I've got a table Accounts
AMOUNT| ID_CLIENT | ID_BRANCH
250 1 1
250 1 3
100 1 4
300 2 1
300 2 3
450 3 2
100 3 2
225 4 1
225 4 2
225 4 4
225 4 5
I need to find clients who have the same amount in every branch (like ID_CLIENT = 2 and ID_CLIENT = 4). I have no idea how can I implement this ( Could anyone help me, please?
Use two levels of aggregation:
select client
from (select client, branch, sum(amount) as amount
from t
group by client, branch
) cb
group by client
having min(amount) = max(amount);
I can't tell if you can have multiple rows per client/branch. If not, you just need:
select client
from t
group by client
having min(amount) = max(amount);
You can use analytical functions to achieve the same:
Demo
with CTE1 as
(
SELECT A.*, DENSE_RANK() OVER (PARTITION BY ID_CLIENT ORDER BY AMOUNT) DN,
COUNT(*) OVER (PARTITION BY ID_CLIENT) TOTAL_COUNT
FROM TABLE1 A ORDER BY ID_CLIENT
)
SELECT ID_CLIENT FROM
(
SELECT ID_CLIENT, SUM(DN), TOTAL_COUNT
FROM CTE1
GROUP BY ID_CLIENT, TOTAL_COUNT
HAVING SUM(DN) = TOTAL_COUNT
);
By using First_value and Last_value:
Demo
SELECT DISTINCT ID_CLIENT FROM
(
SELECT A.*,
FIRST_VALUE(AMOUNT) OVER(PARTITION BY ID_CLIENT ORDER BY AMOUNT ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FST_VAL,
LAST_VALUE(AMOUNT) OVER(PARTITION BY ID_CLIENT ORDER BY AMOUNT ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) LST_VAL
FROM TABLE1 A
) X WHERE FST_VAL = LST_VAL ;
Related
Is it possible to rank item by partition without use CTE method
Expected Table
item
value
ID
A
10
1
A
20
1
B
30
2
B
40
2
C
50
3
C
60
3
A
70
4
A
80
4
By giving id to the partition to allow agitated function to work the way I want.
item
MIN
MAX
ID
A
10
20
1
B
30
40
2
C
50
60
3
A
70
80
4
SQL Version: Microsoft SQL Sever 2017
Assuming that the value column provides the intended ordering of the records which we see in your question above, we can try using the difference in row numbers method here. Your problem is a type of gaps and islands problem.
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (ORDER BY value) rn1,
ROW_NUMBER() OVER (PARTITION BY item ORDER BY value) rn2
FROM yourTable
)
SELECT item, MIN(value) AS [MIN], MAX(value) AS [MAX], MIN(ID) AS ID
FROM cte
GROUP BY item, rn1 - rn2
ORDER BY MIN(value);
Demo
If you don't want to use a CTE here, for whatever reason, you may simply inline the SQL code in the CTE into the bottom query, as a subquery:
SELECT item, MIN(value) AS [MIN], MAX(value) AS [MAX], MIN(ID) AS ID
FROM
(
SELECT *, ROW_NUMBER() OVER (ORDER BY value) rn1,
ROW_NUMBER() OVER (PARTITION BY item ORDER BY value) rn2
FROM yourTable
) t
GROUP BY item, rn1 - rn2
ORDER BY MIN(value);
You can generate group IDs by analyzing the previous row item value that could be obtained with the LAG function and finally use GROUP BY to get the minimum and maximum value in item groups.
SELECT
item,
MIN(value) AS "min",
MAX(value) AS "max",
group_id + 1 AS id
FROM (
SELECT
*,
SUM(CASE WHEN item = prev_item THEN 0 ELSE 1 END) OVER (ORDER BY value) AS group_id
FROM (
SELECT
*,
LAG(item, 1, item) OVER (ORDER BY value) AS prev_item
FROM t
) items
) groups
GROUP BY item, group_id
Query produces output
item
min
max
id
A
10
20
1
B
30
40
2
C
50
60
3
A
70
80
4
You can check a working demo here
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 Tabel MatchResults
id | player_win_id
------------------
1 | 1
2 | 1
3 | 3
4 | 1
5 | 2
6 | 3
7 | 3
8 | 1
9 | 1
10 | 1
I need to find out for each player ID the highest number of consecutive victories. I use MS SQL Server.
Expected Result
PLAYER_ID | WIN_COUNT
------------------
1 | 3
2 | 1
3 | 2
This is a type of gaps-and-islands problem. One solution uses the difference of row numbers. So, to get all streaks:
select player_win_id, count(*)
from (select t.*,
row_number() over (order by id) as seqnum,
row_number() over (partition by player_win_id order by id) as seqnum_p
from MatchResults t
) t
group by player_win_id, (seqnum - seqnum_p);
Why this works is a little tricky to explain. But if you look at the results of the subquery, you'll probably see how the difference between the row number values captures adjacent rows with the same player win id.
For the maximum, the simplest is probably just an aggregation query:
select player_win_id, max(cnt)
from (select player_win_id, count(*) as cnt
from (select t.*,
row_number() over (order by id) as seqnum,
row_number() over (partition by player_win_id order by id) as seqnum_p
from MatchResults t
) t
group by player_win_id, (seqnum - seqnum_p)
) p
group by player_win_id;
Now I understand the previous comment. The code for my table is:
select player_win_id, max(cnt)
from (select player_win_id, count(*) as cnt
from (select *,
row_number() over (order by id) as seqnum,
row_number() over (partition by player_win_id order by id) as seqnum_p
from MatchResults ) t
group by player_win_id, (seqnum - seqnum_p)
) p
group by player_win_id;
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
The source table:
id num
-------------------
1 1
2 1
3 1
4 2
5 2
6 1
The output:(appear at least 2 times)
num times
--------------
1 3
2 2
Based on the addition logic defined in the comments it appears this is what you're after:
WITH YourTable AS(
SELECT V.id,
V.num
FROM (VALUES(1,1),
(2,1),
(3,1),
(4,2),
(5,2),
(6,1),
(7,1))V(id,num)), --Added extra row due to logic defined in comments
Grps AS(
SELECT YT.id,
YT.num,
ROW_NUMBER() OVER (ORDER BY id) -
ROW_NUMBER() OVER (PARTITION BY Num ORDER BY id) AS Grp
FROM YourTable YT),
Counts AS(
SELECT num,
COUNT(num) AS Times
FROM grps
GROUP BY grp,
num)
SELECT num,
MAX(times) AS times
FROM Counts
GROUP BY num;
This uses a CTE and ROW_NUMBER to define the groups, and then an additional CTE to get the COUNT per group. Finally you can then get the MAX COUNT per num.
I would adress this with a gaps-and-islands technique:
select num, max(cnt)
from (
select num, count(*) cnt
from (
select
id,
num,
row_number() over(order by id) rn1,
row_number() over(partition by num order by id) rn2
from mytable
) t
group by num, rn1 - rn2
) t
group by num
The most inner query computes row numbers over the whole table and within num groups; the difference between the row numbers gives you the group of adjacent records that each record belong to (you can run that subquery independently and follow how the difference evolves to understand more).
Then, the next level count the number of records in each group of adjacent records. The most outer query takes the maximum count of adjacent records in for each num.
Demo on DB Fiddle:
num | (No column name)
--: | ---------------:
1 | 3
2 | 2
this will work for you
select num,count(num) times from Tabl
group by num