I have a table with duplicate rows and I need to extract these duplicate rows alone. Below is an example of the table I have:
my_table:
ID Offer
1 10
2 10
1 12
1 10
2 20
2 10
What I want next is to count the occurrence of the offer for each ID. i.e, my final result should be:
ID Offer Count
1 10 1
2 10 1
1 12 1
1 10 2
2 20 1
2 10 2
As you can see, the count should increase based on the number of times the offer shows up per ID.
I tried something like:
select id,offer,count(offer) over (partition by id);
But this just gives the total count of that particular offer for that ID and is not the result I am looking for.
Any help is much appreciated!
You could use ROW_NUMBER:
select id,offer,ROW_NUMBER() over (partition by id, offer order by rownum)
from tab
Related
I have a table below
days
balance
user_id
wanted column
2022/08/01
10
1
1
2022/08/02
11
1
1
2022/08/03
10
1
1
2022/08/03
0
2
1
2022/08/05
3
2
2
2022/08/06
3
2
2
2022/08/07
3
3
3
2022/08/08
0
2
3
since I'm new to SQL couldn't aggregate over window by clauses, correctly.
which means; I want to find unique users that have balance>0 per day.
thanks
update:
exact output wanted:
days
unque users
2022/08/01
1
2022/08/02
1
2022/08/03
1
2022/08/05
2
2022/08/06
2
2022/08/07
3
2022/08/08
3
update: how if I want to accumulate the number of unique users over time? with consideration of new users [means: counting users who didn't exist before], and the balance > 0
everyones help is appreaciated deeply :)
SELECT
*,
COUNT(DISTINCT CASE WHEN balance > 0 THEN USER_ID END) OVER (ORDER BY days)
FROM
your_table
View of a table
ID
kWh
1
3
1
10
1
8
1
11
2
12
2
4
2
7
2
8
3
3
3
4
3
5
I want to recive
ID
kWh
1
32
2
31
3
12
The table itself is more complex and larger. But the point is this. How can this be done? And I can't know in advance the ID numbers of the first column.
SELECT T.ID,SUM(T.KWH)SUM_KWH
FROM YOUR_TABLE T
GROUP BY T.ID
Do you need this one?
Let's assume your database name is 'testdb' and table name is 'table1'.
SELECT * FROM testdb.table1;
SELECT id, SUM(kwh) AS "kwh2"
FROM stack.table1
WHERE id = 1
keep running the query will all (ids). you will get output.
By following this query you will get desired output.
Hope this helps.
I currently have two tables, users and coupons
id
first_name
1
Roberta
2
Oliver
3
Shayna
4
Fechin
id
discount
user_id
1
20%
1
2
40%
2
3
15%
3
4
30%
1
5
10%
1
6
70%
4
What I want to do is select from the coupons table until I've selected X users.
so If I chose X = 2 the resulting table would be
id
discount
user_id
1
20%
1
2
40%
2
4
30%
1
5
10%
1
I've tried using both dense_rank and row_number but they return the count of occurrences of each user_id not it's order.
SELECT id,
discount,
user_id,
dense_rank() OVER (PARTITION BY user_id)
FROM coupons
I'm guessing I need to do it in multiple subqueries (which is fine) where the first subquery would return something like
id
discount
user_id
order_of_occurence
1
20%
1
1
2
40%
2
2
3
15%
3
3
4
30%
1
1
5
10%
1
1
6
70%
4
4
which I can then use to filter by what I need.
PS: I'm using postgresql.
You've stated that you want to parameterize the query so that you can retrieve X users. I'm reading that as all coupons for the first X distinct user_ids in coupon id column order.
It appears your attempt was close. dense_rank() is the right idea. Since you want to look over the entire table you can't use partition by. And a sorting column is also required to determine the ranking.
with data as (
select *,
dense_rank() over (order by id) as dr
from coupons
)
select * from data where dr <= <X>;
I have data that looks like this:
ID num_of_days
1 0
2 0
2 8
2 9
2 10
2 15
3 10
3 20
I want to add another column that increments in value only if the num_of_days column is divisible by 5 or the ID number increases so my end result would look like this:
ID num_of_days row_num
1 0 1
2 0 2
2 8 2
2 9 2
2 10 3
2 15 4
3 10 5
3 20 6
Any suggestions?
Edit #1:
num_of_days represents the number of days since the customer last saw a doctor between 1 visit and the next.
A customer can see a doctor 1 time or they can see a doctor multiple times.
If it's the first time visiting, the num_of_days = 0.
SQL tables represent unordered sets. Based on your question, I'll assume that the combination of id/num_of_days provides the ordering.
You can use a cumulative sum . . . with lag():
select t.*,
sum(case when prev_id = id and num_of_days % 5 <> 0
then 0 else 1
end) over (order by id, num_of_days)
from (select t.*,
lag(id) over (order by id, num_of_days) as prev_id
from t
) t;
Here is a db<>fiddle.
If you have a different ordering column, then just use that in the order by clauses.
I have the following table structure.
ITEM TOTAL
----------- -----------------
ID | TITLE ID |ITEMID|VALUE
1 A 1 2 6
2 B 2 1 4
3 C 3 3 3
4 D 4 3 8
5 E 5 1 2
6 F 6 5 4
7 4 5
8 2 8
9 2 7
10 1 3
11 2 2
12 3 6
I am using Apache Derby DB. I need to perform the average calculation in SQL. I need to show the list of item IDs and their average total of the last 3 records.
That is, for ITEM.ID 1, I will go to TOTAL table and select the last 3 records of the rows which are associated with the ITEMID 1. And take average of them. In Derby database, I am able to do this for a given item ID but I cannot make it without giving a specific ID. Let me show you what I've done it.
SELECT ITEM.ID, AVG(VALUE) FROM ITEM, TOTAL WHERE TOTAL.ITEMID = ITEM.ID GROUP BY ITEM.ID
This SQL gives the average of all items in a list. But this calculates for all values of the total tables. I need last 3 records only. So I changed the SQL to this:
SELECT AVG(VALUE) FROM (SELECT ROW_NUMBER() OVER() AS ROWNUM, TOTAL.* FROM TOTAL WHERE ITEMID = 1) AS TR WHERE ROWNUM > (SELECT COUNT(ID) FROM TOTAL WHERE ITEMID = 1) - 3
This works if I supply the item ID 1 or 2 etc. But I cannot do this for all items without giving an item ID.
I tried to do the same thing in ORACLE using partition and it worked. But derby does not support partitioning. There is WINDOW but I could not make use of it.
Oracle one
SELECT ITEMID, AVG(VALUE) FROM(SELECT ITEMID, VALUE, COUNT(*) OVER (PARTITION BY ITEMID) QTY, ROW_NUMBER() OVER (PARTITION BY ITEMID ORDER BY ID) IDX FROM TOTAL ORDER BY ITEMID, ID) WHERE IDX > QTY -3 GROUP BY ITEMID ORDER BY ITEMID
I need to use derby DB for its portability.
The desired output is this
RESULT
-----------------
ITEMID | AVERAGE
1 (9/3)
2 (17/3)
3 (17/3)
4 (5/1)
5 (4/1)
6 NULL
As you have noticed, Derby's support for the SQL 2003 "OLAP Operations" support is incomplete.
There was some initial work (see https://wiki.apache.org/db-derby/OLAPOperations), but that work was only partially completed.
I don't believe anyone is currently working on adding more functionality to Derby in this area.
So yes, Derby has a row_number function, but no, Derby does not (currently) have partition by.