3

这是类似于：TSQL select rows by one from 2 conditions 的问题，但我想要的结果不同

我有一张这样的桌子：

ORDER_ID   CODE1   CODE2   CODE3   STATUS    TYPE        SUM      GROUP
1          '001'   'BIGP'  NULL    4         'company'   120      48
2          '002'   'BIGP'  NULL    1         'priv'      100      20
3          '001'    NULL   NULL    6         'priv'      50       49
4          '002'    NULL   'L'     1         'company'   1253     22

第二张桌子像这样：

ADDRESS_ID   ORDER_ID   ZIP       TYPE   ADD_DATE       CATEGORY     VERIFIED
1            1          '15-125'    'K1'   '2010-01-01'   'CLIENT'     1
2            1          '22-022'    'D1'   '2010-01-02'   'SYSTEM'     1
3            2          '16-159'    'D2'   '2010-01-02'   'SYSTEM'     1
4            2          '15-125'    'D2'   '2010-02-01'   'CLIENT'     0

第三和第四个表包含邮政编码和城市名称，如下所示：

ZIP       CITY
'15-125'    'Warszawa'
'22-022'    'Koszalin'
'16-159'    'Krakow'
'15-125'    'Lublin'

对于每一个订单

  • 状态不在 (4,6)
  • '002' 和 '005' 之间的代码 1
  • (code2=null and code3=null) or (code2 in ('BIGA', 'BIGP') and code3=null) or (code2=NULL and code3 = 'L')

如果 code1 ='002' AND group IN (48,59,60,87) 我必须选择一个地址
（非常感谢 Nikola Markovinović）：

SELECT TOP 1000 o.order_Id
              , a.Address_Id
              , a.Zip
            --, *
FROM orders o
CROSS APPLY
(
 select TOP 1
        a.Address_Id,
        a.Zip
   from address a
  WHERE a.order_Id = o.order_Id
  ORDER BY case a.Type 
                when 'D2' then 1 
                when 'K1' then 2 
                else 3 
            end,
        a.ADD_DATE
) a
WHERE
 o.Status NOT IN (4, 6)
 AND code1='002'
 AND group IN (48,59,60,87)
 AND ((code2 IS NULL AND code3 IS NULL) OR (code2 IN ('BIGA', 'BIGP') AND code3 IS NULL) OR (code2 IS NULL AND code3 = 'L'))

对于所有其他符合最高标准并获得 code1 ='002' AND group NOT IN (48,59,60,87) 的订单，我必须为那些已验证 =1 的订单选择所有地址

收集这些地址后，我将能够检查特定邮政公司是否可以将我的邮件递送到这些地址（我将检查另一个包含邮政编码的表格）

我正在考虑全部联合，首先选择并与第二个进行联合，这将返回 code1 ='002' 和 group NOT IN (48,59,60,87) 的所有地址。

但也许没有联合就可以做到吗？

这是我想要得到的最终结果：

CODE1        TYPE        COUNT_OF_ORDERS        COUNT_OF_ADDRESSES     COMPANY1  OTHER
'001'        'NORMAL'    125                    150                    110       40
'002'        'NORMAL'    100                    122                    100       22
'003'        'NORMAL'    150                    110                    100       10
'004'        'NORMAL'    200                    220                    220       0
'005'        'NORMAL'    220                    240                    210       30
'005'        'PRIORITY'  100                    110                    110       0
'SX1'        'PRIORITY'  100                    100                    20        80

因此，如果我的类型是“正常”，我必须检查该订单地址是否存在于具有正常邮政编码的表格中，如果它具有“优先”类型，我必须检查具有优先代码的表格。

如果代码存在于特定表中，我将 +1 添加到 COMPANY1 列，如果不添加到 OTHER，则这些列的总和必须是我的地址的总和。

这是我设法完成的查询（在@Nikola Markovinović 的帮助下）

SELECT TOP 1000 o.order_Id
              , a.Address_Id
              , a.Zip
            --, *
FROM orders o
CROSS APPLY
(
 select TOP 1
        a.Address_Id,
        a.Zip
   from address a
  WHERE a.order_Id = o.order_Id
    AND code1='002'
    AND o.[group] IN (48,59,60,87)
  ORDER BY case a.Type 
                when 'D2' then 1 
                when 'K1' then 2 
                else 3 
            end,
        a.ADD_DATE
  UNION ALL
 select 
        a.Address_Id,
        a.Zip
   from address a
  WHERE a.order_Id = o.order_Id
    AND ((code1='002' AND o.[group] NOT IN (48,59,60,87)) OR code1 IN ('001', '003', '004', '005'))
    --I'm not shure of that top line, it work's but mayby it con de written better
    AND Verified = 1
) a
WHERE
 o.Status NOT IN (4, 6)
 AND ((code2 IS NULL AND code3 IS NULL) 
    OR (code2 IN ('BIGA', 'BIGP') AND code3 IS NULL) 
    OR (code2 IS NULL AND code3 = 'L'))
4

1 回答 1

1

您可以轻松过滤地址 ( [group] IN (48,59,60,87) OR Verified = 1)，但调整 TOP 1 会使事情变得荒谬 ( TOP (case when [group] IN (48,59,60,87) then 1 else (select count(*) from addresses where order_Id = o.order_Id) end)。所以我建议您这样做，union all但仅针对地址：

SELECT TOP 1000 o.order_Id
              , a.Address_Id
              , a.Zip
            --, *
FROM orders o
CROSS APPLY
(
 select TOP 1
        a.Address_Id,
        a.Zip
   from address a
  WHERE a.order_Id = o.order_Id
    AND o.[group] IN (48,59,60,87)
  ORDER BY case a.Type 
                when 'D2' then 1 
                when 'K1' then 2 
                else 3 
            end,
        a.ADD_DATE
  UNION ALL
 select 
        a.Address_Id,
        a.Zip
   from address a
  WHERE a.order_Id = o.order_Id
    AND o.[group] NOT IN (48,59,60,87)
    AND Verified = 1
) a
WHERE
 o.Status NOT IN (4, 6)
 AND code1='002'
 AND ((code2 IS NULL AND code3 IS NULL) 
    OR (code2 IN ('BIGA', 'BIGP') AND code3 IS NULL) 
    OR (code2 IS NULL AND code3 = 'L'))

PS 如果订单可能没有地址，请将 CROSS APPLY 替换为 OUTER APPLY。

于 2012-04-24T15:05:07.547 回答