[英]How to get the previous Id value for a given table?
I have two input tables: 我有两个输入表:
1) Site: 1)网站:
site_id|| site_name|| site_location
1000 || abc || XYZ_123
1001 || tyu || ERD_123
1002 || iok || FTR_678
1003 || okn || YHU_987
1004 || ybg || OLP_008
1005 || qwe || PLM_126
2) 2)
product: 产品:
Product_id|| product_name||start_date||end_date
212 || sme1 ||2014-12-25||2017-03-13
250 || try1 ||2013-12-15|| 2017-03-13
267 || inu1 || 2015-03-27|| 2017-03-17
I need to check how many times the id is repeated and order it like the output table. 我需要检查ID重复多少次,并像输出表一样对其进行排序。
The previousId is to be caluclated if the preoduct_id is repeated and its less than the end date. 如果preoduct_id重复且小于结束日期,则会计算previousId。
This is what I have done 'till now. 这就是我现在所做的。
select top 1 d2.PRODUCT_id, d2.last_date, d1.*
from output_table d1,
output_table d2
where d1.SITE_id = d2.Site_id
and d1.START_DATE >= d2.end_DATE
and d1.Site_id=1001
and d1.PRODUCT_id = 250
order by d2.End_date desc
any help is appreciated. 任何帮助表示赞赏。
i need to output as my table. 我需要输出为我的桌子。
this is my output table structure : 这是我的输出表结构:
Id|| site_id|| product_id|| previous_id|| start_date|| end_date ||Previous_site_id || repeated_times || Previous_id
1 || 1000 ||250 || null || 2015-01-01||2017-03-13 || 1001 ||3 || 2
2 || 1001 ||250 || 1 || 2014-12-25||2015-01-01 || 1002 ||3 || 3
3 || 1002 ||250 || 2 || 2013-12-15||2014-12-25 || Null ||3 || Null
4 || 1003 ||267 || null || 2015-03-27|| 2017-03-17|| Null ||0 || Null
5 ||1004 ||212 || null || 2016-01-01||2017-03-13 ||1004 ||2 || 6
6 || 1005 ||212 || 1 || 2014-12-25||2015-12-30 ||Null ||2 || Null
Since you provided a pretty poor description of your schema, I've made a lot by guess. 由于您对架构的描述很差,因此我做了很多猜测。 Considering you use MSSQL 2012, there should be
LAG
windowed function that could be used in some way similar to the following: 考虑到您使用的是MSSQL 2012,应该有
LAG
窗口函数,该函数可以类似于以下内容的方式使用:
WITH CountedResult AS
(
SELECT
ROW_NUMBER() OVER(ORDER BY product_id, site_id) AS Id, -- use some id instead, if you have
site_id,
product_id,
[start_date],
[end_date],
COUNT(*) OVER(PARTITION BY product_id) AS repeated_times
FROM output_table
WHERE
site_id=1001
AND product_id = 250
)
SELECT
Id, site_id, product_id, [start_date], [end_date],
LAG(site_id, 1, 0) OVER (PARTITION BY product_id ORDER BY [start_date] DESC) AS previous_site_id,
repeated_times,
LAG(Id, 1, 0) OVER (PARTITION BY product_id ORDER BY [start_date] DESC) AS previous_id,
FROM CountedResult
ORDER BY product_id, site_id
Read more about LAG
on MSDN: https://msdn.microsoft.com/en-us/library/hh231256.aspx 在MSDN上阅读有关
LAG
更多信息: https : //msdn.microsoft.com/en-us/library/hh231256.aspx
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