I have data like following
ID SalesTime Qty Unit Price Item
1 01/01/2021 08:10:00 10 10 A
2 01/01/2021 11:30:00 2 9 B
3 01/01/2021 11:59:50 1 8 C
4 01/02/2021 13:00:00 5 15 D
5 01/03/2021 10:00:00 4 10 A
6 01/03/2021 12:00:00 5 9 B
7 01/03/2021 12:50:00 6 15 D
8 01/04/2021 10:50:00 5 8 C
9 01/04/2021 11:10:00 2 10 A
10 ............
I wanna summarize the total into the form,
for example:
Mon Tue Wed Thu Fri Sat Sun
08:00~09:59 20 21 50 100 60 70 210
10:00~11:59 60 25 60 90 75 80 200
12:00~13:59 100 10 50 60 70 50 150
How to do that in MS SQL, thanks a lot.
You can extract the hour and divide by two for the rows. And then use conditional aggregation for the columns. Assuming you want the total of the price times quantity:
select convert(time, dateadd(hour, 2 * (datepart(hour, salestime) / 2), 0)) as hh,
sum(case when datename(weekday, salestime) = 'Monday' then qty * unit_price end) as mon,
sum(case when datename(weekday, salestime) = 'Tuesday' then qty * unit_price end) as tue,
. . .
from t
group by datepart(hour, salestime) / 2
order by min(salestime);
Note: This just returns the beginning of the time period, rather than the full range.
Related
I have the following table and I want to get the specidic Amount per loan_ID that corresponds to the earliest observation with greater than or equal to 10 dpd per month.
Loan_ID date dpd Amount
1 1/1/2017 1 55
1 1/2/2017 2 100
1 1/3/2017 3 5000
1 1/4/2017 5 6000
1 1/5/2017 10 50000
1 1/6/2017 15 50001
1 1/9/2017 31 50004
1 1/10/2017 55 50005
1 1/11/2017 59 50006
1 1/12/2017 65 50007
1 1/13/2017 70 80000
1 1/20/2017 85 900000
1 1/29/2017 92 100000
1 1/30/2017 93 10000
2 1/1/2017 0 522
2 1/2/2017 8 5444
2 1/3/2017 12 8784
2 1/6/2017 15 6221
2 1/12/2017 18 2220
2 1/13/2017 20 177
2 1/29/2017 35 5151
2 1/30/2017 60 40000
2 1/31/2017 61 5500
The expected output:
Loan_ID Month Amount
1 1 50000
2 1 8784
SELECT DISTINCT ON ("Loan_ID", date_trunc('month', "date"))
"Loan_ID",
date_trunc('month', "date")::date as month,
"Amount"
FROM
loans
WHERE
dpd >= 10
ORDER BY
"Loan_ID",
date_trunc('month', "date"),
"date"
;
Returns:
Loan_ID
month
Amount
1
2017-01-01
50000
2
2017-01-01
8784
You can find test case in db<>fiddle
Hmmm . . . if you want the amount per month and the first date that matches the condition, then you want conditional aggregation:
select loan_id, date_trunc('month', date) as mon,
sum(dpd),
min(case when dpd >= 10 then dpd end) as first_dpd_10
from t
group by load_id, mon;
Edit: Based on your comment, you can use distinct on:
select distinct on (loan_id, date_trunc('month', date)) t.*
min(case when dpd >= 10 then dpd end) as first_dpd_10
from t
where dpd >= 10
order by load_id, date_trunc('month', date), date
I have a running balance sheet showing customer balances after inflows and (outflows) by date. It looks something like this:
ID DATE AMOUNT RUNNING AMOUNT
-- ---------------- ------- --------------
10 27/06/2019 14:30 100 100
10 29/06/2019 15:26 -100 0
10 03/07/2019 01:56 83 83
10 04/07/2019 17:53 15 98
10 05/07/2019 15:09 -98 0
10 05/07/2019 15:53 98.98 98.98
10 05/07/2019 19:54 -98.98 0
10 07/07/2019 01:36 90.97 90.97
10 07/07/2019 13:02 -90.97 0
10 07/07/2019 16:32 39.88 39.88
10 08/07/2019 13:41 50 89.88
20 08/01/2019 09:03 890.97 890.97
20 09/01/2019 14:47 -91.09 799.88
20 09/01/2019 14:53 100 899.88
20 09/01/2019 14:59 -399 500.88
20 09/01/2019 18:24 311 811.88
20 09/01/2019 23:25 50 861.88
20 10/01/2019 16:18 -861.88 0
20 12/01/2019 16:46 894.49 894.49
20 25/01/2019 05:40 -871.05 23.44
I have attempted using lag() but I seem not to understand how to use it yet.
SELECT ID, MEDIAN(DIFF) MEDIAN_AGE
FROM
(
SELECT *, DATEDIFF(day, Lag(DATE, 1) OVER(ORDER BY ID), DATE
)AS DIFF
FROM TABLE 1
WHERE RUNNING AMOUNT = 0
)
GROUP BY ID;
The expected result would be:
ID MEDIAN_AGE
-- ----------
10 1
20 2
Please help in writing out the query that gives the expected result.
As already pointed out, you are using syntax that isn't valid for Oracle, including functions that don't exist and column names that aren't allowed.
You seem to want to calculate the number of days between a zero running-amount and the following non-zero running-amount; lead() is probably easier than lag() here, and you can use a case expression to only calculate it when needed:
select id, date_, amount, running_amount,
case when running_amount = 0 then
lead(date_) over (partition by id order by date_) - date_
end as diff
from your_table;
ID DATE_ AMOUNT RUNNING_AMOUNT DIFF
---------- -------------------- ---------- -------------- ----------
10 2019-06-27 14:30:00 100 100
10 2019-06-29 15:26:00 -100 0 3.4375
10 2019-07-03 01:56:00 83 83
10 2019-07-04 17:53:00 15 98
10 2019-07-05 15:09:00 -98 0 .0305555556
10 2019-07-05 15:53:00 98.98 98.98
10 2019-07-05 19:54:00 -98.98 0 1.2375
10 2019-07-07 01:36:00 90.97 90.97
10 2019-07-07 13:02:00 -90.97 0 .145833333
10 2019-07-07 16:32:00 39.88 39.88
10 2019-07-08 13:41:00 50 89.88
20 2019-01-08 09:03:00 890.97 890.97
20 2019-01-09 14:47:00 -91.09 799.88
20 2019-01-09 14:53:00 100 899.88
20 2019-01-09 14:59:00 -399 500.88
20 2019-01-09 18:24:00 311 811.88
20 2019-01-09 23:25:00 50 861.88
20 2019-01-10 16:18:00 -861.88 0 2.01944444
20 2019-01-12 16:46:00 894.49 894.49
20 2019-01-25 05:40:00 -871.05 23.44
Then use the median() function, rounding if desired to get your expected result:
select id, median(diff) as median_age, round(median(diff)) as median_age_rounded
from (
select id, date_, amount, running_amount,
case when running_amount = 0 then
lead(date_) over (partition by id order by date_) - date_
end as diff
from your_table
)
group by id;
ID MEDIAN_AGE MEDIAN_AGE_ROUNDED
---------- ---------- ------------------
10 .691666667 1
20 2.01944444 2
db<>fiddle
I have a list of customers with a date of joining and a date of leaving,
I have to know each month by year how many joined and how many left and what the summary
id Join left
1 01/01/2017 08/03/2017
2 02/01/2017 25/03/2017
3 03/01/2017 06/03/2017
4 04/01/2017
5 30/01/2017
6 31/01/2017 05/05/2017
7 01/02/2017
8 02/02/2017 22/03/2017
9 04/02/2017 29/04/2017
10 05/02/2017 09/04/2017
11 06/02/2017 08/04/2017
12 07/02/2017 13/03/2017
13 04/03/2017 21/05/2017
14 05/03/2017
15 06/03/2017
16 07/03/2017
17 09/03/2017
18 10/03/2017 03/06/2017
19 11/03/2017 14/04/2017
20 12/03/2017 31/05/2017
21 07/04/2017 06/07/2017
22 08/04/2017 16/06/2017
23 09/04/2017 10/05/2017
24 04/03/2018 26/05/2018
25 24/03/2018 01/06/2018
26 25/03/2018 15/06/2018
27 26/03/2018 05/05/2018
28 27/03/2018 02/07/2018
29 04/04/2018
30 05/04/2018 13/06/2018
And that is how the desired result appears
total left join month year
6 0 6 1 2017
6 0 6 2
3 5 8 3
-1 4 3 4
-4 4 0 5
-2 2 0 6
-1 1 0 7
3 2 5 3 2018
2 0 2 4
0 0 0 5
-3 3 0 6
-1 1 0 7
You can try this if your database is either MySQL or SQL server. For other databases, you can use the logic/idea.
SELECT
SUM(CASE WHEN Type = 'J' THEN C ELSE 0 END) - SUM(CASE WHEN Type = 'L' THEN C ELSE 0 END) AS [Total],
SUM(CASE WHEN Type = 'L' THEN C ELSE 0 END) AS [left],
SUM(CASE WHEN Type = 'J' THEN C ELSE 0 END) AS [join],
M Month,
Y Year
FROM
(
SELECT 'J' AS [Type],
MONTH(CONVERT(DATETIME, [Join], 103)) M,
YEAR(CONVERT(DATETIME, [Join], 103)) Y,
COUNT(ID) C
FROM customers
GROUP BY MONTH(CONVERT(DATETIME, [Join], 103)), YEAR(CONVERT(DATETIME, [Join], 103))
UNION ALL
SELECT 'L',
MONTH(CONVERT(DATETIME, [left], 103)) M,
YEAR(CONVERT(DATETIME, [left], 103)) Y,
COUNT(ID) C
FROM customers
GROUP BY MONTH(CONVERT(DATETIME, [left], 103)), YEAR(CONVERT(DATETIME, [left], 103))
)A
WHERE Y IS NOT NULL
GROUP BY M,Y
I am trying to add the total hours where flag =1. Here is how my data look like in the table. The hours are 30 minutes interval
ID FullDateTime Flag
22 2015-02-26 05:30:00.000 1
44 2015-02-26 05:00:00.000 1
25 2015-02-26 04:30:00.000 0
23 2015-02-26 04:00:00.000 1
74 2015-02-26 03:30:00.000 1
36 2015-02-26 03:00:00.000 0
here is what i tried but not working:
select DATEDIFF(minute, sum(FullDatetime), sum(FullDatetime)) / 60.0 as hours
from myTable
where flag = 1
I am expecting the results to be 2 hours.
If the total number of hours is just a function of the number of half hour periods flagged as 1 then a simple count(*) of the rows matching flag 1 multiplied by 0.5 (for the half hour) should do it:
select count(*) * 0.5 from myTable where flag = 1
I have a table in SQLite called param_vals_breaches that looks like the following:
id param queue date_time param_val breach_count
1 c a 2013-01-01 00:00:00 188 7
2 c b 2013-01-01 00:00:00 156 8
3 c c 2013-01-01 00:00:00 100 2
4 d a 2013-01-01 00:00:00 657 0
5 d b 2013-01-01 00:00:00 23 6
6 d c 2013-01-01 00:00:00 230 12
7 c a 2013-01-01 01:00:00 100 0
8 c b 2013-01-01 01:00:00 143 9
9 c c 2013-01-01 01:00:00 12 2
10 d a 2013-01-01 01:00:00 0 1
11 d b 2013-01-01 01:00:00 29 5
12 d c 2013-01-01 01:00:00 22 14
13 c a 2013-01-01 02:00:00 188 7
14 c b 2013-01-01 02:00:00 156 8
15 c c 2013-01-01 02:00:00 100 2
16 d a 2013-01-01 02:00:00 657 0
17 d b 2013-01-01 02:00:00 23 6
18 d c 2013-01-01 02:00:00 230 12
I want to write a query that will show me a particular queue (e.g. "a") with the average param_val and breach_count for each param on an hour by hour basis. So transposing the data to get something that looks like this:
Results for Queue A
Hour 0 Hour 0 Hour 1 Hour 1 Hour 2 Hour 2
param avg_param_val avg_breach_count avg_param_val avg_breach_count avg_param_val avg_breach_count
c xxx xxx xxx xxx xxx xxx
d xxx xxx xxx xxx xxx xxx
is this possible? I'm not sure how to go about it. Thanks!
SQLite does not have a PIVOT function but you can use an aggregate function with a CASE expression to turn the rows into columns:
select param,
avg(case when time = '00' then param_val end) AvgHour0Val,
avg(case when time = '00' then breach_count end) AvgHour0Count,
avg(case when time = '01' then param_val end) AvgHour1Val,
avg(case when time = '01' then breach_count end) AvgHour1Count,
avg(case when time = '02' then param_val end) AvgHour2Val,
avg(case when time = '02' then breach_count end) AvgHour2Count
from
(
select param,
strftime('%H', date_time) time,
param_val,
breach_count
from param_vals_breaches
where queue = 'a'
) src
group by param;
See SQL Fiddle with Demo