如何获取空值的先前值答案

【问题标题】：How to get Previous Value for Null Values如何获取空值的先前值
【发布时间】：2013-05-16 04:44:30
【问题描述】：

我的表中有以下数据。

   | Id  |  FeeModeId   |Name        |   Amount|
   ---------------------------------------------
   | 1   |  NULL        | NULL       |   20    |
   | 2   |  1           | Quarter-1  |   5000  |
   | 3   |  NULL        | NULL       |   2000  |    
   | 4   |  2           | Quarter-2  |   8000  |
   | 5   |  NULL        | NULL       |   5000  |
   | 6   |  NULL        | NULL       |   2000  |
   | 7   |  3           | Quarter-3  |   6000  |
   | 8   |  NULL        | NULL       |   4000  |

如何编写这样的查询以获得以下输出...

   | Id  |  FeeModeId   |Name        |   Amount|
   ---------------------------------------------
   | 1   |  NULL        | NULL       |   20    |
   | 2   |  1           | Quarter-1  |   5000  |
   | 3   |  1           | Quarter-1  |   2000  |    
   | 4   |  2           | Quarter-2  |   8000  |
   | 5   |  2           | Quarter-2  |   5000  |
   | 6   |  2           | Quarter-2  |   2000  |
   | 7   |  3           | Quarter-3  |   6000  |
   | 8   |  3           | Quarter-3  |   4000  |

【问题讨论】：

什么版本的 SQL Server？
谢谢，它很有帮助，但在我的情况下，以前的值差异不是固定的，有些时间 2 记录，有些时间 3，它可以是任何范围。
看起来好像在 Oracle 中这个 could be done 和 LAST_VALUE ... IGNORE NULLS 但 SQL Server 不支持 IGNORE NULLS 选项。
@Randy "...使用 LAG 函数..." 如果您能发布一个示例，那就太好了。

标签： sql sql-server tsql sql-server-2012

【解决方案1】：

由于您使用的是 SQL Server 2012...这里有一个使用它的版本。它可能比其他解决方案更快，但您必须在数据上进行测试。

sum() over() 将执行由Id 排序的运行总和，当列中有值时添加1 并保留null 值的当前值。然后使用计算的运行总和将结果划分为first_value() over()。 Id 为运行总和生成的每个“组”行排序的第一个值具有您想要的值。

select T.Id,
       first_value(T.FeeModeId) 
          over(partition by T.NF 
               order by T.Id 
               rows between unbounded preceding and current row) as FeeModeId,
       first_value(T.Name)      
          over(partition by T.NS 
               order by T.Id 
               rows between unbounded preceding and current row) as Name,
       T.Amount
from (
     select Id,
            FeeModeId,
            Name,
            Amount,
            sum(case when FeeModeId is null then 0 else 1 end) 
              over(order by Id) as NF,
            sum(case when Name is null then 0 else 1 end) 
              over(order by Id) as NS
     from YourTable
     ) as T

SQL Fiddle

在 SQL Server 2012 之前可以使用的东西：

select T1.Id,
       T3.FeeModeId,
       T2.Name,
       T1.Amount
from YourTable as T1
  outer apply (select top(1) Name
               from YourTable as T2
               where T1.Id >= T2.Id and
                     T2.Name is not null
               order by T2.Id desc) as T2
  outer apply (select top(1) FeeModeId
               from YourTable as T3
               where T1.Id >= T3.Id and
                     T3.FeeModeId is not null
               order by T3.Id desc) as T3

SQL Fiddle

【讨论】：

【解决方案2】：

请尝试：

select 
    a.ID,
    ISNULL(a.FeeModeId, x.FeeModeId) FeeModeId,
    ISNULL(a.Name, x.Name) Name,
    a.Amount
from tbl a
outer apply
(select top 1 FeeModeId, Name 
    from tbl b 
    where b.ID<a.ID and 
        b.Amount is not null and 
        b.FeeModeId is not null and 
        a.FeeModeId is null order by ID desc)x

或

select 
    ID,
    ISNULL(FeeModeId, bFeeModeId) FeeModeId,
    ISNULL(Name, bName) Name,
    Amount
From(
    select 
        a.ID , a.FeeModeId, a.Name, a.Amount, 
        b.ID bID, b.FeeModeId bFeeModeId, b.Name bName,
        MAX(b.FeeModeId) over (partition by a.ID) mx
    from tbl a left join tbl b on b.ID<a.ID
    and b.FeeModeId is not null
)x 
where bFeeModeId=mx or mx is null

【讨论】：

【解决方案3】：

试试这个 -

SELECT Id,
       CASE
         WHEN Feemodeid IS NOT NULL THEN
          Feemodeid
         ELSE
          (SELECT Feemodeid
             FROM Table_Name t_2
            WHERE t_2.Id = (SELECT MAX(Id)
                              FROM Table_Name t_3
                             WHERE t_3.Id < t_1.Id
                               AND Feemodeid IS NOT NULL))
        END Feemodeid,
       CASE
         WHEN NAME IS NOT NULL THEN
          NAME
         ELSE
          (SELECT NAME
             FROM Table_Name t_2
            WHERE t_2.Id = (SELECT MAX(Id)
                              FROM Table_Name t_3
                             WHERE t_3.Id < t_1.Id
                               AND NAME IS NOT NULL))
       END NAME,
       Amount
  FROM Table_Name t_1

【讨论】：

【解决方案4】：

SELECT
  T.ID,
  ISNULL(T.FeeModeId, 
    (SELECT TOP 1 FeeModeId 
        FROM TableName AS T1 
        WHERE ID < T.ID AND FeeModeId IS NOT NULL 
        ORDER BY ID DESC)) AS FeeModeId,
  ISNULL(Name,
    (SELECT TOP 1 Name 
    FROM TableName 
    WHERE ID < T.ID AND Name IS NOT NULL 
    ORDER BY ID DESC)) AS Name,
  T.Amount
FROM
  TableName AS T

【讨论】：

【解决方案5】：

id  name
1   toto
2   NULL
3   NULL
4   titi
5   NULL
6   NULL
7   tutu
8   NULL
9   NULL


SELECT 
       id_table
       ,name
FROM 
       (
             SELECT
                    T_01.id AS 'id_table'
                    ,max(T_02.id) AS 'id_name'
             FROM
                    names AS T_01
                    cross join  
                    (
                          SELECT
                                 id
                                 ,name
                          FROM
                                 names
                          WHERE
                                 name IS NOT NULL
                    ) AS T_02 
             WHERE 
                    T_02.id <= T_01.id
             GROUP BY 
                    T_01.id
       ) AS tt02
       left join names
       ON names.id = tt02.id_name


id_table    name
1           toto
2           toto
3           toto
4           titi
5           titi
6           titi
7           tutu
8           tutu
9           tutu

【讨论】：