Question

我在下面有一些示例记录。我有兴趣统计ID记录中只有X值的地方。例如，我需要计算ID = 10720和11120，因为两者都只有X值。如果它们具有X和其他类似XY或AB的值，例如ID = 10586，我将不计算在内。

当前，我正在SQL和电子表格中使用这两个步骤来获取我的结果，但效率不高。

ID  LocationCode        Dates
10720   VA  X       3/20/2012 1:00:00 PM
10586   DC  X       7/12/2003 7:00:00 AM
10586   DC  X       8/2/2003 4:44:25 AM
10586   DC  XY      2/21/2019 8:00:00 AM
10892   NY  X       5/3/2009 4:00:00 PM
10892   NY  X       5/5/2009 6:30:00 AM
10892   NY  X       5/7/2009 11:45:00 AM
10892   CA  AB      4/5/2016 8:40:00 AM
10932   CA  AB      8/3/2009 4:00:00 AM
10932   CA  AB      8/11/2009 5:30:00 PM
10932   CA  X       5/8/2010 4:00:00 AM
11120   FL  X       11/25/2010 10:27:00 AM
11120   FL  X       12/8/2010 9:02:00 PM
11120   FL  X       12/28/2010 10:30:00 AM

步骤1：

select 
  location,
  string_agg(code, '') as CODE,
  count(distinct ID) as count
from TEMP
group by location

步骤2：将结果存入电子表格并过滤出唯一的X值。

SQL Server中是否可以直接产生结果？

Answer 1

只需：

select count(distinct id)
from t
where not exists (select 1 from t t2 where t2.id = t.id and t2.code <> 'X');

或者，如果愿意，可以分为两个层次：

select count(*)
from (select id
      from t
      group by id
      having min(code) = max(code) and min(code) = 'X'
     ) x;

Answer 2

另一种方式：

select count(*)  
from     
(
select id table where code = 'X'    
EXCEPT
select id from where code != 'X'    
) tmp

Answer 3

一种可能的解决方案：

select ID 
from TEMP
where Code = 'X'
group by ID
having count(distinct Code) = 1

Answer 4

对结果求和将得出'X'的总数。

类似

    select code, sum(count) as count 
    from (
        select 
        location,
        string_agg(code, '') as CODE,
        count(distinct ID) as count
        from TEMP
        group by location
    ) codes_aggregated
    group by code
    having code = 'X'

在SQL中填充唯一记录

4 个答案: