I am new to SQL and was not able to solve the following problem: I've got a column of names [name]), a column of integer values that I wanna sum up ([Values]) and another column of integer values ([Day]). I want to sum up the values grouped by name for each day. So for example if there is a name "Chris" with value 4 on day 1 and there is another entry "Chris" with value 2 on day 3, I want to show the sum of chris on day_1 (4) and on day_2 (4+2=6).
As in the example above ("chris") I wanna sum them up, showing the sum for each name on each day (the sum from day 1 until day x).
I was only able to sum up the values for each name per day (see code below) but this is not what I am searching for since I need to keep the structure of the database. Therefore I need to show the sum for each value in every row in a further column.
select name, day,
sum(value) over (partition by name order by day) total
from tablename
There is a table below showing what I want to achieve.
With sum() window function:
select *,
sum(value) over (partition by name order by day) SumValue,
sum(value2) over (partition by name2 order by day) SumValue2
from tablename
order by day
or:
select name, day, value, name2, value2,
sum(value) over (partition by name order by day) SumValue,
sum(value2) over (partition by name2 order by day) SumValue2
from (select *, row_number() over (order by day) rn from tablename) as t
order by rn
if you want to preserve the original sequence of the rows.
See the demo.
Results:
> name | day | value | name2 | value2 | SumValue | SumValue2
> :---- | --: | ----: | :---- | -----: | -------: | --------:
> Chris | 1 | 2 | Paul | 5 | 2 | 5
> Alice | 1 | 5 | Ken | 4 | 5 | 4
> Paul | 2 | 8 | Alice | 1 | 8 | 1
> Ken | 2 | 4 | Chris | 2 | 4 | 2
> Alice | 3 | 3 | Ken | 3 | 8 | 7
> Chris | 3 | 6 | Paul | 0 | 8 | 5
Related
I have a SQL table on Impala that contains ID, dt (monthly basis with no skipped month), and status of each person ID. I want to check how long that each ID is in each status (my expected answer is shown on expected column)
I tried to solve this problem on the value column by using
count(status) over (partition by ID, status order by dt)
but it doesn't reset the value when the status is changed.
+------+------------+--------+-------+----------+
| ID | dt | status | value | expected |
+------+------------+--------+-------+----------+
| 0001 | 01/01/2020 | 0 | 1 | 1 |
| 0001 | 01/02/2020 | 0 | 2 | 2 |
| 0001 | 01/03/2020 | 1 | 1 | 1 |
| 0001 | 01/04/2020 | 1 | 2 | 2 |
| 0001 | 01/05/2020 | 1 | 3 | 3 |
| 0001 | 01/06/2020 | 0 | 3 | 1 |
| 0001 | 01/07/2020 | 1 | 4 | 1 |
| 0001 | 01/08/2020 | 1 | 5 | 2 |
+------+------------+--------+-------+----------+
Is there anyway to reset the counter when the status is changed?
When you partition by ID and status, two groups are formed for the values 0 and 1 in status field. So, the months 1, 2, 6 go into first group with 0 status and the months 3, 4, 5, 7, 8 go into the second group with 1 status. Then, the count function counts the number of statuses individually in those groups. Thus the first group has counts from 1 to 3 and the second group has counts from 1 to 5. This query so far doesn't account for the change in statuses rather just simply divide the record set as per different status values.
One approach would be to divide the records into different blocks where each status change starts a new block. The below query follows this approach and gives the expected result:
SELECT ID,dt,status,
COUNT(status) OVER(PARTITION BY ID,block_number ORDER BY dt) as value
FROM (
SELECT ID,dt,status,
SUM(change_in_status) OVER(PARTITION BY ID ORDER BY dt) as block_number
FROM(
SELECT ID,dt,status,
CASE WHEN
status<>LAG(status) OVER(PARTITION BY ID ORDER BY dt)
OR LAG(status) OVER(PARTITION BY ID ORDER BY dt) IS NULL
THEN 1
ELSE 0
END as change_in_status
FROM statuses
) derive_status_changes
) derive_blocks;
Here is a working example in DB Fiddle.
I am new to SQL and trying to write a statement similar to a 'for loop' in other languages and am stuck. I want to filter out rows of the table where for all of attribute 1, attribute2=attribute3 without using functions.
For example:
| Year | Month | Day|
| 1 | 1 | 1 |
| 1 | 2 | 2 |
| 1 | 4 | 4 |
| 2 | 3 | 4 |
| 2 | 3 | 3 |
| 2 | 4 | 4 |
| 3 | 4 | 4 |
| 3 | 4 | 4 |
| 3 | 4 | 4 |
I would only want the row
| Year | Month | Day|
|:---- |:------:| -----:|
| 3 | 4 | 4 |
because it is the only where month and day are equal for all of the values of year they share.
So far I have
select year, month, day from dates
where month=day
but unsure how to apply the constraint for all of year
-- month/day need to appear in aggregate functions (since they are not in the GROUP BY clause),
-- but the HAVING clause ensure we only have 1 month/day value (per year) here, so MIN/AVG/SUM/... would all work too
SELECT year, MAX(month), MAX(day)
FROM my_table
GROUP BY year
HAVING COUNT(DISTINCT (month, day)) = 1;
year
max
max
3
4
4
View on DB Fiddle
So one way would be
select distinct [year], [month], [day]
from [Table] t
where [month]=[day]
and not exists (
select * from [Table] x
where t.[year]=x.[year] and t.[month] <> x.[month] and t.[day] <> x.[day]
)
And another way would be
select distinct [year], [month], [day] from (
select *,
Lead([month],1) over(partition by [year] order by [month])m2,
Lead([day],1) over(partition by [year] order by [day])d2
from [table]
)x
where [month]=m2 and [day]=d2
The following is a snippet of my table...
My table has a lot of more users and higher order_rank
I'm trying to get the number of visits leading up to that order_rank in postgres.
So the result I'm trying to generate looks like...
I would address this as a gaps-and-island problem, where each island ends with a visit. You want the end of each island, along with the count of preceding records in the same island.
You can define the group with a window count of non-null values that starts from the end of the table. Then, just use that information to count how many records belong to each group:
select *
from (
select t.*,
count(*) over(partition by customer_id, grp) - 1 as number_of_visits
from (
select t.*,
count(order_rank) over(partition by customer_id order by visit_time desc) grp
from mytable t
) t
) t
where order_rank is not null
Demo on DB Fiddle:
customer_id | visit_time | txn_flag | order_rank | grp | number_of_visits
----------: | :--------- | -------: | ---------: | --: | ---------------:
123 | 2020-01-04 | 1 | 1 | 3 | 3
123 | 2020-01-06 | 1 | 2 | 2 | 1
123 | 2020-01-11 | 1 | 3 | 1 | 4
Apologies for the confusing title, I was unsure how to phrase it.
Below is my dataset:
+----+-----------------------------+--------+
| Id | Date | Amount |
+----+-----------------------------+--------+
| 1 | 2019-02-01 12:14:08.8056282 | 10 |
| 1 | 2019-02-04 15:23:21.3258719 | 10 |
| 1 | 2019-02-06 17:29:16.9267440 | 15 |
| 1 | 2019-02-08 14:18:14.9710497 | 10 |
+----+-----------------------------+--------+
It is an example of a bank trying to collect money from a debtor, where first, 10% of the owed sum is attempted to be collected, if a card is managed to be charged 15% is attempted, if that throws an error (for example insufficient funds), 10% is attempted again.
The desired output would be:
+----+--------+---------+
| Id | Amount | Attempt |
+----+--------+---------+
| 1 | 10 | 1 |
| 1 | 15 | 2 |
| 1 | 10 | 3 |
+----+--------+---------+
I have tried:
SELECT Id, Amount
FROM table1
GROUP BY Id, Amount
I am struggling to create a new column based on when value changes in the Amount column as I assume that could be used as another grouping variable that could fix this.
If you just want when a value changes, use lag():
select t.id, t.amount,
row_number() over (partition by id order by date) as attempt
from (select t.*, lag(amount) over (partition by id order by date) as prev_amount
from table1 t
) t
where prev_amount is null or prev_amount <> amount
I'm trying to rank the rows in the following table that looks like this:
| ID | Key | Date | Row|
*****************************
| P175 | 5 | 2017-01| 2 |
| P175 | 5 | 2017-02| 2 |
| P175 | 5 | 2017-03| 2 |
| P175 | 12 | 2017-03| 1 |
| P175 | 12 | 2017-04| 1 |
| P175 | 12 | 2017-05| 1 |
This person has two Keys at once during 2017-03, but I want the formula to put '1' for the rows where Key=12 since it reflects the most recent records.
I want the same formula to also work for the people who don't have overlapping Keys, putting '1' for the most recent records:
| ID | Key | Date | Row|
*****************************
| P170 | 8 | 2017-01| 2 |
| P170 | 8 | 2017-02| 2 |
| P170 | 8 | 2017-03| 2 |
| P170 | 6 | 2017-04| 1 |
| P170 | 6 | 2017-05| 1 |
I've tried variations of ROW_NUMBER() OVER PARTITION BY and DENSE_RANK but cannot figure out the correct formula. Thanks for your help.
First calculate the max date for the key. Then use dense_rank():
select t.*,
dense_rank() over (partition by id order by max_date desc, key) as row
from (select t.*, max(date) over (partition by id, key) as max_date
from t
) t;
If the ranges for each key did not overlap, you could do this with a cumulative count distinct:
select t.*, count(distinct key) over (partition by id order by date desc) as rank
from t;
However, this would not work in the first case. I just find it interesting that this does almost the same thing as the first query.
I guess you are looking for something like this
select personid, mykey, month,
dense_rank() over (partition by personid order by mykey desc) rown
from personkeys
order by month
see the example
http://sqlfiddle.com/#!15/cf751/8