如果值存在且带有日期，SQL 获取 1

Question

我有两张桌子 表1

个人号码	其他值
1
2

表2

个人	日期
1	2023-09-1
2	2023-09-1
2	2023-09-2

我想要一个选择，为我提供一个表，如果当天有条目存在，则设置 1，如果不存在，则设置 0 预计会是

个人号码	2023-09-01	2023-09-02
1	1	0
2	1	1

我尝试了以下方法，但得到的 1 和 0 不正确

    SELECT t1.personalnr,
      case when EXISTS (  select* from table2 
        where date between '2023-01-09' and '2023-02-09' and t1.personalnr = t2.personalnr) then '1' else '0'
      end
case when EXISTS (  select* from table2 
    where date between '2023-02-09' and '2023-03-09' and t1.personalnr = t2.personalnr) then '1' else '0'
  end
    FROM [BKLatWORK].[dbo].[Table1] t1
    left join table2 t2 on t2.PersonalNr = t1.PersonalNr
    group by t1.PersonalNr, t2.personalnr

Answer 1

如果您不寻找动态解决方案，可以使用条件聚合：

select t1.Personalnr, 
       count(case when t2.Date = '2023-09-01' then 1 end) as '2023-09-01',
       count(case when t2.Date = '2023-09-02' then 1 end) as '2023-09-02'
from table1 t1
inner join table2 t2 on t1.Personalnr = t2.Personalnr
group by t1.Personalnr