Question

我有下表，我想生成下面描述的数据“想要”我尝试了几个SQL分析函数（rank（）以上），但是我似乎总是遇到障碍。对于解决此任务的任何见解，我将不胜感激

CREATE TABLE TABLEA (
id VARCHAR(2) ,
val1  VARCHAR(2),
val2 VARCHAR(2),
val3 VARCHAR(2),
dt_val VARCHAR(8)
)
;

-- --data
INSERT INTO TABLEA
(id, val1, val2, val3,  dt_val)
VALUES
('1', '2', '3', '4', '20151011'),
('1', '2', '', '4', '20151012'),
('1', '2', '3', '4', '20151013'),
('2', '4', '3', '4', '20151101'),
('2', '4', '3', '4', '20151102'),
('2', '4', '', '', '20151103'),
('2', '4', '3', '4', '20151104'),
('3', '4', '3', '4', '20151110'),
('4', '4', '3', '4', '20151110'),
('4', '4', '3', '4', '20151111'),
('4', '4', '', '4', '20151112'),
('4', '4', '', '4', '20151113'),
('5', '4', '3', '4', '20151111'),
('5', '4', '3', '4', '20151112'),
('5', '4', '3', '4', '20151113'),
('5', '4', '3', '4', '20151114'),
;

想要我想要以下结果。如果连续2行相同，我想选择最少的（dt_val）。

    Id     val1   val2   val3    dt_val 
  ------ ------ ------ ------    ----------------
    1      2        3         4         20151011
    1      2       (null)     4         20151012
    1      2       3          4         20151013
    2      4       3          4         20151101
    2      4       (null)   (null)      20151103
    2      4       3          4         20151104
    3      4       3          4         20151110
    4      4       3          4         20151110
    4      4       (null)     4         20151112
    5      4       3          4         20151111

Answer 1

此查询给了我想要的行：

select id, val1, val2, val3, min(dt_val) min_dt
  from (select t.*, 
               dt_val - row_number() over (partition by val1, val2, val3 order by dt_val) diff 
          from tablea t)
  group by id, val1, val2, val3, diff
  order by id, min(dt_val)

^demo

Answer 2

您可以使用Tabibitosan根据连续的行为每个ID获取组：

select a.*,
  row_number() over (partition by id order by dt_val)
    - row_number() over (partition by id, val1, val2, val3 order by dt_val) as grp
from tablea a
order by id, dt_val;

，然后应用聚合函数：

select id, val1, val2, val3, min(dt_val) as dt_val
from (
  select a.*,
    row_number() over (partition by id order by dt_val)
      - row_number() over (partition by id, val1, val2, val3 order by dt_val) as grp
  from tablea a
)
group by id, val1, val2, val3, grp
order by id, dt_val;

ID VAL1 VAL2 VAL3 DT_VAL  
-- ---- ---- ---- --------
1  2    3    4    20151011
1  2         4    20151012
1  2    3    4    20151013
2  4    3    4    20151101
2  4              20151103
2  4    3    4    20151104
3  4    3    4    20151110
4  4    3    4    20151110
4  4         4    20151112
5  4    3    4    20151111

db<>fiddle demo。

（我想与Ponder的想法基本相同）

如果它们实际上是日期而不是字符串-db<>fiddle，这也将起作用。（由于日期运算，Ponder也会如此！）

Answer 3

-- Oracle 12c+
with s (id, val1, val2, val3,  dt_val) as (
select 1, '2', '3', '4', '20151011' from dual union all
select 1, '2', '' , '4', '20151012' from dual union all
select 1, '2', '3', '4', '20151013' from dual union all
select 2, '4', '3', '4', '20151101' from dual union all
select 2, '4', '3', '4', '20151102' from dual union all
select 2, '4', '' , '' , '20151103' from dual union all
select 2, '4', '3', '4', '20151104' from dual union all
select 3, '4', '3', '4', '20151110' from dual union all
select 4, '4', '3', '4', '20151110' from dual union all
select 4, '4', '3', '4', '20151111' from dual union all
select 4, '4', '' , '4', '20151112' from dual union all
select 4, '4', '' , '4', '20151113' from dual union all
select 5, '4', '3', '4', '20151111' from dual union all
select 5, '4', '3', '4', '20151112' from dual union all
select 5, '4', '3', '4', '20151113' from dual union all
select 5, '4', '3', '4', '20151114' from dual)
select *
from s
match_recognize (
order by id
measures 
  v.id            as id, 
  v.val1          as val1, 
  v.val2          as val2, 
  v.val3          as val3,
  first(v.dt_val) as dt_val 
pattern (v+)
define v as 
     decode(v.id  , first(id  ), 0) = 0 and
     decode(v.val1, first(val1), 0) = 0 and
     decode(v.val2, first(val2), 0) = 0 and
     decode(v.val3, first(val3), 0) = 0
);

 ID VAL1 VAL2 VAL3 DT_VAL  
--- ---- ---- ---- --------
  1 2    3    4    20151011
  1 2         4    20151012
  1 2    3    4    20151013
  2 4    3    4    20151101
  2 4              20151103
  2 4    3    4    20151104
  3 4    3    4    20151110
  4 4    3    4    20151110
  4 4         4    20151112
  5 4    3    4    20151111

    10 rows selected.

仅选择组中2个连续行的最小唯一值

3 个答案: