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
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)
2、插入测试数据
INSERT INTO #T0 (GRADE,CLASS,NAME,COURSE,RESULT)SELECT '2019','CLASS1','9A01','C#',100UNIONSELECT '2019','CLASS1','9A02','C#',100UNIONSELECT '2019','CLASS2','9B01','C#',100UNIONSELECT '2019','CLASS2','9B02','C#',100UNIONSELECT '2018','CLASS1','8A01','JAVA',100UNIONSELECT '2018','CLASS1','8A02','JAVA',100UNIONSELECT '2018','CLASS2','8B01','JAVA',100UNIONSELECT '2018','CLASS2','8B02','JAVA',100
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。
SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULTFROM #T0GROUP BY GRADE,CLASS,NAME,COURSESELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULTFROM #T0GROUP BY GRADE,CLASS,NAME,COURSEORDER BY GRADE,CLASS,NAME,COURSE
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:
INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY GRADE,CLASS,NAME,COURSEINSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)SELECT 'ZZ' GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY CLASS,NAME,COURSEINSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)SELECT 'ZZ' GRADE,'ZZ' CLASS,NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY NAME,COURSEINSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)SELECT 'ZZ' GRADE,'ZZ' CLASS,'ZZ' NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY COURSEINSERT 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,GRADEUPDATE #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'
WITH CUBE的结果:
SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULTFROM #T0GROUP BY GRADE,CLASS,NAME,COURSEWITH 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:
TRUNCATE TABLE #T1TRUNCATE TABLE #T2INSERT INTO #T1 (GRADE,CLASS,NAME,COURSE,RESULT)SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULT
FROM #T0
GROUP BY GRADE,CLASS,NAME,COURSEINSERT 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,NAMEINSERT 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,CLASSINSERT 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 GRADEINSERT 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,COURSEUPDATE #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'
WITH ROLLUP的结果:
SELECT GRADE,CLASS,NAME,COURSE,SUM(RESULT) RESULTFROM #T0GROUP BY GRADE,CLASS,NAME,COURSEWITH 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 #T0GROUP BY GRADE,CLASS,NAME,COURSEWITH 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。
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) RESULTFROM #T0GROUP BY GRADE,CLASS,NAME,COURSEWITH ROLLUP
好了,原理测试及应用就到这里结束了。