如何将 GroupID 列添加到 SQL 中的查询?
How do you add a GroupID Column to a Query in SQL?
我试图即时创建 GroupID 列。下面是我需要的样子。如何动态添加基于 ParentJob 的 GroupID?
ParentJob ParentPart ParentDescription ParentCompleteDate GroupID
00000111 360120 4/1 Gal-Egg Food Color 2002-01-13 1
00000111 360120 4/1 Gal-Egg Food Color 2002-01-13 1
00000111 360120 4/1 Gal-Egg Food Color 2002-01-13 1
00000111 360120 4/1 Gal-Egg Food Color 2002-01-13 1
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000415 360120 4/1 Gal-Egg Food Color 2002-01-17 3
00000415 360120 4/1 Gal-Egg Food Color 2002-01-17 3
00000415 360120 4/1 Gal-Egg Food Color 2002-01-17 3
00000415 360120 4/1 Gal-Egg Food Color 2002-01-17 3
select ..., dense_rank() over (order by ParentJob) as GroupId
from ....
我试图即时创建 GroupID 列。下面是我需要的样子。如何动态添加基于 ParentJob 的 GroupID?
ParentJob ParentPart ParentDescription ParentCompleteDate GroupID
00000111 360120 4/1 Gal-Egg Food Color 2002-01-13 1
00000111 360120 4/1 Gal-Egg Food Color 2002-01-13 1
00000111 360120 4/1 Gal-Egg Food Color 2002-01-13 1
00000111 360120 4/1 Gal-Egg Food Color 2002-01-13 1
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000155 360120 4/1 Gal-Egg Food Color 2002-01-11 2
00000415 360120 4/1 Gal-Egg Food Color 2002-01-17 3
00000415 360120 4/1 Gal-Egg Food Color 2002-01-17 3
00000415 360120 4/1 Gal-Egg Food Color 2002-01-17 3
00000415 360120 4/1 Gal-Egg Food Color 2002-01-17 3
select ..., dense_rank() over (order by ParentJob) as GroupId
from ....