SELECT和用於具有2列

Question

我的表

ID Name Addr tDate 
------------------------------- 
| 1 | Aa | street | 20151231 
| 2 | Aa | street | 20130202 
| 2 | Aa | street | 20120101 
| 3 | Aa | way | 20150821 
| 4 | Bb | street | 20150821 
| 7 | Xb | street | 20150821 
| 5 | Cc | way | 20150821 
| 5 | Cc | way | 20150821 
| 6 | Cc | no way | 20150821 

結果相同的值的唯一ID創建新的ID 01

ID Name Addr | tDate 
------------------------------- 
| 1 | Aa | street | 20151231 
| 2 | Aa | street | 20130202 
| 2 | Aa | street | 20120101 

將創建一個新的nID 應當原樣複製或合併如果名稱和地址相同並且選擇了ID和最新的tDate

結果02

ID Name Addr  tDate  nID 
------------------------------------ 
| 1 | Aa | street | 20151231 | 1 
| 2 | Aa | street | 20120101 | 1 <-- nID != ID 
| 2 | Aa | street | 20130202 | 1 <-- nID != ID 
| 3 | Aa | way | 20150821 | 3 
| 4 | Bb | street | 20150821 | 4 
| 7 | Xb | street | 20150821 | 7 
| 5 | Cc | way | 20150821 | 5 
| 5 | Cc | way | 20150821 | 5 
| 6 | Cc | no way | 20150821 | 6 

我試過這個。不知道它是否正確。

SELECT DISTINCT dr.* 
       FROM MyTable dr 
       inner join(
        SELECT ID, Name, Addr 
        FROM MyTable 
        GROUP BY ID, Name, Addr 
       ) ss on dr.Name = ss.Name and 
         dr.Addr = ss.Addr and 
         dr.ID <> ss.ID 
       order by Name 

Answer 1

編輯：你除了tDate後徹底的改變，以及需要兩個結果集

結果集1：

SELECT 
    id, Name, Addr, tDate 
FROM 
(
    SELECT 
     *, 
     COUNT(*) OVER (PARTITION BY Name, Addr) AS occurrences 
    FROM 
     MyTable 
) 
    AS parsed 
WHERE 
    occurrences > 1 

結果集2：

SELECT 
    *, 
    FIRST_VALUE(ID) OVER (PARTITION BY Name, Addr 
           ORDER BY tDate DESC 
           ROWS UNBOUNDED PRECEDING) AS nID 
FROM 
    MyTable 
ORDER BY 
    ID 

例子：http://sqlfiddle.com/#!6/9285ae/9

Answer 2

的dense_rank窗函數應該做的伎倆：

SELECT ID, Name, Addr 
     DENSE_RANK() OVER (PARTITION BY Name, Addr ORDER BY ID) AS nID 
FROM  mytable 
ORDER BY 1, 4 

Answer 3

您應該使用RANK不分區

SELECT ID, Name, Addr, 
     RANK() OVER (ORDER BY Id,name,addr) AS nID 
FROM  table