING的是COURSE字段,有NULL值就是WITH ROLLUP额外添加的,GROUPING结果值为1。

    有了GROUPING,那做小计、总计就方便了。

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SELECT 
    GRADE,
    CASE WHEN GROUPING(GRADE)=1 AND GROUPING(CLASS)=1 THEN '总计' WHEN GROUPING(GRADE)=0 AND GROUPING(CLASS)=1 THEN '小计' ELSE CLASS END CLASS,
    NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY GRADE,CLASS,NAME,COURSE
WITH ROLLUP
复制代码

     好了,原理测试及应用就到这里结束了。

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