Question

我有下面的问题，我必须在SQL中解决。该数据库在SQL-Server 2005上运行。对于一组（年份和数字），该组应该是唯一的，例如在下面的集合中我应该有4组（1 - 4）而不是组（1-2）。我不被允许更新表格，我无法修复该应用程序。

我可以为我的套装计算4个不同的组，还是有不同的方法来解决问题？

year    |number |row    |group<br>
2004    |1000   |1  |1<br>
2004    |1000   |2  |1<br>
2004    |1000   |3  |1<br>
2004    |1000   |4  |1<br>
2004    |1000   |5  |2<br>
2004    |1000   |6  |2<br>
2004    |1000   |7  |2<br>
2004    |1000   |8  |2<br>
2004    |1000   |9  |1<br>
2004    |1000   |10 |1<br>
2004    |1000   |11 |1<br>
2004    |1000   |12 |1<br>
2004    |1000   |13 |2<br>
2004    |1000   |14 |2<br>
2004    |1000   |15 |2<br>
2004    |1000   |16 |2<br>

Answer 1

如果我理解正确，你可以使用窗口功能。对于此特定问题，您可以对行号使用算术。对于更一般的解决方案，这是一个“间隙和岛屿”问题。在SQL Server 2005中，您可以使用窗口函数来解决此问题：

select t.*,
       dense_rank() over (partition by year, number order by group, seqnum_r - seqnum_rg) as grp
from (select t.*,
             row_number() over (partition by year, number order by row) as seqnum_r,
             row_number() over (partition by year, number, group order by row) as seqnum_rg
      from t
     ) t;

了解其工作原理有点棘手。重要的是，对于具有相似值的“岛”，两个序列号之间的差异是恒定的。 dense_rank()然后只提供最终值。

以上将产生四组，但它们的顺序可能不正确。为此，使用另一轮窗口函数来获得最小id：

select t.*,
       dense_rank() over (partition by year, number order by minid) as grp
from (select t.*, min(id) over (partition by year, number, group, seqnum_r - seqnum_rg) as minid
      from (select t.*,
                   row_number() over (partition by year, number order by row) as seqnum_r,
                   row_number() over (partition by year, number, group order by row) as seqnum_rg
            from t
           ) t
     ) t;

我可以在SQL中为我的集合计算4个不同的组吗？

1 个答案: