Question

我试图运行一个select语句来对具有相似ID的记录进行分组，但也为每个主ID的另一列计算值。所以例如下面。每行的结果将是第一个实例唯一ID，并且每个记录显示的2个名称由分号分隔。提前谢谢。

当前设置

ID                               Name              Cnt
-------------------------------- ----------------- ---
0001D72BA5F664BE129B6AB5744E2BE0 Talati, Shilpa    1
0001D72BA5F664BE129B6AB5744E2BE0 Weaver, Larry     1
0007EAB7CE9A3F2F95D2D63D0BBD08A9 St-Hilaire, Edith 1
0007EAB7CE9A3F2F95D2D63D0BBD08A9 Talati, Shilpa    1

结果：

0001D72BA5F664BE129B6AB5744E2BE0 Talati, Shilpa; Weaver, Larry

Answer 1

在SQL Server中解决此问题的最简单方法是：

select masterId, min(name) + '; ' + max(name)
from table t
group by masterId;

Answer 2

这是使用递归公用表表达式的一种方式。给出这样一个表：

create table dbo.Fizzbuzz
(
  id       int         not null identity(1,1) primary key clustered ,
  group_id int         not null ,
  name     varchar(50) not null ,
  cnt      int         not null ,
)

包含此数据

id group_id name   cnt
-- -------- ------ ---
1  1        Bob      3
2  1        Carol    5
3  1        Ted      6
4  1        Alice   16
5  2        Harold  72
6  2        Maude   28

此查询

with recursive_cte as
(
  select group_id = t.group_id ,
         row      = t.row      ,
         name     = convert(varchar(8000),t.name) ,
         cnt      = t.cnt
  from ( select * ,
                row = row_number() over (
                        partition by group_id
                        order by id
                        )
         from dbo.Fizzbuzz
       ) t
  where t.row = 1
  UNION ALL
  select group_id = prv.group_id ,
         row      = nxt.row      ,
         name     = convert(varchar(8000), prv.name + ' and ' + nxt.name ) ,
         cnt      = prv.cnt  + nxt.cnt
  from recursive_cte prv
  join ( select * ,
                row = row_number() over (
                        partition by group_id
                        order by id
                        )
         from dbo.Fizzbuzz
       ) nxt on nxt.group_id = prv.group_id
            and nxt.row      = prv.row + 1
)
select group_id = t.group_id ,
       total    = t.cnt      ,
       names    = t.name
from ( select * ,
              rank = rank() over (
                       partition by group_id
                       order by row desc
                     )
       from recursive_cte
     ) t
where rank = 1
order by group_id

产生以下输出

group_id cnt name
-------- --- -------------------------------
    1     30 Bob and Carol and Ted and Alice
    2    100 Harold and Maude

但是应该注意，递归的深度在SQL Server中是有限的。

Answer 3

SELECT  
      t1.ID, 
      (SELECT Name + '; ' 
       FROM yourtable t2 
       WHERE t1.ID = t2.ID 
       for xml path('')) as Name
FROM yourtable t1 
GROUP BY t1.ID

SQL中每条记录的分组和计数值

3 个答案: