Question

我的查询输出如下：

ID      customer_name_Now    customer_name_Before       MOVEMENT
123451  Rustle Bock ltd      N                          £2,121
123451  N                    Rustle Bock ltd           -£25,666,899
123452  Little Garage Ltd    N                          £6,987
123453  N                    The Big Shop               £15,850

故事是，我有2个月的数据。在这两个月中，客户可能会或可能不会根据上个月的客户情况或现在是客户的情况进行调整。很多情况下它在两个月都是客户，因此我得到了如上所述的2行。

理想的输出应该是：

ID      customer_name_Now   customer_name_Before      MOVEMENT
123451  Rustle Bock ltd     Rustle Bock ltd          -£25,664,778
123452  Little Garage Ltd   N                         £6,987
123453  N                   The Big Shop              £15,850

因此，假设客户在两个月内都有关系，那么移动应该总和给我几个月的实际移动，客户的名字应该在两列中。

请你帮我解决这个问题，非常紧急。

@DMK我用来得到初始输出的查询是：

select /*+ NO_REWRITE */
customer_id,
customer_name_now,
customer_name_before,
movement

from
    (select /*+ NO_REWRITE */
    main.customer_id,
    main.customer_name_now,
    main.customer_name_before,
    main.limits_before,
    main.limits_now,
    sum(main.limits_now-main.limits_before) as movement

    from
        (select /*+ NO_REWRITE */
        customer_id,
        (customer_name_before) as customer_name_before,
        (customer_name_now) as customer_name_now,
        sum(limits_current) as limits_now,
        sum(limits_previous) as limits_before

        from
             (select /*+ NO_REWRITE */
             sub.customer_id,
             sub.customer_name_now,
             sub.customer_name_before,
             sub.limits_current,
             sub.limits_previous
             from
                 (select /*+ NO_REWRITE */
                 T2.customer_ID,
                 (T2.customer_name) customer_name_now,
                 'N' customer_name_before,

                 sum(T26.AGREED_LIMIT) limits_current,
                 0 limits_previous
                 from 

                 DWH_customer_HISTORY T2,
                 DWH_TIME_DIM T25,
                 DWH_FACILITY_MONTHLY T2
                 where  
                 ---some internal filters are applied here, i habe ot shown coz of security reasons----
                 and 
                 T25.MONTH_END = '2012-11-30' and 

                 group by 
                 T2.customer_ID,
                 T2.customer_name,
                 ) sub

             union all

             select /*+ NO_REWRITE */
             sub.customer_id,
             sub.customer_name_now,
             sub.customer_name_before,
             sub.limits_current,
             sub.limits_previous
             from
                 (select /*+ NO_REWRITE */
                 T2.customer_ID,
                 'N' as customer_name_now,
                 (T2.customer_name)customer_name_before,

                 0 limits_current,
                 sum(T2.AGREED_LIMIT) limits_previous,

                from 
                DWH_customer_HISTORY T2,
                DWH_TIME_DIM T25,
                DWH_FACILITY_MONTHLY T2
                where  
                ---some internal filters are applied here, i habe ot shown coz of security reasons----
                and 
                T25.MONTH_END = '2012-10-31'
                group by 
                T2.customer_ID,
                T2.customer_name,) sub
            ) un
        group by
        customer_id,
        customer_name_now,
        customer_name_before,) main

    group by 
    main.customer_id,
    main.customer_name_now,
    main.customer_name_before,
    main.limits_before,
    main.limits_now)

Answer 1

我假设你使用的是SqlServer，不过下面的查询也适用于MySql。

Select c1.ID, c1.customer_name_Now, c2.customer_name_Before, Total
from Customers c1
left Join Customers c2 
on c2.ID = c1.ID
left join
    (select ID as ID2, sum(MOVEMENT) as Total, count(*) as Cnt
    from Customers 
    group by ID) t1
on ID2 = c1.ID
where (c1.customer_name_Now <> 'N' and c2.customer_name_Before <> 'N') 
or CNT = 1

如果您不确定，请查看以下演示

SqlFiddle

查看您刚刚添加的查询后，上面应该仍然有效。你需要

用查询替换我的表Customers
或者将您的查询结果移至临时表，并将我的表Customers替换为临时表

我会选择第二个。保存重新运行相同的查询。

Answer 2

在SQLServer2005 +中使用OVER子句

SELECT DISTINCT 
  ID,MAX(customer_name_Now) OVER (PARTITION BY ID) AS customer_name_Now, 
  MAX(customer_name_Before) OVER (PARTITION BY ID) AS customer_name_Before,
  SUM(MOVEMENT) OVER (PARTITION BY ID) AS MOVEMENT 
FROM your_table

OR

SELECT ID, MAX(customer_name_Now) AS customer_name_Now,
       MAX(customer_name_Before) AS customer_name_Before,
       SUM(MOVEMENT) AS MOVEMENT
FROM your_table
GROUP BY ID

SQLFiddle上的演示

组合2个或更多行具有字符和其他具有货币的行

2 个答案:

SqlFiddle