oracle 分析函数
发布日期:2021-08-28 04:40:40 浏览次数:2 分类:技术文章

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一、总体介绍12.1 分析函数如何工作语法 FUNCTION_NAME(
<参数>
,…) OVER (
>
) PARTITION子句 ORDER BY子句 WINDOWING子句 缺省时相当于RANGE UNBOUNDED PRECEDING 1. 值域窗(RANGE WINDOW) RANGE N PRECEDING 仅对数值或日期类型有效,选定窗为排序后当前行之前,某列(即排序列)值大于/小于(当前行该列值 –/+ N)的所有行,因此与ORDER BY子句有关系。 2. 行窗(ROW WINDOW) ROWS N PRECEDING 选定窗为当前行及之前N行。 还可以加上BETWEEN AND 形式,例如RANGE BETWEEN m PRECEDING AND n FOLLOWING 函数 AVG(
eXPr) 一组或选定窗中表达式的平均值 CORR(expr, expr) 即COVAR_POP(exp1,exp2) / (STDDEV_POP(expr1) * STDDEV_POP(expr2)),两个表达式的互相关,-1(反相关) ~ 1(正相关),0表示不相关 COUNT(
<*>
) 计数 COVAR_POP(expr, expr) 总体协方差 COVAR_SAMP(expr, expr) 样本协方差 CUME_DIST 累积分布,即行在组中的相对位置,返回0 ~ 1 DENSE_RANK 行的相对排序(与ORDER BY搭配),相同的值具有一样的序数(NULL计为相同),并不留空序数 FIRST_VALUE 一个组的第一个值 LAG(expr,
,
) 访问之前的行,OFFSET是缺省为1 的正数,表示相对行数,DEFAULT是当超出选定窗范围时的返回值(如第一行不存在之前行) LAST_VALUE 一个组的最后一个值 LEAD(expr,
,
) 访问之后的行,OFFSET是缺省为1 的正数,表示相对行数,DEFAULT是当超出选定窗范围时的返回值(如最后行不存在之前行) MAX(expr) 最大值 MIN(expr) 最小值 NTILE(expr) 按表达式的值和行在组中的位置编号,如表达式为4,则组分4份,分别为1 ~ 4的值,而不能等分则多出的部分在值最小的那组 PERCENT_RANK 类似CUME_DIST,1/(行的序数 - 1) RANK 相对序数,答应并列,并空出随后序号 RATIO_TO_REPORT(expr) 表达式值 / SUM(表达式值) ROW_NUMBER 排序的组中行的偏移 STDDEV(expr) 标准差 STDDEV_POP(expr) 总体标准差 STDDEV_SAMP(expr) 样本标准差 SUM(expr) 合计 VAR_POP(expr) 总体方差 VAR_SAMP(expr) 样本方差 VARIANCE(expr) 方差 REGR_ xxxx(expr, expr) 线性回归函数 REGR_SLOPE:返回斜率,等于COVAR_POP(expr1, expr2) / VAR_POP(expr2)REGR_INTERCEPT:返回回归线的y截距,等于AVG(expr1) - REGR_SLOPE(expr1, expr2) * AVG(expr2)REGR_COUNT:返回用于填充回归线的非空数字对的数目REGR_R2:返回回归线的决定系数,计算式为:If VAR_POP(expr2) = 0 then return NULLIf VAR_POP(expr1) = 0 and VAR_POP(expr2) != 0 then return 1If VAR_POP(expr1) > 0 and VAR_POP(expr2 != 0 then return POWER(CORR(expr1,expr),2)REGR_AVGX:计算回归线的自变量(expr2)的平均值,去掉了空对(expr1, expr2)后,等于AVG(expr2)REGR_AVGY:计算回归线的应变量(expr1)的平均值,去掉了空对(expr1, expr2)后,等于AVG(expr1)REGR_SXX: 返回值等于REGR_COUNT(expr1, expr2) * VAR_POP(expr2)REGR_SYY: 返回值等于REGR_COUNT(expr1, expr2) * VAR_POP(expr1)REGR_SXY: 返回值等于REGR_COUNT(expr1, expr2) * COVAR_POP(expr1, expr2)首先:创建表及接入测试数据create table students(id number(15,0),area varchar2(10),stu_type varchar2(2),score number(20,2));insert into students values(1, '111', 'g', 80 );insert into students values(1, '111', 'j', 80 );insert into students values(1, '222', 'g', 89 );insert into students values(1, '222', 'g', 68 );insert into students values(2, '111', 'g', 80 );insert into students values(2, '111', 'j', 70 );insert into students values(2, '222', 'g', 60 );insert into students values(2, '222', 'j', 65 );insert into students values(3, '111', 'g', 75 );insert into students values(3, '111', 'j', 58 );insert into students values(3, '222', 'g', 58 );insert into students values(3, '222', 'j', 90 );insert into students values(4, '111', 'g', 89 );insert into students values(4, '111', 'j', 90 );insert into students values(4, '222', 'g', 90 );insert into students values(4, '222', 'j', 89 );commit;二、具体应用:1、分组求和:1)GROUP BY子句 --A、GROUPING SETSselect id,area,stu_type,sum(score) score from studentsgroup by grouping sets((id,area,stu_type),(id,area),id)order by id,area,stu_type;/*--------理解grouping setsselect a, b, c, sum( d ) from tgroup by grouping sets ( a, b, c )等效于select * from (select a, null, null, sum( d ) from t group by aunion allselect null, b, null, sum( d ) from t group by b union allselect null, null, c, sum( d ) from t group by c )*/--B、ROLLUPselect id,area,stu_type,sum(score) score from studentsgroup by rollup(id,area,stu_type)order by id,area,stu_type;/*--------理解rollupselect a, b, c, sum( d )from tgroup by rollup(a, b, c);等效于select * from (select a, b, c, sum( d ) from t group by a, b, c union allselect a, b, null, sum( d ) from t group by a, bunion allselect a, null, null, sum( d ) from t group by aunion allselect null, null, null, sum( d ) from t)*/--C、CUBEselect id,area,stu_type,sum(score) score from studentsgroup by cube(id,area,stu_type)order by id,area,stu_type;/*--------理解cubeselect a, b, c, sum( d ) from tgroup by cube( a, b, c)等效于select a, b, c, sum( d ) from tgroup by grouping sets( ( a, b, c ), ( a, b ), ( a ), ( b, c ), ( b ), ( a, c ), ( c ), () )*/--D、GROUPING/*从上面的结果中我们很容易发现,每个统计数据所对应的行都会出现null,如何来区分到底是根据那个字段做的汇总呢,grouping函数判断是否合计列!*/select decode(grouping(id),1,'all id',id) id,decode(grouping(area),1,'all area',to_char(area)) area,decode(grouping(stu_type),1,'all_stu_type',stu_type) stu_type,sum(score) scorefrom studentsgroup by cube(id,area,stu_type)order by id,area,stu_type; 二、OVER()函数的使用1、统计名次——DENSE_RANK(),ROW_NUMBER()1)允许并列名次、名次不间断,DENSE_RANK(),结果如122344456……将score按ID分组排名:dense_rank() over(partition by id order by score desc)将score不分组排名:dense_rank() over(order by score desc)select id,area,score,dense_rank() over(partition by id order by score desc) 分组id排序,dense_rank() over(order by score desc) 不分组排序from students order by id,area;2)不允许并列名次、相同值名次不重复,ROW_NUMBER(),结果如123456……将score按ID分组排名:row_number() over(partition by id order by score desc)将score不分组排名:row_number() over(order by score desc)select id,area,score,row_number() over(partition by id order by score desc) 分组id排序,row_number() over(order by score desc) 不分组排序from students order by id,area;3)允许并列名次、复制名次自动空缺,rank(),结果如12245558……将score按ID分组排名:rank() over(partition by id order by score desc)将score不分组排名:rank() over(order by score desc)select id,area,score,rank() over(partition by id order by score desc) 分组id排序,rank() over(order by score desc) 不分组排序from students order by id,area;4)名次分析,cume_dist()——-最大排名/总个数 函数:cume_dist() over(order by id)select id,area,score,cume_dist() over(order by id) a, --按ID最大排名/总个数 cume_dist() over(partition by id order by score desc) b, --ID分组中,scroe最大排名值/本组总个数row_number() over (order by id) 记录号from students order by id,area;5)利用cume_dist(),允许并列名次、复制名次自动空缺,取并列后较大名次,结果如22355778……将score按ID分组排名:cume_dist() over(partition by id order by score desc)*sum(1) over(partition by id)将score不分组排名:cume_dist() over(order by score desc)*sum(1) over()select id,area,score,sum(1) over() as 总数,sum(1) over(partition by id) as 分组个数,(cume_dist() over(partition by id order by score desc))*(sum(1) over(partition by id)) 分组id排序,(cume_dist() over(order by score desc))*(sum(1) over()) 不分组排序from students order by id,area2、分组统计--sum(),max(),avg(),RATIO_TO_REPORT()select id,area,sum(1) over() as 总记录数, sum(1) over(partition by id) as 分组记录数,sum(score) over() as 总计 , sum(score) over(partition by id) as 分组求和,sum(score) over(order by id) as 分组连续求和,sum(score) over(partition by id,area) as 分组ID和area求和,sum(score) over(partition by id order by area) as 分组ID并连续按area求和,max(score) over() as 最大值,max(score) over(partition by id) as 分组最大值,max(score) over(order by id) as 分组连续最大值,max(score) over(partition by id,area) as 分组ID和area求最大值,max(score) over(partition by id order by area) as 分组ID并连续按area求最大值,avg(score) over() as 所有平均,avg(score) over(partition by id) as 分组平均,avg(score) over(order by id) as 分组连续平均,avg(score) over(partition by id,area) as 分组ID和area平均,avg(score) over(partition by id order by area) as 分组ID并连续按area平均,RATIO_TO_REPORT(score) over() as "占所有%",RATIO_TO_REPORT(score) over(partition by id) as "占分组%",score from students;3、LAG(COL,n,default)、LEAD(OL,n,default) --取前后边N条数据取前面记录的值:lag(score,n,x) over(order by id)取后面记录的值:lead(score,n,x) over(order by id) 参数:n表示移动N条记录,X表示不存在时填充值,iD表示排序列select id,lag(score,1,0) over(order by id) lg,score from students;select id,lead(score,1,0) over(order by id) lg,score from students;4、FIRST_VALUE()、LAST_VALUE()取第起始1行值:first_value(score,n) over(order by id)取第最后1行值:LAST_value(score,n) over(order by id)select id,first_value(score) over(order by id) fv,score from students;select id,last_value(score) over(order by id) fv,score from students;
sum(...) over ...【功能】连续求和分析函数【参数】具体参示例【说明】Oracle分析函数NC示例:select bdcode,sum(1) over(order by bdcode) aa from bd_bdinfo 【示例】1.原表信息: SQL> break on deptno skip 1  -- 为效果更明显,把不同部门的数据隔段显示。SQL> select deptno,ename,sal   2   from emp   3   order by deptno;DEPTNO ENAME          SAL---------- ---------- ----------       10 CLARK          2450          KING          5000          MILLER           1300       20 SMITH          800          ADAMS          1100          FORD          3000          SCOTT          3000          JONES          2975       30 ALLEN          1600          BLAKE          2850          MARTIN           1250          JAMES          950          TURNER           1500          WARD          12502.先来一个简单的,注意over(...)条件的不同,使用 sum(sal) over (order by ename)... 查询员工的薪水“连续”求和,注意over (order   by ename)如果没有order by 子句,求和就不是“连续”的,放在一起,体会一下不同之处:SQL> select deptno,ename,sal,   2   sum(sal) over (order by ename) 连续求和,   3   sum(sal) over () 总和,                -- 此处sum(sal) over () 等同于sum(sal)   4   100*round(sal/sum(sal) over (),4) "份额(%)"   5   from emp   6   /DEPTNO ENAME          SAL 连续求和    总和 份额(%)---------- ---------- ---------- ---------- ---------- ----------       20 ADAMS          1100    1100    29025    3.79       30 ALLEN          1600    2700    29025    5.51       30 BLAKE          2850    5550    29025    9.82       10 CLARK          2450    8000    29025    8.44       20 FORD          3000    11000    29025    10.34       30 JAMES          950    11950    29025    3.27       20 JONES          2975    14925    29025    10.25       10 KING          5000    19925    29025    17.23       30 MARTIN           1250    21175    29025    4.31       10 MILLER           1300    22475    29025    4.48       20 SCOTT          3000    25475    29025    10.34       20 SMITH          800    26275    29025    2.76       30 TURNER           1500    27775    29025    5.17       30 WARD          1250    29025    29025    4.313.使用子分区查出各部门薪水连续的总和。注意按部门分区。注意over(...)条件的不同,sum(sal) over (partition by deptno order by ename) 按部门“连续”求总和sum(sal) over (partition by deptno) 按部门求总和sum(sal) over (order by deptno,ename) 不按部门“连续”求总和sum(sal) over () 不按部门,求所有员工总和,效果等同于sum(sal)。SQL> select deptno,ename,sal,   2   sum(sal) over (partition by deptno order by ename) 部门连续求和,--各部门的薪水"连续"求和   3   sum(sal) over (partition by deptno) 部门总和,   -- 部门统计的总和,同一部门总和不变   4   100*round(sal/sum(sal) over (partition by deptno),4) "部门份额(%)",   5   sum(sal) over (order by deptno,ename) 连续求和, --所有部门的薪水"连续"求和   6   sum(sal) over () 总和,   -- 此处sum(sal) over () 等同于sum(sal),所有员工的薪水总和   7   100*round(sal/sum(sal) over (),4) "总份额(%)"   8   from emp   9   /DEPTNO ENAME SAL 部门连续求和 部门总和 部门份额(%) 连续求和 总和   总份额(%)------ ------ ----- ------------ ---------- ----------- ---------- ------ ----------10 CLARK 2450       2450    8750       28    2450   29025    8.44   KING 5000       7450    8750    57.14    7450   29025    17.23   MILLER   1300       8750    8750    14.86    8750   29025    4.4820 ADAMS 1100       1100    10875    10.11    9850   29025    3.79   FORD 3000       4100    10875    27.59    12850   29025    10.34   JONES 2975       7075    10875    27.36    15825   29025    10.25   SCOTT 3000        10075    10875    27.59    18825   29025    10.34   SMITH 800        10875    10875        7.36    19625   29025    2.7630 ALLEN 1600       1600    9400    17.02    21225   29025    5.51   BLAKE 2850       4450    9400    30.32    24075   29025    9.82   JAMES 950       5400    9400    10.11    25025   29025    3.27   MARTIN   1250       6650    9400        13.3    26275   29025    4.31   TURNER   1500       8150    9400    15.96    27775   29025    5.17   WARD 1250       9400    9400        13.3    29025   29025    4.314.来一个综合的例子,求和规则有按部门分区的,有不分区的例子SQL> select deptno,ename,sal,sum(sal) over (partition by deptno order by sal) dept_sum,   2   sum(sal) over (order by deptno,sal) sum   3   from emp;DEPTNO ENAME          SAL DEPT_SUM        SUM---------- ---------- ---------- ---------- ----------       10 MILLER           1300    1300    1300          CLARK          2450    3750    3750          KING          5000    8750    8750       20 SMITH          800        800    9550          ADAMS          1100    1900    10650          JONES          2975    4875    13625          SCOTT          3000    10875    19625          FORD          3000    10875    19625       30 JAMES          950        950    20575          WARD          1250    3450    23075          MARTIN           1250    3450    23075          TURNER           1500    4950    24575          ALLEN          1600    6550    26175          BLAKE          2850    9400    290255.来一个逆序的,即部门从大到小排列,部门里各员工的薪水从高到低排列,累计和的规则不变。SQL> select deptno,ename,sal,   2   sum(sal) over (partition by deptno order by deptno desc,sal desc) dept_sum,   3   sum(sal) over (order by deptno desc,sal desc) sum   4   from emp;DEPTNO ENAME          SAL DEPT_SUM        SUM---------- ---------- ---------- ---------- ----------       30 BLAKE          2850    2850    2850          ALLEN          1600    4450    4450          TURNER           1500    5950    5950          WARD          1250    8450    8450          MARTIN           1250    8450    8450          JAMES          950    9400    9400       20 SCOTT          3000    6000    15400          FORD          3000    6000    15400          JONES          2975    8975    18375          ADAMS          1100    10075    19475          SMITH          800    10875    20275       10 KING          5000    5000    25275          CLARK          2450    7450    27725          MILLER           1300    8750    290256.体会:在"... from emp;"后面不要加order   by 子句,使用的分析函数的(partition by deptno order by sal)里已经有排序的语句了,如果再在句尾添加排序子句,一致倒罢了,不一致,结果就令人费劲了。如:SQL> select deptno,ename,sal,sum(sal) over (partition by deptno order by sal) dept_sum,   2   sum(sal) over (order by deptno,sal) sum   3   from emp   4   order by deptno desc;DEPTNO ENAME          SAL DEPT_SUM        SUM---------- ---------- ---------- ---------- ----------       30 JAMES          950        950    20575          WARD          1250    3450    23075          MARTIN           1250    3450    23075          TURNER           1500    4950    24575          ALLEN          1600    6550    26175          BLAKE          2850    9400    29025       20 SMITH          800        800    9550          ADAMS          1100    1900    10650          JONES          2975    4875    13625          SCOTT          3000    10875    19625          FORD          3000    10875    19625       10 MILLER           1300    1300    1300          CLARK          2450    3750    3750          KING          5000    8750    8750
RANK()dense_rank()【语法】RANK ( ) OVER ( [query_partition_clause] order_by_clause )	dense_RANK ( ) OVER ( [query_partition_clause] order_by_clause )【功能】聚合函数RANK 和 dense_rank 主要的功能是计算一组数值中的排序值。【参数】dense_rank与rank()用法相当,【区别】dence_rank在并列关系是,相关等级不会跳过。rank则跳过rank()是跳跃排序,有两个第二名时接下来就是第四名(同样是在各个分组内) dense_rank()l是连续排序,有两个第二名时仍然跟着第三名。【说明】Oracle分析函数【示例】聚合函数RANK 和 dense_rank 主要的功能是计算一组数值中的排序值。    在9i版本之前,只有分析功能(analytic ),即从一个查询结果中计算每一行的排序值,是基于order_by_clause子句中的value_exprs指定字段的。    其语法为:    RANK ( ) OVER ( [query_partition_clause] order_by_clause )    在9i版本新增加了合计功能(aggregate),即对给定的参数值在设定的排序查询中计算出其排序值。这些参数必须是常数或常值表达式,且必须和ORDER BY子句中的字段个数、位置、类型完全一致。    其语法为:    RANK ( expr [, expr]... ) WITHIN GROUP  ( ORDER BY  expr [ DESC | ASC ] [NULLS { FIRST | LAST }]  [, expr [ DESC | ASC ] [NULLS { FIRST | LAST }]]...  )    例子1:    有表Table内容如下    COL1 COL2    1 1    2 1    3 2    3 1    4 1    4 2    5 2    5 2    6 2    分析功能:列出Col2分组后根据Col1排序,并生成数字列。比较实用于在成绩表中查出各科前几名的信息。    SELECT a.*,RANK() OVER(PARTITION BY col2 ORDER BY col1) "Rank" FROM table a;    结果如下:    COL1 COL2 Rank    1 1   1    2 1   2    3 1   3    4 1   4    3 2   1    4 2   2    5 2   3    5 2   3    6 2   5    例子2:    TABLE:A (科目,分数)    数学,80  语文,70  数学,90  数学,60  数学,100  语文,88  语文,65  语文,77    现在我想要的结果是:(即想要每门科目的前3名的分数)    数学,100  数学,90  数学,80  语文,88  语文,77  语文,70    那么语句就这么写:    select * from (select rank() over(partition by 科目 order by 分数 desc) rk,a.* from a) t  where t.rk<=3;    例子3:    合计功能:计算出数值(4,1)在Orade By Col1,Col2排序下的排序值,也就是col1=4,col2=1在排序以后的位置    SELECT RANK(4,3) WITHIN GROUP (ORDER BY col1,col2) "Rank" FROM table;    结果如下:  Rank  4    dense_rank与rank()用法相当,但是有一个区别:dence_rank在并列关系是,相关等级不会跳过。rank则跳过    例如:表    A      B      C  a     liu     wang  a     jin     shu  a     cai     kai  b     yang     du  b     lin     ying  b     yao     cai  b     yang     99    例如:当rank时为:    select m.a,m.b,m.c,rank() over(partition by a order by b) liu from test3 m     A     B       C     LIU   a     cai      kai     1   a     jin      shu     2   a     liu      wang     3   b     lin      ying     1   b     yang     du      2   b     yang     99      2   b     yao      cai     4    而如果用dense_rank时为:    select m.a,m.b,m.c,dense_rank() over(partition by a order by b) liu from test3 m     A     B       C     LIU   a     cai     kai     1   a     jin     shu     2   a     liu     wang     3   b     lin     ying     1   b     yang     du      2   b     yang     99      2   b     yao     cai     3
ROW_NUMBER()【语法】ROW_NUMBER() OVER (PARTITION BY COL1 ORDER BY COL2) 【功能】表示根据COL1分组,在分组内部根据 COL2排序,而这个值就表示每组内部排序后的顺序编号(组内连续的唯一的) row_number() 返回的主要是“行”的信息,并没有排名【参数】【说明】Oracle分析函数主要功能:用于取前几名,或者最后几名等【示例】表内容如下:name | seqno | descriptionA | 1 | testA | 2 | testA | 3 | testA | 4 | testB | 1 | testB | 2 | testB | 3 | testB | 4 | testC | 1 | testC | 2 | testC | 3 | testC | 4 | test我想有一个sql语句,搜索的结果是 A | 1 | testA | 2 | testB | 1 | testB | 2 | testC | 1 | testC | 2 | test实现: select name,seqno,description from(select name,seqno,description,row_number() over (partition by name order by seqno) idfrom table_name) where id<=3;
lag()和lead()【语法】lag(EXPR,
,
)LEAD(EXPR,
,
)【功能】表示根据COL1分组,在分组内部根据 COL2排序,而这个值就表示每组内部排序后的顺序编号(组内连续的唯一的) lead () 下一个值 lag() 上一个值【参数】EXPR是从其他行返回的表达式 OFFSET是缺省为1 的正数,表示相对行数。希望检索的当前行分区的偏移量DEFAULT是在OFFSET表示的数目超出了分组的范围时返回的值。【说明】Oracle分析函数【示例】-- Create tablecreate table LEAD_TABLE( CASEID VARCHAR2(10), STEPID VARCHAR2(10), ACTIONDATE DATE)tablespace COLM_DATA pctfree 10 initrans 1 maxtrans 255 storage ( initial 64K minextents 1 maxextents unlimited );insert into LEAD_TABLE values('Case1','Step1',to_date('20070101','yyyy-mm-dd'));insert into LEAD_TABLE values('Case1','Step2',to_date('20070102','yyyy-mm-dd'));insert into LEAD_TABLE values('Case1','Step3',to_date('20070103','yyyy-mm-dd'));insert into LEAD_TABLE values('Case1','Step4',to_date('20070104','yyyy-mm-dd'));insert into LEAD_TABLE values('Case1','Step5',to_date('20070105','yyyy-mm-dd'));insert into LEAD_TABLE values('Case1','Step4',to_date('20070106','yyyy-mm-dd'));insert into LEAD_TABLE values('Case1','Step6',to_date('20070101','yyyy-mm-dd'));insert into LEAD_TABLE values('Case1','Step1',to_date('20070201','yyyy-mm-dd'));insert into LEAD_TABLE values('Case2','Step2',to_date('20070202','yyyy-mm-dd'));insert into LEAD_TABLE values('Case2','Step3',to_date('20070203','yyyy-mm-dd'));commit; 结果如下:Case1 Step1 2007-1-1 Step2 2007-1-2 Case1 Step2 2007-1-2 Step3 2007-1-3 Step1 2007-1-1Case1 Step3 2007-1-3 Step4 2007-1-4 Step2 2007-1-2Case1 Step4 2007-1-4 Step5 2007-1-5 Step3 2007-1-3Case1 Step5 2007-1-5 Step4 2007-1-6 Step4 2007-1-4Case1 Step4 2007-1-6 Step6 2007-1-7 Step5 2007-1-5Case1 Step6 2007-1-7 Step4 2007-1-6Case2 Step1 2007-2-1 Step2 2007-2-2 Case2 Step2 2007-2-2 Step3 2007-2-3 Step1 2007-2-1Case2 Step3 2007-2-3 Step2 2007-2-2还可以进一步统计一下两者的相差天数select caseid,stepid,actiondate,nextactiondate,nextactiondate-actiondate datebetween from (select caseid,stepid,actiondate,lead(stepid) over (partition by caseid order by actiondate) nextstepid,lead(actiondate) over (partition by caseid order by actiondate) nextactiondate,lag(stepid) over (partition by caseid order by actiondate) prestepid,lag(actiondate) over (partition by caseid order by actiondate) preactiondatefrom lead_table) 结果如下:Case1 Step1 2007-1-1 2007-1-2 1Case1 Step2 2007-1-2 2007-1-3 1Case1 Step3 2007-1-3 2007-1-4 1Case1 Step4 2007-1-4 2007-1-5 1Case1 Step5 2007-1-5 2007-1-6 1Case1 Step4 2007-1-6 2007-1-7 1Case1 Step6 2007-1-7 Case2 Step1 2007-2-1 2007-2-2 1Case2 Step2 2007-2-2 2007-2-3 1Case2 Step3 2007-2-3 每一条记录都能连接到上/下一行的内容lead () 下一个值 lag() 上一个值select caseid,stepid,actiondate,lead(stepid) over (partition by caseid order by actiondate) nextstepid,lead(actiondate) over (partition by caseid order by actiondate) nextactiondate,lag(stepid) over (partition by caseid order by actiondate) prestepid,lag(actiondate) over (partition by caseid order by actiondate) preactiondatefrom lead_table

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哈哈,博客排版真的漂亮呢~
[***.90.31.176]2024年04月24日 15时16分07秒