oracle 10g函数大全--分析函数

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 NULL
If VAR_POP(expr1) = 0 and VAR_POP(expr2) != 0 then return 1
If 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 SETS

select id,area,stu_type,sum(score) score 
from students
group by grouping sets((id,area,stu_type),(id,area),id)
order by id,area,stu_type;

/*--------理解grouping sets
select a, b, c, sum( d ) from t
group by grouping sets ( a, b, c )

等效于

select * from (
select a, null, null, sum( d ) from t group by a
union all
select null, b, null, sum( d ) from t group by b 
union all
select null, null, c, sum( d ) from t group by c 
)
*/

--B、ROLLUP

select id,area,stu_type,sum(score) score 
from students
group by rollup(id,area,stu_type)
order by id,area,stu_type;

/*--------理解rollup
select a, b, c, sum( d )
from t
group by rollup(a, b, c);

等效于

select * from (
select a, b, c, sum( d ) from t group by a, b, c 
union all
select a, b, null, sum( d ) from t group by a, b
union all
select a, null, null, sum( d ) from t group by a
union all
select null, null, null, sum( d ) from t
)
*/

--C、CUBE

select id,area,stu_type,sum(score) score 
from students
group by cube(id,area,stu_type)
order by id,area,stu_type;

/*--------理解cube
select a, b, c, sum( d ) from t
group by cube( a, b, c)

等效于

select a, b, c, sum( d ) from t
group 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) score
from students
group 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,area

2、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          1250

2.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    4.31

3.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    4.48

20 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    2.76

30 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    4.31


4.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    29025

5.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    29025


6.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 | description
A | 1 | test
A | 2 | test
A | 3 | test
A | 4 | test
B | 1 | test
B | 2 | test
B | 3 | test
B | 4 | test
C | 1 | test
C | 2 | test
C | 3 | test
C | 4 | test

我想有一个sql语句,搜索的结果是 
A | 1 | test
A | 2 | test
B | 1 | test
B | 2 | test
C | 1 | test
C | 2 | test
实现: 
select name,seqno,description 
from(select name,seqno,description,row_number() over (partition by name order by seqno) id
from 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 table
create 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 2007-1-1
Case1 Step3 2007-1-3 Step4 2007-1-4 Step2 2007-1-2
Case1 Step4 2007-1-4 Step5 2007-1-5 Step3 2007-1-3
Case1 Step5 2007-1-5 Step4 2007-1-6 Step4 2007-1-4
Case1 Step4 2007-1-6 Step6 2007-1-7 Step5 2007-1-5
Case1 Step6 2007-1-7 Step4 2007-1-6
Case2 Step1 2007-2-1 Step2 2007-2-2 
Case2 Step2 2007-2-2 Step3 2007-2-3 Step1 2007-2-1
Case2 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) preactiondate
from lead_table) 
结果如下:

Case1 Step1 2007-1-1 2007-1-2 1
Case1 Step2 2007-1-2 2007-1-3 1
Case1 Step3 2007-1-3 2007-1-4 1
Case1 Step4 2007-1-4 2007-1-5 1
Case1 Step5 2007-1-5 2007-1-6 1
Case1 Step4 2007-1-6 2007-1-7 1
Case1 Step6 2007-1-7 
Case2 Step1 2007-2-1 2007-2-2 1
Case2 Step2 2007-2-2 2007-2-3 1
Case2 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) preactiondate
from lead_table

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