我有两个表[Product1]和[Product2_transform]
create table [product1](
[product_id] nvarchar(256)
,[product_rev] nvarchar(256)
,[product_name] nvarchar(256)
,[product_description] nvarchar(256)
,[product_owner] nvarchar(256)
,[product_group] nvarchar(256)
,[product_type] nvarchar(256)
);
create table [product2_transform](
[product_id] nvarchar(256)
,[product_rev] nvarchar(256)
,[product_name] nvarchar(256)
,[product_description] nvarchar(256)
,[product_owner] nvarchar(256)
,[product_group] nvarchar(256)
,[product_type] nvarchar(256)
);
insert into [product1] values
('111' ,'AAA', 'CAR' , 'SPARE PARTS' , 'Nissan' , 'Nissan' , 'AUTOMOTIVE' )
,('222' ,'BBB', 'MODEL' , 'DESGIN' , 'NIFT' , 'NIFT' , 'FASHION' )
,('333' ,'CCC', 'REAR' , 'REAR MATERIAL' , 'KLM' , 'KLM GROUP' , 'MANUFACTURING' )
,('444' ,'DDD', 'FINACLE 2.0' , 'BANKING PRODUCT' , 'IFLEX' , 'ORACLE' , 'SOFTWARE' )
,('555' ,'EEE', 'TYRE' , 'CEAR TYRES' , 'TATA MOTORS' , 'TATA' , 'AUTOMOTIVE' );
insert into [product2_transform] values
('111', 'AAA', 'CAR' , 'SPARE PARTS Ford' , 'Ford' , 'Nissan ' , 'AUTOMOTIVE LTD')
,('222', 'BBB', 'MODEL' , 'DESGIN' , 'NIFTY' , 'NIFT' , 'FASHION' )
,('333', 'CCC', 'REAR Head Left' , 'REAR MATERIAL' , 'KLM' , 'KLM GROUP' , 'MANUFACTURING' )
,('444', 'DDD', 'FINACLE 2.5' , 'BANKING PRODUCT' , 'Oracle IFLEX' , 'ORACLE' , 'SOFTWARE' )
,('555', 'EEE', 'SEAT TYRE' , 'CEAR TYRES' , 'TATA BANCS' , 'TATA' , 'AUTOMOTIVE' );
我想要[product1]和[product2]中的差值并将其放在[Product_post_validation]表格中。
create table [product_post_validation](
[product_id] nvarchar(256)
,[product_rev] nvarchar(256)
,[validation_column] nvarchar(256)
,[value_in_transform] nvarchar(256)
,[value_in_output] nvarchar(256)
);
[product_post_validation]中的预期结果是： Product_id / Product_rev / validation_column / value_in_transform / value_in_output
111 AAA Product_description SPARE PARTS SPARE PARTS Ford
111 AAA Product_owner Nissan Ford
111 AAA Product_TYPE AUTOMOTIVE AUTOMOTIVE LTD
333 CCC Product_Name REAR REAR Head Left
444 DDD Product_Name FINACLE 2.0 FINACLE 2.5
444 DDD Product_owner IFLEX Oracle IFLEX
555 EEE Product_Name TYRE SEAT TYRE
555 EEE Product_owner TATA MOTORS TATA BANCS
请帮助撰写查询...
其实我在oracle查询中写的工作正常，现在我想在MS sql server中编写，所以请协助。
with comp as (
select /*+ qb_name(CDC_GROUP) */
"PRODUCT_ID", "PRODUCT_REV", "PRODUCT_NAME", "PRODUCT_DESCRIPTION",
"PRODUCT_OWNER", "PRODUCT_GROUP", "PRODUCT_TYPE",
case when count(*) over(partition by "PRODUCT_ID", "PRODUCT_REV") - Z##NEW_CNT <= 1
then 'PRODUCT2_TRANSFORM'
else 'PRODUCT1'
end TBL
FROM (
select /*+ qb_name(COMPARE) NO_MERGE */
"PRODUCT_ID", "PRODUCT_REV", "PRODUCT_NAME", "PRODUCT_DESCRIPTION",
"PRODUCT_OWNER", "PRODUCT_GROUP", "PRODUCT_TYPE",
sum(Z##NEW_CNT) Z##NEW_CNT
FROM (
select /*+ qb_name(old) */
"PRODUCT_ID", "PRODUCT_REV", "PRODUCT_NAME", "PRODUCT_DESCRIPTION",
"PRODUCT_OWNER", "PRODUCT_GROUP", "PRODUCT_TYPE",
-1 Z##NEW_CNT
from PRODUCT1 O
union all
select /*+ qb_name(new) */
"PRODUCT_ID", "PRODUCT_REV", "PRODUCT_NAME", "PRODUCT_DESCRIPTION",
"PRODUCT_OWNER", "PRODUCT_GROUP", "PRODUCT_TYPE",
1 Z##NEW_CNT
from product2_transform N
)
group by
"PRODUCT_ID", "PRODUCT_REV", "PRODUCT_NAME", "PRODUCT_DESCRIPTION",
"PRODUCT_OWNER", "PRODUCT_GROUP", "PRODUCT_TYPE"
having sum(Z##NEW_CNT) != 0
)
)
select * from comp
unpivot(val for col in(PRODUCT_NAME, PRODUCT_DESCRIPTION, PRODUCT_OWNER, PRODUCT_GROUP, PRODUCT_TYPE))
pivot(max(val) for tbl in ('PRODUCT1' PRODUCT1, 'PRODUCT2_TRANSFORM' PRODUCT2_TRANSFORM))
where decode(PRODUCT1, PRODUCT2_TRANSFORM, 0, 1) = 1
order by 1,2,3;
答案 0 :(得分：1)
我会通过取消数据然后比较来做到这一点：
with p1 as (
select p.product_id, p.product_rev, v.col, v.val
from product1 p cross apply
(values ('product_name', product_name),
('product_description', product_description),
('product_owner', product_owner),
('product_group', product_group),
('product_type', product_type)
) v(col, val)
),
p2 as (
select p.product_id, p.product_rev, v.col, v.val
from Product2_Transform p cross apply
(values ('product_name', product_name),
('product_description', product_description),
('product_owner', product_owner),
('product_group', product_group),
('product_type', product_type)
) v(col, val)
)
select p1.product_id, p2.product_id, p1.which as validation_column,
p2.val as val_in_transform, p1.val as val_in_product
from p1 join
p2
on p1.product_id = p2.product_id and
p1.product_rev = p2.product_rev and
p1.col = p2.col and
(p1.val <> p2.val or p1.val is null and p2.val is not null or p1.val is not null and p2.val is null);
unpivot的一大问题是类型需要兼容。它们似乎都是字符串，所以这应该没问题。
请注意，这会处理NULL值。它不处理两个表之间缺少的产品/转速对。在full join中使用coalesce()然后select即可轻松处理。如果这是一个优先事项，您的问题不明确。
答案 1 :(得分：0)
首先：感谢您提出有趣的挑战：）
可能的解决方案是：
;with [data] as (
select
[product_id] = isnull([p1].[product_id], [p2].[product_id])
,[product_rev] = isnull([p1].[product_rev], [p2].[product_rev])
,[validation_column1] = [p1_c].[value]('(local-name(.))[1]', 'nvarchar(256)')
,[validation_column2] = [p2_c].[value]('(local-name(.))[1]', 'nvarchar(256)')
,[validation_value1] = iif([p1_c].[value]('(@xsi:nil)[1]', 'bit') = 'true', null, [p1_c].[value]('(.)[1]', 'nvarchar(256)'))
,[validation_value2] = iif([p2_c].[value]('(@xsi:nil)[1]', 'bit') = 'true', null, [p2_c].[value]('(.)[1]', 'nvarchar(256)'))
from
[product1] as [p1]
full join
[product2_transform] as [p2]
on
[p1].[product_id] = [p2].[product_id]
and [p1].[product_rev] = [p2].[product_rev]
cross apply
(
select [data] = convert(xml, (select [p1].* for xml path(''), elements xsinil))
) as [p1x]
cross apply
(
select [data] = convert(xml, (select [p2].* for xml path(''), elements xsinil))
) as [p2x]
cross apply
[p1x].[data].[nodes]('*') as [p1_t]([p1_c])
cross apply
[p2x].[data].[nodes]('*') as [p2_t]([p2_c])
)
insert into [product_post_validation](
[product_id]
,[product_rev]
,[validation_column]
,[value_in_transform]
,[value_in_output]
)
select
[product_id] = [product_id]
,[product_rev] = [product_rev]
,[validation_column] = isnull([validation_column1], [validation_column2])
,[value_in_transform] = [validation_value1]
,[value_in_output] = [validation_value2]
from
[data]
where
isnull(nullif([validation_column1], [validation_column2]), nullif([validation_column2], [validation_column1])) is null
and isnull(nullif([validation_value1], [validation_value2]), nullif([validation_value2], [validation_value1])) is not null
order by
[product_id] asc
,[product_rev] asc;