计算不同数据源Oracle SQL的差异

Question

我正试图根据从两个差异来源获得的数据创建一个方差列表。该数据包含一个日期、一系列引用和包含数字计数等的列。

这背后的想法是检查数据源1的数据是否与数据源2的数据具有相同的数字计数，然后记录每5分钟间隔的方差。

这里我有创建简化场景的表和样本数据所需的代码

Create Table ABP_PROFILE 
(  ABP_DATE            Date          not Null, 
  ABP_SOURCE_UID      Number(10)  not Null, 
  ABP_REFERENCE_1     Varchar2(30)  not Null, 
  ABP_CHARGE          Number(18,6), 
  ABP_COUNT           Number(18)
);
insert into ABP_PROFILE (ABP_DATE, ABP_SOURCE_UID, ABP_REFERENCE_1, ABP_CHARGE, ABP_COUNT)
values (to_date('15-06-2015 00:05:00', 'dd-mm-yyyy hh24:mi:ss'), 1, 'Another Reference', 757.500000, 101);
insert into ABP_PROFILE (ABP_DATE, ABP_SOURCE_UID, ABP_REFERENCE_1, ABP_CHARGE, ABP_COUNT)
values (to_date('15-06-2015 00:05:00', 'dd-mm-yyyy hh24:mi:ss'), 1, 'Some Reference', 2954.000000, 211);
insert into ABP_PROFILE (ABP_DATE, ABP_SOURCE_UID, ABP_REFERENCE_1, ABP_CHARGE, ABP_COUNT)
values (to_date('15-06-2015 00:05:00', 'dd-mm-yyyy hh24:mi:ss'), 2, 'Another Reference', 757.500000, 101);
insert into ABP_PROFILE (ABP_DATE, ABP_SOURCE_UID, ABP_REFERENCE_1, ABP_CHARGE, ABP_COUNT)
values (to_date('15-06-2015 00:05:00', 'dd-mm-yyyy hh24:mi:ss'), 2, 'Some Reference', 2954.000000, 211);
insert into ABP_PROFILE (ABP_DATE, ABP_SOURCE_UID, ABP_REFERENCE_1, ABP_CHARGE, ABP_COUNT)
values (to_date('15-06-2015 00:10:00', 'dd-mm-yyyy hh24:mi:ss'), 1, 'Another Reference', 5300.250000, 191);
insert into ABP_PROFILE (ABP_DATE, ABP_SOURCE_UID, ABP_REFERENCE_1, ABP_CHARGE, ABP_COUNT)
values (to_date('15-06-2015 00:10:00', 'dd-mm-yyyy hh24:mi:ss'), 1, 'Some Reference', 9568.000000, 208);
insert into ABP_PROFILE (ABP_DATE, ABP_SOURCE_UID, ABP_REFERENCE_1, ABP_CHARGE, ABP_COUNT)
values (to_date('15-06-2015 00:10:00', 'dd-mm-yyyy hh24:mi:ss'), 2, 'Another Reference', 5300.250000, 5555);
insert into ABP_PROFILE (ABP_DATE, ABP_SOURCE_UID, ABP_REFERENCE_1, ABP_CHARGE, ABP_COUNT)
values (to_date('15-06-2015 00:10:00', 'dd-mm-yyyy hh24:mi:ss'), 2, 'Some Reference', 1111.000000, 208);

在这里，我创建了一个BASIC SQL查询，以了解我想做什么

With A_DATA As (
  Select  ABP_DATE              As A_DATE,
          ABP_REFERENCE_1       As A_REFERENCE_1, 
          ABP_CHARGE            As A_CHARGE, 
          ABP_COUNT             As A_COUNT
  From    ABP_PROFILE           
  Where   ABP_SOURCE_UID = 1
), B_DATA As (
  Select  ABP_DATE              As B_DATE,
          ABP_REFERENCE_1       As B_REFERENCE_1, 
          ABP_CHARGE            As B_CHARGE, 
          ABP_COUNT             As B_COUNT
  From    ABP_PROFILE           
  Where   ABP_SOURCE_UID = 2
)
Select  A_DATE,
        A_REFERENCE_1,
        B_CHARGE  - A_CHARGE  As ChargeDifference,
        B_COUNT   - A_COUNT   As CountDifference
From    A_DATA,
        B_DATA
Where   A_DATE        = B_DATE
And     A_REFERENCE_1 = B_REFERENCE_1
;

这将根据两个数据源中的日期和引用进行联接。我需要一个更通用的解决方案，如果一方的数据丢失，也需要显示差异，是的，可以使用完整的外部连接，但我想探索其他选项。

我一直在研究分析函数，我相信有一个可以做我想做的事情。我想知道是否有Oracle SQL专家有任何想法可以在这里提供帮助。

仅供参考，我正在运行11gR2 Enterprise

Answer 1

使用FULL JOIN的解决方案似乎更具可读性，但如果您搜索具有函数lag()和lead():的替代方案

with data as (
  select abp_date dt, abp_source_uid id, abp_reference_1 ref, 
      abp_charge charge, abp_count cnt,
      lag(abp_source_uid) over (partition by abp_date, abp_reference_1 
                                order by abp_source_uid) lgid,
      lead(abp_charge)    over (partition by abp_date, abp_reference_1 
                                order by abp_source_uid) ldcharge,
      lead( abp_count)    over (partition by abp_date, abp_reference_1 
                                order by abp_source_uid) ldcnt
    from abp_profile a)
select dt, ref, 
    case when id = 1 then nvl(ldcharge, 0) - charge else charge end chrg_diff,
    case when id = 1 then nvl(ldcnt, 0) - cnt else cnt end cnt_diff
  from data
  where id = 1 or id = 2 and lgid is null;

以及您修改后的查询转换为full join版本，我制作该版本是为了比较结果：

With A_DATA As (
  Select  ABP_DATE              As A_DATE,
          ABP_REFERENCE_1       As A_REFERENCE_1, 
          ABP_CHARGE            As A_CHARGE, 
          ABP_COUNT             As A_COUNT
  From    ABP_PROFILE           
  Where   ABP_SOURCE_UID = 1
), B_DATA As (
  Select  ABP_DATE              As B_DATE,
          ABP_REFERENCE_1       As B_REFERENCE_1, 
          ABP_CHARGE            As B_CHARGE, 
          ABP_COUNT             As B_COUNT
  From    ABP_PROFILE           
  Where   ABP_SOURCE_UID = 2
)
Select  nvl(A_DATE, b_date) dt,
        nvl(A_REFERENCE_1, b_reference_1) ref,
        nvl(B_CHARGE, 0) - nvl(A_CHARGE, 0)  As Chrg_Diff,
        nvl(B_COUNT, 0)   - nvl(A_COUNT, 0)  As Cnt_Diff
From    A_DATA 
full join B_DATA on A_DATE = B_DATE and A_REFERENCE_1 = B_REFERENCE_1;

SQLFiddle

两个查询都得到相同的结果。在示例中，我添加了两行来展示如何处理丢失的数据。在这里，我使用了nvl（…，0），但当然可以保留null或添加列来通知这种情况。