ORACLE高级SQL之分析函数

首先来看一个例子：

--删除emp表
drop table emp purge;

--创建emp表
CREATE TABLE emp
(
emp_id    NUMBER(6),
ename  VARCHAR2(45),
dept_id   NUMBER(4),
hire_date DATE,
sal    NUMBER(8,2)
);

--创建emp数据
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (101, 'Tom',    20,  TO_DATE('21-09-1989', 'DD-MM-YYYY'), 2000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (102, 'Mike',   20,  TO_DATE('13-01-1993', 'DD-MM-YYYY'), 8000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (120, 'John',   50,  TO_DATE('18-07-1996', 'DD-MM-YYYY'), 1000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (121, 'Joy',    50,  TO_DATE('10-04-1997', 'DD-MM-YYYY'), 4000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (122, 'Rich',   50,  TO_DATE('01-05-1995', 'DD-MM-YYYY'), 3000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (123, 'Kate',   50,  TO_DATE('10-10-1997', 'DD-MM-YYYY'), 5000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (124, 'Jess',   50,  TO_DATE('16-11-1999', 'DD-MM-YYYY'), 6000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (100, 'Stev',   10,  TO_DATE('01-01-1990', 'DD-MM-YYYY'), 7000);

COMMIT;

查询结果：

SELECT
emp_id,ename,dept_id,hire_date,sal,
SUM(sal) OVER (PARTITION BY dept_id ORDER BY hire_date) sum_sal,
SUM(sal) OVER (PARTITION BY dept_id ) sum_sal2,
SUM(sal) OVER ( ) sum_sal3
FROM emp;

SELECT
  emp_id,ename,dept_id,hire_date,sal,
  SUM(sal) OVER (PARTITION BY dept_id ORDER BY hire_date) sum_sal1,
  SUM(sal) OVER (PARTITION BY dept_id ORDER BY hire_date DESC) sum_sal2,      --desc是从下往上累加
  SUM(sal) OVER (PARTITION BY dept_id ORDER BY hire_date DESC nulls LAST) sum_sal3,
  SUM(sal) OVER (PARTITION BY dept_id ) sum_sal4,
  SUM(sal) OVER ( ) sum_sal5
FROM emp;

查询结果：

开窗函数

1、开窗函数之ROWS：

SELECT emp_id,ename,dept_id,hire_date,sal,
-- 以下均为首先按dept_id进行分组，其次按照hire_date进行排序，且所有统计不能跨越其所在分区，故不再重复
-- 窗口范围为该分区的第一行到该分区的最后一行，与sum_sal_part等同
SUM(sal) OVER (PARTITION BY dept_id ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) sum_1_to_last,
-- 窗口范围为该分区的第一行到本行，与sum_sal_part_order等同
SUM(sal) OVER (PARTITION BY dept_id ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) sum_1_to_cur,
-- 窗口范围为该分区的第一行到本行前一行，统计的是第一行到本行前一行薪资的累计
SUM(sal) OVER (PARTITION BY dept_id ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1/*value_expr*/ PRECEDING) sum_1_to_curbef1,
-- 窗口范围为该分区的第一行到本行后一行，统计的是第一行到本行后一行薪资的累计
SUM(sal) OVER (PARTITION BY dept_id ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 FOLLOWING) sum_1_to_curaft1
FROM emp 
order by dept_id,hire_date;

查询结果：

2、开窗函数之RANGE：

RANGE窗口中只能对NUMBER、DATE起作用。

RANGE窗口中，ORDER BY中只能有一列；但是在ROWS窗口中，ORDER BY中可以有多列。

SELECT emp_id,ename,dept_id,hire_date,sal,
-- 后面均为以dept_id分组，再按hire_date排序，且所有统计不能跨分区，由于是逻辑范围，因此PRECEDING和FOLLOWING表达式有符号
-- 窗口范围为该分区的第一行到该分区的最后一行，与sum_sal_part等同,在非条件表达式中等同于ROWS
SUM(sal) OVER (PARTITION BY dept_id ORDER BY sal RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) sum_1_to_last,
-- 窗口范围为该分区的第一行到本行，与sum_sal_part_order等同,在非条件表达式中等同于ROWS
SUM(sal) OVER (PARTITION BY dept_id ORDER BY sal RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) sum_1_to_cur,
-- 窗口范围为该分区内小于（当前行sal - 2500）的所有的薪资累计
SUM(sal) OVER (PARTITION BY dept_id ORDER BY sal RANGE BETWEEN UNBOUNDED PRECEDING AND 2500/*value_expr*/ PRECEDING) sum1,
-- 窗口范围为该分区内小于（当前行sal + 2500）的所有薪资累计
SUM(sal) OVER (PARTITION BY dept_id ORDER BY sal RANGE BETWEEN UNBOUNDED PRECEDING AND 2500/*value_expr*/ FOLLOWING) sum2
FROM emp;

查询结果：

3、开窗函数之KEEP

KEEP为聚合函数的特殊关键字。（弄不明白这个函数）
当聚合函数MIN、MAX、SUM、AVG、COUNT、VARIANCE和STDDEV使用KEEP。
当KEEP和DENSE_RANK FIRST/DENSE_RANK LAST 一起使用时，获取一组中排名第一或者排名最后的记录。必须由order by子句来排序。后面也可以接over()分析函数部分。
Min(col2) keep (dense_rank first order by coll)保留按coll排名第一的col2的最小值。
Min(col2) keep (dense_rank first order by co11) over (partition by co 13 ）按col3分组保留按coll排名各组第一的col2 的最小值。

SELECT emp_id,ename,dept_id,hire_date,sal,
DENSE_RANK() OVER(PARTITION BY dept_id ORDER BY sal) DENSE_RANK,
MIN(hire_date) KEEP (DENSE_RANK FIRST ORDER BY sal) OVER(PARTITION BY dept_id) min_first,
MIN(hire_date) KEEP (DENSE_RANK LAST ORDER BY sal) OVER(PARTITION BY dept_id) min_last,
MAX(hire_date) KEEP (DENSE_RANK FIRST ORDER BY sal) OVER(PARTITION BY dept_id) max_first,
MAX(hire_date) KEEP (DENSE_RANK LAST ORDER BY sal) OVER(PARTITION BY dept_id) max_last
FROM emp;

查询结果：

 

函数部分

1、排名函数：ROW_NUMBER、RANK、DENSE_RANK

这三个函数都是根据PARTITION BY后面的字段分组，然后根据ORDER BY后面的字段进行排序后再进行排名

SELECT
    emp_id,ename,dept_id,hire_date,sal,
    ROW_NUMBER() OVER (PARTITION BY dept_id ORDER BY sal) AS row_number,
    RANK() OVER (PARTITION BY dept_id ORDER BY sal) AS rank,
    DENSE_RANK() OVER (PARTITION BY dept_id ORDER BY sal) AS dense_rank
FROM emp;

查询结果：

SELECT
  emp_id,ename,dept_id,hire_date,sal,
  MIN(sal) KEEP (DENSE_RANK FIRST ORDER BY hire_date) OVER (PARTITION BY dept_id) Worst,
  MAX(sal) KEEP (DENSE_RANK LAST ORDER BY hire_date) OVER (PARTITION BY dept_id) Best
FROM emp;

查询结果：

2、FIRST_VALUE与LAST_VALUE

FIRST_VALUE 返回一个排序数据集合的第一行，而LAST_VALUE 则返回一个排序数据集合的最后一行。

SELECT
  emp_id,ename,dept_id,hire_date,sal,
  FIRST_VALUE(ename) OVER(PARTITION BY dept_id ORDER BY sal ) AS fir_val,
  FIRST_VALUE(ename) OVER(PARTITION BY dept_id ORDER BY sal DESC) AS fir_val_desc,
  LAST_VALUE(ename) OVER(PARTITION BY dept_id ORDER BY sal ) AS last_val,
  LAST_VALUE(ename) OVER(PARTITION BY dept_id ORDER BY sal DESC) AS last_val_desc
FROM emp;

查询结果：

3、LAG与LEAD

lag与lead函数是跟偏移量相关的两个分析函数，通过这两个函数可以在一次查询中取出同一字段的前N行的数据(lag)和后N行的数据(lead)作为独立的列,从而更方便地进行进行数据过滤。这种操作可以代替表的自联接，并且LAG和LEAD有更高的效率。

over()表示 lag()与lead()操作的数据都在over()的范围内，它里面可以使用partition by语句（用于分组） order by 语句（用于排序）。partition by a order by b表示以a字段进行分组，再以b字段进行排序，对数据进行查询。

例如：lead(field, num, defaultvalue) field需要查找的字段，num往后查找的num行的数据，defaultvalue没有符合条件的默认值。

SELECT
  emp_id,ename,dept_id,hire_date,sal,
  LAG(sal) OVER (ORDER BY hire_date) AS prev_sal1,
  LEAD(sal) OVER (ORDER BY hire_date) AS next_sal1,
  LAG(sal, 1, 0) OVER (ORDER BY hire_date) AS prev_sal2,
  LEAD(sal, 1,0) OVER (ORDER BY hire_date) AS next_sal2,
  LAG(sal, 1, 0) OVER (partition BY dept_id ORDER BY hire_date) AS prev_sal3,
  LEAD(sal, 1,0) OVER (partition BY dept_id ORDER BY hire_date) AS next_sal3,
  LAG(sal, 2, 999) OVER (partition BY dept_id ORDER BY hire_date) AS prev_sal4,
  LEAD(sal, 2,999) OVER (partition BY dept_id ORDER BY hire_date) AS next_sal4
FROM emp;

查询结果：

案例分析

1、排名论次

环境准备：

drop table emp purge;
CREATE TABLE emp
(
emp_id    NUMBER(6),
ename  VARCHAR2(45),
dept_id   NUMBER(4),
hire_date DATE,
sal    NUMBER(8,2)
);
--创建emp数据
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (101, 'Tom',    20,  TO_DATE
('21-09-1989', 'DD-MM-YYYY'), 2000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (102, 'Mike',   20,  TO_DATE
('13-01-1993', 'DD-MM-YYYY'), 8000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (120, 'John',   50,  TO_DATE
('18-07-1996', 'DD-MM-YYYY'), 1000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (121, 'Joy',    50,  TO_DATE
('10-04-1997', 'DD-MM-YYYY'), 1000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (122, 'Rich',   50,  TO_DATE
('01-05-1995', 'DD-MM-YYYY'), 4000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (123, 'Kate',   50,  TO_DATE
('10-10-1997', 'DD-MM-YYYY'), 4000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (124, 'Jess',   50,  TO_DATE
('16-11-1999', 'DD-MM-YYYY'), 7000);
INSERT INTO emp (emp_id, ename, dept_id, hire_date, sal) VALUES (100, 'Stev',   10,  TO_DATE
('01-01-1990', 'DD-MM-YYYY'), 7000);
COMMIT;

SQL> select * from emp;
 
 EMP_ID ENAME DEPT_ID HIRE_DATE          SAL
------- ----- ------- ----------- ----------
    101 Tom        20 1989-09-21     2000.00
    102 Mike       20 1993-01-13     8000.00
    120 John       50 1996-07-18     1000.00
    121 Joy        50 1997-04-10     1000.00
    122 Rich       50 1995-05-01     4000.00
    123 Kate       50 1997-10-10     4000.00
    124 Jess       50 1999-11-16     7000.00
    100 Stev       10 1990-01-01     7000.00

用普通语句查询每个部门内收入最低的人：

WITH t as
(SELECT dept_id, min(sal) as min_sal FROM emp GROUP BY dept_id)
select emp.emp_id, emp.ename, emp.dept_id, emp.hire_date,emp.sal
from emp, t
where emp.dept_id = t.dept_id
and emp.sal = t.min_sal;

查询结果：
EMP_ID ENAME DEPT_ID HIRE_DATE          SAL
------- ----- ------- ----------- ----------
    101 Tom        20 1989-09-21     2000.00
    121 Joy        50 1997-04-10     1000.00
    120 John       50 1996-07-18     1000.00
    100 Stev       10 1990-01-01     7000.00

用dense_rand分析函数查询每个部门内收入最低的人（效率更高）：

SELECT  emp_id, ename, dept_id, hire_date,sal
FROM (SELECT emp.*,
                       dense_rank() OVER(PARTITION BY dept_id ORDER BY sal ) AS  N
              FROM emp)
WHERE N = 1;

查询结果：
EMP_ID ENAME DEPT_ID HIRE_DATE          SAL
------- ----- ------- ----------- ----------
    100 Stev       10 1990-01-01     7000.00
    101 Tom        20 1989-09-21     2000.00
    120 John       50 1996-07-18     1000.00
    121 Joy        50 1997-04-10     1000.00

2、数据去重

环境准备：

DROP TABLE t purge ;
CREATE TABLE t AS SELECT * FROM dba_objects WHERE rownum<=10;
UPDATE t SET object_id=rownum;
UPDATE t SET object_id=3 WHERE object_id<=3;
UPDATE t SET object_id=4 WHERE object_id>=4 AND object_id<=6;
COMMIT;

普通的去重写法（随便删除，保留rowid最大的一条）：

delete from t
where rowid <
         (select max(rowid) 
            from t t2
          where t.object_id = t2.object_id
         );

数据去重的分析函数写法（保留最新的一条记录）：

delete t
 where rowid in (select rid
                   from (select rowid rid,
                                row_number() over(partition by object_id order by created desc) rn
                           from t)
                  where rn > 1);

日常在处理数据去重的时候，如果数据量太大，可以考虑先把不重复的数据移出到另外一个表中，然后只针对重复数据进行处理。

3、占比应用 ratio_to_report

SELECT
  emp_id,ename,dept_id,hire_date,sal,
  ratio_to_report(sal) OVER () as pct1l,
  ratio_to_report(sal) OVER (partition by dept_id) as pct2
FROM emp;

查询结果：

4、连续值判定

环境准备：

drop table t purge;
create table t (id1 int,id2 int ,id3 int);
insert into t (id1 ,id2,id3) values (1,45,89);
insert into t (id1 ,id2,id3) values (2,45,89);
insert into t (id1 ,id2,id3) values (3,45,89);
insert into t (id1 ,id2,id3) values (8,45,89);
insert into t (id1 ,id2,id3) values (12,45,89);
insert into t (id1 ,id2,id3) values (36,45,89);
insert into t (id1 ,id2,id3) values (22,45,89);
insert into t (id1 ,id2,id3) values (23,45,89);
insert into t (id1 ,id2,id3) values (89,45,89);
insert into t (id1 ,id2,id3) values (92,45,89);
insert into t (id1 ,id2,id3) values (91,45,89);
insert into t (id1 ,id2,id3) values (90,45,89);
commit;

SQL> select * from t;
 
 ID1  ID2  ID3
---- ---- ----
   1   45   89
   2   45   89
   3   45   89
   8   45   89
  12   45   89
  36   45   89
  22   45   89
  23   45   89
  89   45   89
  92   45   89
  91   45   89
  90   45   89

需要实现的需求：

1、将ID1连续的数据查找出来，效果如下：

ID1  ID2  ID3
---- ---- ----
   1   45   89
   2   45   89
   3   45   89
  22   45   89
  23   45   89
  89   45   89
  90   45   89
  91   45   89
  92   45   89

2、要求查出连续数据，并且要算出出最小值和最大值及连续的个数，效果如下：

   1     3    3
  22   23   2
  89   92   4

实现方法：

1、先看如下sql的查询结果，使用分析函数ROW_NUMBER已经把连续的数据找出来了，分成3个组。

SELECT id1,
  id2,
  id3,
  ROW_NUMBER() OVER(ORDER BY id1) - ID1 AS group_id
FROM t;

ID1  ID2  ID3   GROUP_ID
---- ---- ---- ----------
   1   45   89          0
   2   45   89          0
   3   45   89          0
   8   45   89         -4
  12   45   89         -7
  22   45   89        -16
  23   45   89        -16
  36   45   89        -28
  89   45   89        -80
  90   45   89        -80
  91   45   89        -80
  92   45   89        -80

2、需求1的实现方法：

select id1, id2, id3
  from (select id1, id2, id3, count(*) over(partition by group_id) cnt
          from (select id1,
                       id2,
                       id3,
                       row_number() over(order by id1) - id1 as group_id
                  from t)
        )
 where cnt > 1
 order by id1;

查询结果：
ID1  ID2  ID3
---- ---- ----
   1   45   89
   2   45   89
   3   45   89
  22   45   89
  23   45   89
  89   45   89
  90   45   89
  91   45   89
  92   45   89

3、需求2的实现方法：

select min(id1), max(id1), count(*)
  from (select id1,
               id2,
               id3,
               row_number() over(order by id1) - id1 as group_id
          from t)
having count(*) > 1
 group by group_id
 order by 1;

查询结果：
MIN(ID1)   MAX(ID1)   COUNT(*)
---------- ---------- ----------
         1          3          3
        22         23          2
        89         92          4

5、高频数获取

原始数据查询结果：

SELECT sal,COUNT(*) repeat_num
FROM emp
GROUP BY sal;

查询结果：
     SAL      REPEAT_NUM
----------       ----------
   1000.00          1
   4000.00          4    --4000这个数值出现了4次，出现频率最高
   2000.00          1
   8000.00          1
   7000.00          1

普通写法查找：

select sal
  from (select sal, count(*) as repeat_num from emp group by sal) t
 where t.repeat_num =
       (select max(repeat_num)
          from (select sal, count(*) as repeat_num from emp group by sal));

查询结果：
     SAL
----------
   4000.00

分析函数查找：

select sal
  from (select sal, rank() over(order by repeat_num desc) rank_repeat_num
          from (select sal, count(*) repeat_num 
                  from emp 
                  group by sal)
                )
 where rank_repeat_num = 1;

查询结果：
      SAL
----------
   4000.00