Question

我在表格中有这样的数据：

column1   column2
a         1
a         2
b         2
b         3
a         4
c         5

我想要这样的输出：

column1 column2
a       1-2
b       2-3
a       4-0
c       5-0

Answer 1

尝试此查询：

with vw1 as 
(select table1.*,rownum rn from table1),
vw2 as (select col1,col2,rn,rn - col2 dis from vw1),
vw3 as (select col1,min(rn),to_char(min(col2))||' - '||
case when min(col2) = max(col2) then '0' else to_char(max(col2)) end col2 from vw2
group by col1,dis order by min(rn))
select col1,col2 from vw3;

SQL Fiddle

Answer 2

你还没有给出很多信息 - 对于column1中的每个值，第2列中是否始终只有1或2个值？
- column2总是按顺序排列吗？ - 等等？

但是，对于给定的数据，以下内容应该给出您要求的结果。

SELECT
  column1,
  MIN(column2)                                          AS first_column2,
  CASE WHEN COUNT(*) = 1 THEN 0 ELSE MAX(column2) END   AS final_column2
FROM
(
  SELECT
    ROW_NUMBER() OVER (                     ORDER BY column2, column1)  AS sequence_main,
    ROW_NUMBER() OVER (PARTITION BY column1 ORDER BY column2         )  AS sequence_c1,
    *
  FROM
    your_table
)
  AS sequenced_table
GROUP BY
  sequence_main - sequence_c1,
  column1
ORDER BY
  MIN(sequence_main)

示例计算：

column1   column2   |  sequence_main  sequence_c1  main - c1  |  group
a         1         |  1              1            0          |  a1
a         2         |  2              2            0          |  a1
b         2         |  3              1            2          |  b2
b         3         |  4              2            2          |  b2
a         4         |  5              3            2          |  a2
c         5         |  6              1            5          |  c5

Answer 3

尝试此查询...

select * from 
(
select col1 as column1,case when LEAD(col1 , 1, 0) OVER (ORDER BY col2) = col1
then concat( LEAD(col2 , 1, 0) OVER (ORDER BY col2),'-'||col2)   
else (case when lag(col1,1,0) over (ORDER BY col2) <> col1 then 
concat(col2,'-'||'0')else '0' end)
end as column2
from table
order by col2
)
where column2<>'0'
;

Answer 4

你应该小心使用像

这样的东西

 select column1, column2, rownum from table1

为您的数据创建一些唯一的ID。每个定义是在SQL 未定义中没有订单的排序。因此巧合的是

select * from table1

以您在数据库中插入的顺序返回行。随着数据的增长，您将获得此排序的例外情况。因此建议高度将主键列放在数据表中以保留此插入顺序。我为此添加了id列。

使用此pimped数据集，您可以使用此查询获取请求的数据：

with data_aggr as (
select column1,
case 
  when lead(column1,1,' ') over (order by id)<>column1
  and lag(column1,1,' ') over (order by id)<>column1
  then column2 || '-0'
  when lead(column1,1,' ') over (order by id)=column1
  and lag(column1,1,' ') over (order by id)<>column1
  then column2 || '-' || lead(column2,1) over (order by id)
  else null
end aggr_col2
from table1)
select column1, aggr_col2 from data_aggr where not aggr_col2 is null

http://sqlfiddle.com/#!4/cd24d/19

如何查询才能找到这个？

4 个答案:

SQL Fiddle