ere*_*ale 11 sql gaps-and-islands
我有如下数据集
name  date  
x     2014-01-01
x     2014-01-02
y     2014-01-03
x     2014-01-04
而我正试图得到这个结果
name  date           row_num
x     2014-01-01      1
x     2014-01-02      2
y     2014-01-03      1
x     2014-01-04      1
我试图运行此查询
select name,
       date,
       row_number () over (partition by name order by date) as row_num
from myTBL
但不幸的是我得到了这个结果
name  date           row_num
x     2014-01-01      1
x     2014-01-02      2
y     2014-01-03      1
x     2014-01-04      3
请帮忙.
Gor*_*off 20
您需要识别出names一起出现的组.您可以使用行号的不同来执行此操作.然后,使用grpfor分区row_number():
select name, date,
       row_number() over (partition by name, grp order by date) as row_num
from (select t.*,
             (row_number() over (order by date) -
              row_number() over (partition by name order by date)
             ) as grp
      from myTBL t
     ) t
对于您的样本数据:
name  date         1st row_number   2nd      Grp
x     2014-01-01         1           1        0
x     2014-01-02         2           2        0
y     2014-01-03         3           1        2
x     2014-01-04         4           3        1
这应该让你知道它是如何工作的.
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