Question

答案太近了，谢谢但问题是如果要输入太多记录

           |    id   |    name   |    age    |   Tel
------------------------------------------
      1    |    1   |   Frank   |    40     |   null
      2    |    1   |   null    |    50     |   7834xx
      3    |    1   |   Alex    |    null   |   null
      4    |    1   |   null    |    20     |   null
      5    |    2   |   James   |    null   |   4121xx

查询返回最大值像：

           |    id   |   name    |    age    |   Tel
------------------------------------------
      1    |    1    |   Frank   |    50     |  7834xx

我需要像这样选择查询：

           |    id   |   name    |    age    |   Tel
------------------------------------------
      1    |    1    |   Alex    |    20     |  7834xx

我该怎么办？ PLZ？

Answer 1

一种简单的方法是获得以下最大值：

 Select Id, Max(name) as [Name], Max(age) as Age, Max(Tel) as Tel
 from yourtable
 Group by Id

Answer 2

这是一种结合3列的最后一个非空值的迂回方式：

-- Using a table variable for test data
declare @Test table (tableId int identity(1,1), id int, name varchar(100), age int, tel varchar(30));
insert into @Test (id, name, age, tel) values
(1,'Frank',40,null),
(1,null,50,'7834xx'),
(1,'Alex',null,null),
(1,null,20,null),
(2,'James',null,'4121xx');

select n.id, n.name, a.age, t.tel
from (
    select top(1) with ties id, name 
    from @Test 
    where name is not null
    order by row_number() over (partition by id order by tableId desc)
) n
inner join (
    select top(1) with ties id, age 
    from @Test 
    where age is not null
    order by row_number() over (partition by id order by tableId desc)
) a on (n.id = a.id)
inner join (
    select top(1) with ties id, tel 
    from @Test 
    where tel is not null
    order by row_number() over (partition by id order by tableId desc)
) t on (n.id = t.id);

或重新使用CTE

;with CTE AS (
    select * ,
    row_number() over (partition by id, iif(name is not null,1,0) order by tableId desc) as rn_name,
    row_number() over (partition by id, iif(age is not null,1,0) order by tableId desc) as rn_age,
    row_number() over (partition by id, iif(tel is not null,1,0) order by tableId desc) as rn_tel
    from @Test
 )
select n.id, n.name, a.age, t.tel
from CTE n
join CTE a on (a.id = n.id and a.age is not null and a.rn_age = 1)
join CTE t on (t.id = n.id and t.tel is not null and t.rn_tel = 1)
where (n.name is not null and n.rn_name = 1);

结果：

╔════╦══════╦═════╦════════╗
║ id ║ name ║ age ║  tel   ║
╠════╬══════╬═════╬════════╣
║  1 ║ Alex ║  20 ║ 7834xx ║
╚════╩══════╩═════╩════════╝

一年多以后看了这个答案您也可以使用窗口函数first_value。
不使用任何连接。

select Id, name, age, tel
from
(
    select Id
    , row_number() over (partition by id order by tableId desc) as rn
    , first_value(name) over (partition by id order by iif(name is null,1,0), tableId desc) as name
    , first_value(age) over (partition by id order by iif(age is null,1,0), tableId desc) as age
    , first_value(tel) over (partition by id order by iif(tel is null,1,0), tableId desc) as tel
    from @Test
) q
where rn = 1
and name is not null and age is not null and tel is not null;

Answer 3

感谢您的回答和帮助。我找到了这个问题：

选择TOP 10 （SELECT top（1）name FROM test1，其中id = 1，name不是autoID desc的null order）作为名称，（选择top（1）age FROM test1，其中id = 1，age不是null命令，autoID desc）作为Age ，（选择top（1）Tel FROM test1，其中id = 1，Tel不是autoID desc的null顺序）作为Telephon 来自[dbo]。[test1] 按ID分组

它的工作!!! 但我还有另一种简单方法，也许就是这样：

选择Id，NotNull（name）作为[Name]，NotNull（age）作为Age，NotNull（Tel）作为Tel 来自Idrtable Group by Id

???

SQL Server中的两行合二为一

3 个答案: