July 16, 2013 at 3:38 pm
Hi Folks,
I need to write a query to find out all the parent items which has a MAX of Qty and Max of Childitem. Here the process is like, first we need to get the MAX Qty per Parent and for this Qty get the MAX of Childitem.
Parent ItemChildItem Qty
62338-87468-0062338-77961-2045.333
62338-87468-0062338-77961-1845.333
62338-87468-0062338-84683-2145.333
62338-87468-0062338-84683-1846
62338-87468-0062338-87135-0146
62338-87468-0062338-87135-0045
From the above example, there are two childitems of the MAX Qty. so the final output should be 62338-87468-00(parent),62338-87135-01(Child), 46(Qty). It means there should be only one row per each parent item.
Help would be greatly appreciated.
Thank you,
Venu Babu,
July 16, 2013 at 3:50 pm
It helps if you provide your data in a re-usable format , such as this:
with data(ParentItem,ChildItem,Qty)
as
(
select '2338-87468-00', '62338-77961-20', '45.333' union all
select '2338-87468-00', '62338-77961-18', '45.333' union all
select '2338-87468-00', '62338-84683-21', '45.333' union all
select '2338-87468-00', '62338-84683-18', '46' union all
select '2338-87468-00', '62338-87135-01', '46' union all
select '2338-87468-00', '62338-87135-00', '45'
),
-- Here is the query you need
RowNumbered(ParentItem,ChildItem,Qty,Rn) as
(
select *,row_number() over(partition by ParentItem order by Qty desc,ChildItem desc) as Rn
from data
)
select ParentItem,ChildItem,Qty
from RowNumbered
where Rn=1
I have done it for you because you are new here 🙂
MM
select geometry::STGeomFromWKB(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
July 18, 2013 at 5:25 am
Thanks Buddy this helped me a lot to complete my Proc.
February 3, 2014 at 4:07 pm
thanks for the info 🙂
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