Min Date from one column multiple rows - sql

Question

My apologies, I should have added every column and complete problem not just portion.
I have a table A which stores all invoices issued(id 1) payments received (id 4) from clients. Sometimes client pay in 2-3 installments. I want to find dateifference between invoice issued and last payment collected for the invoice. My data looks like this
**a.cltid**|**A.Invnum**|A.Cash|A.Date | a.type| a.status
70 |112 |-200 |2012-03-01|4 |P
70 |112 |-500 |2012-03-12|4 |P
90 |124 |-550 |2012-01-20|4 |P
70 |112 |700 |2012-02-20|1 |p
55 |101 |50 |2012-01-15|1 |d
90 |124 |550 |2012-01-15|1 |P
I am running
Select *, Datediff(dd,T.date,P.date)
from (select a.cltid, a.invnumber,a.cash, min(a.date)date
from table.A as A
where a.status<>'d' and a.type=1
group by a.cltid, a.invnumber,a.cash)T
join
Select *
from (select a.cltid, a.invnumber,a.cash, min(a.date)date
from table.A as A
where a.status<>'d' and a.type=4
group by a.cltid, a.invnumber,a.cash)P
on
T.invnumb=P.invnumber and T.cltid=P.cltid
How can I make it work? So it shows me
70|112|-500|2012-03-12|4|P 70|112|700|2012-02-20|1|p|22
90|124|-550|2012-01-20|4|P 90|124|550|2012-01-15|1|P|5
Edited***

Answer 1

You can use row_number to assign sequence number within each cltid in the order of decreasing date and then filter to get the first row for each cltid which will be the row with latest date for that cltid:
select *
from (
select A.*,
row_number() over (
partition by a.cltid order by a.date desc
) rn
from table.A as A
) t
where rn = 1;
It will return one row (with latest date) for each client. If you want to return all the rows which have latest date, use rank() instead.

Answer 2

Use a ranking function to get all the columns:
select a.*
from (select a.*,
row_number() over (partition by cltid order by date desc) as seqnum
from a
) a
where seqnum = 1;
Use aggregation if you only want the date. The issue with your query is that the group by clause has too many columns:
select a.cltid, max(a.date) as date
from table.A as A
group by a.cltid;
And the fact that min() returns the first date not the last date.

Answer 3

There are many ways to do this. Here are some of them:
test setup: http://rextester.com/VGUY60367
with common_table_expression as () using row_number()
with cte as (
select *
, rn = row_number() over (
partition by cltid, Invnum
order by [date] desc
)
from a
)
select cltid, Invnum, Cash, [date]
from cte
where rn = 1
cross apply version:
select distinct
a.cltid
, a.Invnum
, x.Cash
, x.[date]
from a
cross apply (
select top 1
cltid, Invnum
, [date]
, Cash
from a as i
where i.cltid =a.cltid
and i.Invnum=a.Invnum
order by i.[date] desc
) as x;
top with ties version:
select top 1 with ties
*
from a
order by
row_number() over (
partition by cltid, Invnum
order by [date] desc
)
all return:
+-------+--------+---------------------+------+
| cltid | Invnum | date | Cash |
+-------+--------+---------------------+------+
| 70 | 112 | 12.03.2012 00:00:00 | -500 |
| 90 | 124 | 20.01.2012 00:00:00 | -550 |
+-------+--------+---------------------+------+

Answer 4

You can achieve the desired o/p by this:
Select
a.cltid, a.invnumber,a.cash, max(a.date) [date]
from
YourTable a
group by
a.cltid, a.invnumber, a.cash, a.date