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
Related
I have the following table named customerOrders.
ID user order
1 1 2
2 1 3
3 1 1
4 2 1
5 1 5
6 2 4
7 3 1
8 6 2
9 2 2
10 2 3
I want to return to users with most orders. Currently, I have the following QUERY:
SELECT user, COUNT(user) AS UsersWithMostOrders
FROM customerOrders
GROUP BY user
ORDER BY UsersWithMostOrders DESC;
This returns me all the values grouped by total orders like.
user UsersWithMostOrders
1 4
2 4
3 1
6 1
I only want to return the users with most orders. In my case that would be user 1 and 2 since both of them have 4 orders. If I use TOP 1 or LIMIT, it will only return the first user. If I use TOP 2, it will only work in this scenario, it will return invalid data when top two users have different count of orders.
Required Result
user UsersWithMostOrders
1 4
2 4
You can use TOP 1 WITH TIES:
SELECT TOP 1 WITH TIES
[user], COUNT(*) AS UsersWithMostOrders
FROM customerOrders
GROUP BY [user]
ORDER BY UsersWithMostOrders DESC;
See the demo.
Results:
> user | UsersWithMostOrders
> ---: | ------------------:
> 1 | 4
> 2 | 4
Option 1
Should work with most versions of SQL.
select *
from (
select *,
rank() over(order by numOrders desc) as rrank
from (
select `user`, count(*) as numOrders
from customerOrders
group by `user`
) summed
) ranked
where rrank = 1
Play around with the code here
Option 2
If your version of SQL allows window functions (with), here is a much more readable solution which does the same thing
with summed as (
select `user`, count(*) as numOrders
from customerOrders
group by `user`
),
ranked as (
select *,
rank() over(order by numOrders desc) as rrank
from summed
)
select *
from ranked
where rrank = 1
Play around with the code here
You can use a CTE to attain this Req:
;WITH CTE AS(
SELECT [user], COUNT(user) AS UsersWithMostOrders
FROM #T
GROUP BY [user])
SELECT M.* from CTE M
INNER JOIN ( SELECT
MAX(UsersWithMostOrders) AS MaximumOrders FROM CTE) S ON
M.UsersWithMostOrders=S.MaximumOrders
Below Oracle Query can help:
WITH test_table AS
(
SELECT user, COUNT(order) AS total_order , DENSE_RANK() OVER (ORDER BY
total_order desc) AS rank_orders FROM customerOrders
GROUP BY user
)
select * from test_table where rank_orders = 1
I want to calculate unique rankings but I get duplicate rankings
Here's my attempt:
SELECT
TG.EMPCODE,
DENSE_RANK() OVER (ORDER BY TS.COUNT_DEL DESC, TG.COUNT_TG DESC) AS YOUR_RANK
FROM
(SELECT
EmpCode,
SUM(CASE WHEN Tgenerate = 1 THEN 1 ELSE 0 END) AS COUNT_TG
FROM
TBLTGENERATE1
GROUP BY
EMPCODE) TG
INNER JOIN
(SELECT
EMP_CODE,
SUM(CASE WHEN STATUS = 'DELIVERED' THEN 1 ELSE 0 END) AS COUNT_DEL
FROM
TBLSTAT
GROUP BY
EMP_CODE) TS ON TG.EMPCODE = TS.EMP_CODE;
The output I get is like this:
EID Rank
---------
102 1
105 2
101 2
103 3
106 4
There is same rank for 105 and 101.
How do I calculate unique ranking?
Use ROW_NUMBER() instead of DENSE_RANK():
SELECT TG.EMPCODE,
ROW_NUMBER() OVER (ORDER BY TS.COUNT_DEL DESC, TG.COUNT_TG DESC) AS YOUR_RANK
Ties will then be given sequential rankings.
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 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;
I have a query,which is not returning proper result,
i want my query to return total of score for same user_id, so that each user_id will have only one record with sum of all of its score.
My query is this:
SELECT
DENSE_RANK() OVER (ORDER BY score DESC) AS rank,
user_id,
SUM(score) AS total_score
FROM
account_game
GROUP BY
user_id, score
ORDER BY
rank ASC
Query output is :
rank user_id total_score
1 2 4837
2 1 600
2 6 600
3 1 30
4 1 20
There should be three records with user_id 1,2,6
Expected result should be
rank user_id total_score
-------------------------
1 2 4837
2 6 700
3 1 650
Please suggest
As StuartLC commented, you can just remove the score from your GROUP BY and all should be fine:
SELECT DENSE_RANK() OVER (Order by SUM(score) DESC) AS rank,
user_id,
SUM(score) as total_score
FROM
account_game
GROUP BY user_id
ORDER BY rank ASC