I have a table:
table1
tran_id user_id start_date end_date
1 100 01-04-2018 02-04-2018
2 100 14-06-2018 14-06-2018
4 100 12-06-2018 12-06-2018
7 101 05-01-2018 05-01-2018
9 101 08-01-2018 08-01-2018
3 101 03-01-2018 03-01-2018
Date format is DD-MM-YYYY
I need to find the day difference ordered by user_id and start_date. The formula to find day difference is to subtract end_date of the prior record and start_date of the next.
Here is the expected output:
tran_id user_id start_date end_date day_diff
1 100 01-04-2018 02-04-2018 71
4 100 12-06-2018 12-06-2018 2
2 100 14-06-2018 14-06-2018 0
3 101 03-01-2018 03-01-2018 2
7 101 05-01-2018 05-01-2018 3
9 101 08-01-2018 08-01-2018 0
How to get this in an SQL query?
Use function lead() to find next value, then substract dates:
select tran_id, user_id, start_date, end_date, nvl(nsd - start_date, 0) diff
from (
select t.*, lead(start_date) over (partition by user_id
order by start_date) nsd
from table1 t)
demo
You can use lead() for this. Because lead() takes three arguments, you don't need a subquery or to deal with null values:
select t.*,
(lead(start_date, 1, start_date) over (partition by user_id order by start_date) -
start_date
) as diff
from t;
I think this: DATEDIFF function in Oracle in combination with the LAG function should help you out.
A minor correction to the answer given by Ponder Stibbons. Difference highlighted in Bold.
select tran_id, user_id, start_date, end_date, nvl(nsd - end_date, 0) diff
from (
select t.*, lead(start_date) over (partition by user_id
order by start_date) nsd
from date_test t);
Regards
Akash
Related
I am new SQL coding using in SQL developer.
I have a table that has 4 columns: Patient ID (ptid), service date (dt), insurance payment amount (insr_amt), out of pocket payment amount (op_amt). (see table 1 below)
What I would like to do is (1) create two columns "start_dt" and "end_dt" using the "dt" column where if there are no gaps in the date by the patient ID then populate the start and end date with the first and last date by patient ID, however if there is a gap in service date within the patient ID then to create the separate start and end date rows per patient ID, along with (2) summing the two payment amounts by patient ID with in the one set of start and end date visits (see table 2 below).
What would be the way to run this using SQL code in SQL developer?
Thank you!
Table 1:
Ptid
dt
insr_amt
op_amt
A
1/1/2021
30
20
A
1/2/2021
30
10
A
1/3/2021
30
10
A
1/4/2021
30
30
B
1/6/2021
10
10
B
1/7/2021
20
10
C
2/1/2021
15
30
C
2/2/2021
15
30
C
2/6/2021
60
30
Table 2:
Ptid
start_dt
end_dt
total_insr_amt
total_op_amt
A
1/1/2021
1/4/2021
120
70
B
1/6/2021
1/7/2021
30
20
C
2/1/2021
2/2/2021
30
60
C
2/6/2021
2/6/2021
60
30
You didn't mention the specific database so this solution works in PostgreSQL. You can do:
select
ptid,
min(dt) as start_dt,
max(dt) as end_dt,
sum(insr_amt) as total_insr_amt,
sum(op_amt) as total_op_amt
from (
select *,
sum(inc) over(partition by ptid order by dt) as grp
from (
select *,
case when dt - interval '1 day' = lag(dt) over(partition by ptid order by dt)
then 0 else 1 end as inc
from t
) x
) y
group by ptid, grp
order by ptid, grp
Result:
ptid start_dt end_dt total_insr_amt total_op_amt
----- ---------- ---------- -------------- -----------
A 2021-01-01 2021-01-04 120 70
B 2021-01-06 2021-01-07 30 20
C 2021-02-01 2021-02-02 30 60
C 2021-02-06 2021-02-06 60 30
See running example at DB Fiddle 1.
EDIT for Oracle
As requested, the modified query that works in Oracle is:
select
ptid,
min(dt) as start_dt,
max(dt) as end_dt,
sum(insr_amt) as total_insr_amt,
sum(op_amt) as total_op_amt
from (
select x.*,
sum(inc) over(partition by ptid order by dt) as grp
from (
select t.*,
case when dt - 1 = lag(dt) over(partition by ptid order by dt)
then 0 else 1 end as inc
from t
) x
) y
group by ptid, grp
order by ptid, grp
See running example at db<>fiddle 2.
I have the following table,
id status price date
2 complete 10 2020-01-01 10:10:10
2 complete 20 2020-02-02 10:10:10
2 complete 10 2020-03-03 10:10:10
3 complete 10 2020-04-04 10:10:10
4 complete 10 2020-05-05 10:10:10
Required output,
id status_count price ratio
2 0 0 0
2 1 10 0
2 2 30 0.33
I am looking to add the price for previous row. Row 1 is 0 because it has no previous row value.
Find ratio ie 10/30=0.33
You can use analytical function ROW_NUMBER and SUM as follows:
SELECT
id,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) - 1 AS status_count,
COALESCE(SUM(price) OVER (PARTITION BY id ORDER BY date), 0) - price as price
FROM yourTable;
DB<>Fiddle demo
I think you want something like this:
SELECT
id,
COUNT(*) OVER (PARTITION BY id ORDER BY date) - 1 AS status_count,
COALESCE(SUM(price) OVER (PARTITION BY id
ORDER BY date ROWS BETWEEN
UNBOUNDED PRECEDING AND 1 PRECEDING), 0) price
FROM yourTable;
Demo
Please also check another method:
with cte
as(*,ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) - 1 AS status_count,
SUM(price) OVER (PARTITION BY id ORDER BY date) ss from yourTable)
select id,status_count,isnull(ss,0)-price price
from cte
How can we use this data to calculate the activity of each user in each period?
user_id date status
101 2018-01-1 1
101 2018-01-2 0
101 2018-01-3 1
101 2018-01-4 1
101 2018-01-5 1
Output:
user_id start_date end_date status length
101 2018-01-1 2018-01-1 1 1
101 2018-01-2 2018-01-2 0 1
101 2018-01-3 2018-01-5 1 3
This is a gaps-and-islands problem. You can group them together using by subtracting a sequence:
select user_id, status, min(date), max(date),
julianday(max(date)) - julianday(min(date)) as length
from (select t.*,
row_number() over (partition by user_id, status order by date) as seqnum
from t
) t
group by user_id, status, date(date, '-' || seqnum || ' day')
order by user_id, length;
Here is a db<>fiddle.
I have a table that looks like this:
Using the LAG function in SQL, I would like to perform the LAG on only values where star_date=end_date and get the past previous start_date record where start_date=end_date.
That my end table will have an extra column like this:
I hope my question is clear, any help is appreciated.
You can assign a group to these values and use that:
select t.*,
(case when start_date = end_date
then lag(start_date) over (partition by (case when start_date = end_date then 1 else 0 end) order by start_date)
end) as prev_eq_start_date
from t;
Or:
select t.*,
(case when start_date = end_date
then lag(start_date) over (partition by start_date = end_date order by start_date)
end) as prev_eq_start_date
from t;
Note if you data is big and most rows have different dates, then you might have a resources issue. In this case, an additional, unused partition by key can help:
select t.*,
(case when start_date = end_date
then lag(start_date) over (partition by (case when start_date = end_date then 1 else 2 end), (case when start_date <> end_date then start_date end) order by start_date)
end) as prev_eq_start_date
from t;
This has no impact on the result but it can avoid a resources error caused by too many rows with different values.
Below is for BigQuery Standard SQL
#standardSQL
SELECT *, NULL AS lag_result
FROM `project.dataset.table` WHERE start_date != end_date
UNION ALL
SELECT *, LAG(start_date) OVER(ORDER BY start_date)
FROM `project.dataset.table` WHERE start_date = end_date
If to apply to sample data in your question - result is
Row user_id start_date end_date lag_result
1 1 2019-01-01 2019-02-28 null
2 3 2019-02-27 2019-02-28 null
3 4 2019-08-04 2019-09-01 null
4 2 2019-02-01 2019-02-01 null
5 5 2019-08-07 2019-08-07 2019-02-01
6 6 2019-08-27 2019-08-27 2019-08-07
Btw, in case if your start_date and end_date are of STRING data type ('27/02/2019') vs. DATE type ('2019-02-27' as it was assumed in above query) - you should use below one
#standardSQL
SELECT *, NULL AS lag_result
FROM `project.dataset.table` WHERE start_date != end_date
UNION ALL
SELECT *, LAG(start_date) OVER(ORDER BY PARSE_DATE('%d/%m/%Y', start_date))
FROM `project.dataset.table` WHERE start_date = end_date
with result
Row user_id start_date end_date lag_result
1 1 01/01/2019 28/02/2019 null
2 3 27/02/2019 28/02/2019 null
3 4 04/08/2019 01/09/2019 null
4 2 01/02/2019 01/02/2019 null
5 5 07/08/2019 07/08/2019 01/02/2019
6 6 27/08/2019 27/08/2019 07/08/2019
Use JOIN
SQL FIDDLE
SELECT T.*,T1.LAG_Result
FROM TABLE T LEFT JOIN
(
SELECT User_Id,LAG(start_date) OVER(ORDER BY start_date) LAG_Result
FROM TABLE S
WHERE start_date = end_date
) T1 ON T.User_Id = T1.User_Id
Database my_table:
id seq start_date end_date
1 1 01-01-2017 02-01-2017
1 2 07-01-2017 09-01-2017
1 3 11-01-2017 11-01-2017
2 1 20-01-2017 20-01-2017
3 1 01-02-2017 02-02-2017
3 2 03-02-2017 04-02-2017
3 3 08-01-2017 09-02-2017
3 4 09-01-2017 10-02-2017
3 5 10-01-2017 12-02-2017
My requirement is to get the first date (normally seq 1 start date) and end date (normally last seq end date) and the number of dates occurred during all seq for each unique ID.
Date occurred:
id 1 2 3
01-01-2017 20-01-2017 01-02-2017
02-01-2017 02-02-2017
07-01-2017 03-02-2017
08-01-2017 04-02-2017
09-01-2017 08-02-2017
11-01-2017 09-02-2017
10-02-2017
11-02-2017
12-02-2017
total 6 1 9
Here is the result I want:
id start_date end_date num_date
1 01-01-2017 11-01-2017 6
2 20-01-2017 20-01-2017 1
3 01-02-2017 12-02-2017 9
I have tried
SELECT id
, MIN(start_date)
, MAX(end_date)
, SUM(end_date - start_date + 1)
FROM my_table
GROUP BY id
and this SQL statement work fine in id 1 and 2 since there is no overlap date between begin date and end date. But for id 3, the result num_date is 11. Could you please suggest the SQL statement to solve this problem? Thank you.
One more question: The date in database is in datetime format. How do I convert it to date. I tried to use TRUNC function but it sometimes convert date to yesterday instead.
You need to count how many times an end_date equals the following start_date. For this you need to use the lag() or the lead() analytic function. You can use a case expression for the comparison, but alas you can't wrap the case expression within a COUNT or SUM in the same query; you need a subquery and an outer query.
Something like this; not tested, since you didn't provide CREATE TABLE and INSERT statements to recreate your sample data.
select id, min(start_date) as start_date, max(end_date) as end_date,
sum(end_date - start_date + 1 - flag) as num_days
from ( select id, start_date, end_date,
case when start_date = lag(end_date)
over (partition by id order by end_date) then 1
else 0 end as flag
from my_table
)
group by id;
SELECT id,
MIN( start_date ) AS start_date,
MAX( end_date ) AS end_date,
SUM( end_date - start_date + 1 ) AS num_days
FROM (
SELECT id,
GREATEST(
start_date,
COALESCE(
LAG( end_date ) OVER ( PARTITION BY id ORDER BY seq ) + 1,
start_date
)
) AS start_date,
end_date
FROM your_table
)
WHERE start_date <= end_date
GROUP BY id;