这个问题类似于:TSQL select rows by one from 2 conditions,但我想要的结果有所不同
我有一张这样的表:
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
和第二个表如下:
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
第三和第四张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'
对于每个订单
如果code1 ='002'和IN组(48,59,60,87),我必须选择一个地址
(非常感谢NikolaMarkovinović):
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'))
对于所有符合最高标准且得到code1 ='002'IN组且NOT IN(48,59,60,87)的其他订单,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邮件发送到这些地址(我将检查另一个包含邮政编码的表格)
我正在考虑使用union all,首先选择并与第二个进行联合,这将返回code1 ='002'和group NOT(48,59,60,87)的所有地址。
但也许没有工会可以做到这一点吗?
这是我想得到的最终结果:
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”,我必须检查订单中的地址是否存在于具有正常邮政编码的表中,如果它具有“优先级”类型,我必须在表格中检查优先级代码。
如果特定表格中存在代码,我会向COMPANY1列添加+1,如果不存在于OTHER,则这些列的总和必须是我的地址的总和。
这是我设法做的查询(在@NikolaMarkovinović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'))
答案 0 :(得分:1)
您可以轻松过滤地址([group] IN (48,59,60,87) OR Verified = 1)
,但调整TOP 1会让事情变得荒谬(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union all
但只针对地址:
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'))
P.S。如果订单可能没有地址,请用OUTER APPLY替换CROSS APPLY。