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
Related
I'm trying to find repetitions between rows based on the column. I've tried window functions with row_number() / rank() but they group all the values that are found (similar to GROUP BY) which I do not expect.
How can I find repetitions of the values?
I tried to do something like this:
SELECT *, rank() OVER(PARTITION BY customer ORDER BY id) FROM customers ORDER BY id
And got the following result:
id
customer
rank
1
customer_1
1
2
customer_2
1
3
customer_2
2
4
customer_1
2
5
customer_3
1
6
customer_1
3
What I want to do:
id
customer
rank
1
customer_1
1
2
customer_2
1
3
customer_2
2
4
customer_1
1
5
customer_3
1
6
customer_1
1
You are looking for counts within adjacent rows. This is a type of gaps-and-islands problem. You can define the adjacent rows with the difference of row_numbers() and then enumerate them:
SELECT c.*,
ROW_NUMBER() OVER (PARTITION BY customer, seqnum - seqnum_2 ORDER BY id) as ranking
FROM (SELECT c.*,
ROW_NUMBER() OVER (ORDER BY id) as seqnum,
ROW_NUMBER() OVER (PARTITION BY customer ORDER BY id) as seqnum_2
FROM customers c
) c
ORDER BY id
You can use a recursive query:
WITH RECURSIVE repeatitions(id, customer, repeat_count) AS (
SELECT id, customer, 1 as repeat_count
FROM customers
UNION ALL
SELECT c.id, c.customer, r.repeat_count + 1
FROM customers c, repeatitions r
WHERE c.id = r.id + 1 AND c.customer = r.customer
)
SELECT id, customer, repeat_count
FROM repeatitions
ORDER by id
I created a working fiddle to demonstrate it.
I have a table with values
id sales date
1 5 "2015-01-04"
1 3 "2015-01-03"
1 1 "2015-01-01"
1 1 "2015-01-01"
2 7 "2015-01-05"
2 6 "2015-01-04"
2 4 "2015-01-03"
3 11 "2015-01-08"
3 10 "2015-01-07"
3 9 "2015-01-06"
3 8 "2015-01-05"
I want to select top two values of each id as shown in desired output.
Desired output:
id sales date
1 5 "2015-01-04"
1 3 "2015-01-03"
2 7 "2015-01-05"
2 6 "2015-01-04"
3 11 "2015-01-08"
3 10 "2015-01-07"
My attempt:
can someone help me with this. Thank you in advance!
select transactions.salesperson_id, transactions.id, transactions.date
from transactions
ORDER BY transactions.salesperson_id ASC, transactions.date DESC;
This can be done using window functions:
select id, sales, "date"
from (
select id, sales, "date",
dense_rank() over (partition by id order by "date" desc) as rnk
from transactions
) t
where rnk <= 2;
If there are multiple rows on the same date this might return more than two rows for the same ID. If you don't want that, use row_number() instead of dense_rank()
row_number() will get what you want.
select * from
(select row_number() over (partition by id order by date) as rn, sales, date from transactions) t1
where t1.rn <= 2
My dataset name bollywood.csv:
This is my data. I need the actors who have the most lead roles in movies.
I Need the name of the lead actors and the number of films in which they have lead.
My code is:
select lead, count(*) as nos from bollywood group by lead order by nos desc;
And the result is:
Amitabh 3
Akshay 3
John 3
Riteish 2
Shahrukh 2
Sunny 2
Emraan 2
Katrina 2
Nawazuddin 2
Tiger 2
Sharman 2
Manoj 2
Vidya 1
Tusshar 1
Tannishtha 1
Sushant 1
SunnyDeol 1
Sonam 1
Sonakshi 1
Siddarth 1
Shahid 1
Sandeep 1
Salman 1
If you want all actors with most lead roles (possibly multiple records):
select lead, count(*) as nos
from bollywood
group by lead
having count(*) =
(select max(cnt) from
(select lead, count(*) cnt
from bollywood
group by lead ) tblBolly )
Use rownum pseudocolumn with order by (oracle).
select from (
select lead, count(*) as nos
from bollywood
group by lead
order by lead desc
)
where rownum = 1
Use window functions:
select lead, cnt
from (select lead, count(*) as cnt,
rank() over (order by count(*) desc) as rnk
from bollywood
group by lead
) b
where rnk = 1;
The question is: For each day, list the User ID who has read the most number of messages.
user_id msgID read_date
1 1 10
1 2 10
2 2 10
2 2 23
3 2 23
I believe the date is an outer group and user_id is an inner group, but how to do group nesting in sql? Or somehow avoid this?
This is a task for a Window Function:
select *
from
(
select user_id, read_date, count(*) as cnt,
rank()
over (partition by read_date -- each day
order by count(*) desc) as rnk -- maximum number
from tab
group by user_id, read_date
) dt
where rnk = 1
This might return multiple users for one with the same maximum count, if you want just one (randomly) switch to ROW_NUMBER
select user_id
from
(
select user_id,count(msgID)
from table
group by read_date
)
where rownum <= 1;
I have different versions of the charges in a table. I want to grab and sum the last charge grouped by Type.
So I want to add 9.87, 9.63, 1.65.
I want the Parent ID , sum(9.87 + 9.63 + 1.65) as the results of this query.
We use MSSQL
ID ORDER CHARGES TYPE PARENT ID
1 1 6.45 1 1
2 2 1.25 1 1
3 3 9.87 1 1
4 1 6.54 2 1
5 2 5.64 2 1
6 3 0.84 2 1
7 4 9.63 2 1
8 1 7.33 3 1
9 2 5.65 3 1
10 3 8.65 3 1
11 4 5.14 3 1
12 5 1.65 3 1
WITH recordsList
AS
(
SELECT Type, Charges,
ROW_NUMBER() OVER (PArtition BY TYPE
ORDER BY [ORDER] DESC) rn
FROM tableName
)
SELECT SUM(Charges) totalCharge
FROM recordsLIst
WHERE rn = 1
SQLFiddle Demo
Use row_number() to identify the rows to be summed, and then sum them:
select SUM(charges)
from (select t.*,
ROW_NUMBER() over (PARTITION by type order by id desc) as seqnum
from t
) t
where seqnum = 1
Alternatively you could use a window aggregate MAX():
SELECT SUM(Charges)
FROM (
SELECT
[ORDER],
Charges,
MaxOrder = MAX([ORDER]) OVER (PARTITION BY [TYPE])
FROM atable
) s
WHERE [ORDER] = MaxOrder
;
SELECT t.PARENT_ID, SUM(t.CHARGES)
FROM dbo.test73 t
WHERE EXISTS (
SELECT 1
FROM dbo.test73
WHERE [TYPE] = t.[TYPE]
HAVING MAX([ORDER]) = t.[ORDER]
)
GROUP BY t.PARENT_ID
Demo on SQLFiddle