如何從充滿重復項的表中選擇最新記錄

Question

我有下表，需要查詢才能產生預期的結果。

**Table**
  Address Num  My_Date
 Address1 7777 03/NOV/15
 Address2 2222 02/NOV/15
 Address2 3333 02/NOV/15
 Address2 2222 05/NOV/15
 Address2 3333 05/NOV/15
 Address3 8888 01/NOV/15
 Address4 9999 04/NOV/15

預期結果

Address  Num  My_Date
Address1 7777 03/NOV/15
Address2 2222 05/NOV/15
Address2 3333 05/NOV/15
Address3 8888 01/NOV/15
Address4 9999 04/NOV/15

如您所見，我需要帶回所有記錄，如果有重復的Num，則僅帶入具有最新My_Date的記錄。

Answer 1

進行GROUP BY ：

select address, num, max(date)
from tablename
group by address, num

或NOT EXISTS ：

select *
from tablename t1
where not exists (select 1 from tablename t2
                  where t2.address = t1.address
                    and t2.num = t1.num
                    and t2.date > t1.date)

在ANSI SQL中，date是保留字，因此您可能需要用雙引號將其作為分隔標識符，即"date" 。

Answer 2

您還可以使用下面的查詢獲取所需的o / p

select "Address","Num","Date"
from
(
select "Address","Num","Date",rank() over(partition by "Address" order by "Date" desc) r_no from table_name
)
where r_no=1