I am trying for a CTE with a ROW_NUMBER function.
Query:
with Demo as
(
select *, ROW_NUMBER() over (partition by GID, MID order by MMID) as ROWNUMBER from Table1 where DateEntered > '2015-06-13 00:00:00.000'
)
select * from Demo
Here, the result I get is
GID MID ROWNUMBER
1 1 1
1 2 1
1 2 2
1 2 3
2 1 1
2 2 1
2 2 2
2 3 5
2 4 4
Now, I want to get all the rows where combination of GID,MID has max row number value. But a condition here is that for those rows, the combination of GID,MID should also have 1.
In simple terms, get me the rows with max row number value, if that combination of gid,mid has rownumber=1.
The output I desire is
GID MID ROWNUMBER
1 1 1
1 2 3
2 1 1
2 2 2
I hope i did not made it complex. Can anyone pls inform me on how to do this ?
with Demo as
(
select *, ROW_NUMBER() over (partition by GID, MID order by MMID) as RN
from Table1 where DateEntered > '2015-06-13 00:00:00.000'
)
, x as
(select gid, mid
from demo
where RN = 1
)
select demo.gid, demo.mid, max(demo.rn) as rownumb
from demo left join x
on x.gid = demo.gid and x.mid = demo.mid
group by demo.gid, demo.mid;
You can use max to select the highest rownumber per mid, gid combination.
If you don't need the row number value, just use desc instead of asc and a filter:
with Demo as (
select t.*,
ROW_NUMBER() over (partition by GID, MID order by MID DESC) as seqnum
from Table1
where DateEntered > '2015-06-13'
)
select *
from Demo
where seqnum = 1;
If you do want a more meaningful ROWNUMBER in the output, then use two calculations in the CTE.
The max of row_number() is just count.
select GID, MID, COUNT(*) as ROWNUMBER
from Table1
where DateEntered > '2015-06-13 00:00:00.000'
group by GID, MID
This should work for you:
with Demo as
(
select *,
ROW_NUMBER() over (partition by GID, MID order by MMID) as ROWNUMBER
from Table1
where DateEntered > '2015-06-13 00:00:00.000'
)
select GID,
MID,
MAX(ROWNUMBER) as MaxROWNUMBER
from Demo
GROUP BY GID,MID;
Related
How to get the rows of the longest consecutive same value?
Table Learning:
rowID
values
1
1
2
1
3
0
4
0
5
0
6
1
7
0
8
1
9
1
10
1
Longest consecutive value is 1 (rowID 8-10 as rowID 1-2 is 2 and rowID 6-6 is 1). How to query to get the actual rows of consecutive values (not just rowStart and rowEnd values) like :
rowID
values
8
1
9
1
10
1
And for longest consecutive values of both 1 and 0?
DB Fiddle
I think that the simplest approach is to use a window count to define the islands. Then to get the "longest" island, we just need to aggregate, sort and limit:
select min(valueid) grp_start, max(valueid) grp_end
from (select t.*, sum(value = 0) over(order by valueid) grp from testing t) t
where value = 1
group by grp
order by count(*) desc limit 1
In the DB Fiddle that you provided, the query returns:
grp_start
grp_end
8
10
This is a gaps and islands problem, and one approach is to use the difference in row numbers method:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (ORDER BY rowID) rn1,
ROW_NUMBER() OVER (PARTITION BY values ORDER BY rowID) rn2
FROM yourTable
),
cte2 AS (
SELECT *,
MIN(rowID) OVER (PARTITION BY values, rn1 - rn2) AS minRowID,
MAX(rowID) OVER (PARTITION BY values, rn1 - rn2) AS maxRowID
FROM cte1
),
cte3 AS (
SELECT *, RANK() OVER (PARTITION BY values ORDER BY maxRowID - minRowID DESC) rnk
FROM cte2
)
SELECT rowID, values
FROM cte3
WHERE rnk = 1
ORDER BY values, rowID;
I have a query in SQL looks like that:
select fldCustomer, fldTerminal, COUNT(fldbill)
from tblDataBills
group by fldCustomer, fldTerminal
order by fldCustomer
results looks like:
fldCustomer fldTerminal (number of bills)
0 1 19086
0 2 10
0 5 236
1 1 472
1 5 3
1 500 19
2 1 292
2 500 22
how can i get the MAX count of each customer so i get results like
0 1 19086
1 1 472
2 1 292
Thanks in advance!
Use a subquery with row_number():
select fldCustomer, fldTerminal, cnt
from (select fldCustomer, fldTerminal, COUNT(*) as cnt,
row_number() over (partition by fldCustomer order by count(*) desc) as seqnum
from tblDataBills
group by fldCustomer, fldTerminal
) db
where seqnum = 1
order by fldCustomer ;
Note that in the event of ties, this will arbitrarily return one of the rows. If you want all of them, then use rank() or dense_rank().
This might require a little trickery with the RANK() function
SELECT fldCustomer, fldTerminal, [(number of bills)]
FROM (
SELECT fldCustomer, fldTerminal, COUNT(fldbill) [(number of bills)],
RANK() OVER (PARTITION BY fldCustomer ORDER BY COUNT(fldbill) DESC) Ranking
FROM tblDataBills
GROUP BY fldCustomer, fldTerminal
) a
WHERE Ranking = 1
I'm having problem with getting only TOP 2 values for each group (groups are in column).
Example :
ID Group Value
1 A 30
2 A 150
3 A 40
4 A 70
5 B 0
6 B 100
7 B 90
I expect my output to be
ID Group Value
1 A 150
2 A 70
3 B 100
4 B 90
Simply, for each group I want just 2 rows with the highest Value
Most databases support the ANSI standard row_number() function. You would use it as:
select group, value
from (select t.*,
row_number() over (partition by group order by value desc) as seqnum
from t
) t
where seqnum <= 2;
To set the id you can use row_number() in the outer query:
select row_number() over (order by group, value) as id,
group, value
from (select t.*,
row_number() over (partition by group order by value desc) as seqnum
from t
) t
where seqnum <= 2;
However, changing the id seems suspicious.
You can use CTE with rank function ROW_NUMBER() .
Here is query to get your result.
;WITH cte AS
( SELECT Group, value,
ROW_NUMBER() OVER (PARTITION BY Group ORDER BY value DESC) AS rn
FROM test
)
SELECT Group, value FROM cte
WHERE rn <= 2
ORDER BY value
I have a table like so
ID OrdID Value
1 1 0
2 2 0
3 1 1
4 2 1
5 1 1
6 2 0
7 1 0
8 2 0
9 2 1
10 1 0
11 2 0
I want to get the count of consecutive value where the value is 0. Using the example above the result will be 3 (Rows 6, 7 and 8). I am using sql server 2008 r2.
I am going to presume that id is unique and increasing. You can get counts of consecutive values by using the different of row numbers. The following counts all sequences:
select grp, value, min(id), max(id), count(*) as cnt
from (select t.*,
(row_number() over (order by id) - row_number() over (partition by value order by id)
) as grp
from table t
) t
group by grp, value;
If you want the longest sequence of 0s:
select top 1 grp, value, min(id), max(id), count(*) as cnt
from (select t.*,
(row_number() over (order by id) - row_number() over (partition by value order by id)
) as grp
from table t
) t
group by grp, value
having value = 0
order by count(*) desc
A query using not exists to find consecutive 0s
select top 1 min(t2.id), max(t2.id), count(*)
from mytable t
join mytable t2 on t2.id <= t.id
where not exists (
select 1 from mytable t3
where t3.id between t2.id and t.id
and t3.value <> 0
)
group by t.id
order by count(*) desc
http://sqlfiddle.com/#!3/52989/3
I need to select data base upon three conditions
Find the latest date (StorageDate Column) from the table for each record
See if there is more then one entry for date (StorageDate Column) found in first step for same ID (ID Column)
and then see if DuplicateID is = 2
So if table has following data:
ID |StorageDate | DuplicateTypeID
1 |2014-10-22 | 1
1 |2014-10-22 | 2
1 |2014-10-18 | 1
2 |2014-10-12 | 1
3 |2014-10-11 | 1
4 |2014-09-02 | 1
4 |2014-09-02 | 2
Then I should get following results
ID
1
4
I have written following query but it is really slow, I was wondering if anyone has better way to write it.
SELECT DISTINCT(TD.RecordID)
FROM dbo.MyTable TD
JOIN (
SELECT T1.RecordID, T2.MaxDate,COUNT(*) AS RecordCount
FROM MyTable T1 WITH (nolock)
JOIN (
SELECT RecordID, MAX(StorageDate) AS MaxDate
FROM MyTable WITH (nolock)
GROUP BY RecordID)T2
ON T1.RecordID = T2.RecordID AND T1.StorageDate = T2.MaxDate
GROUP BY T1.RecordID, T2.MaxDate
HAVING COUNT(*) > 1
)PT ON TD.RecordID = PT.RecordID AND TD.StorageDate = PT.MaxDate
WHERE TD.DuplicateTypeID = 2
Try this and see how the performance goes:
;WITH
tmp AS
(
SELECT *,
RANK() OVER (PARTITION BY ID ORDER BY StorageDate DESC) AS StorageDateRank,
COUNT(ID) OVER (PARTITION BY ID, StorageDate) AS StorageDateCount
FROM MyTable
)
SELECT DISTINCT ID
FROM tmp
WHERE StorageDateRank = 1 -- latest date for each ID
AND StorageDateCount > 1 -- more than 1 entry for date
AND DuplicateTypeID = 2 -- DuplicateTypeID = 2
You can use analytic function rank , can you try this query ?
Select recordId from
(
select *, rank() over ( partition by recordId order by [StorageDate] desc) as rn
from mytable
) T
where rn =1
group by recordId
having count(*) >1
and sum( case when duplicatetypeid =2 then 1 else 0 end) >=1