I have the below table and I need to aggregate
Id Month Days Hours Audit
1 201803 20 30 Yes
1 201803 20 15 Yes
1 201802 19 4 No
2 201803 20 5 Yes
Expected output:
Id Month Days Hours Audit Total
1 201803 20 2 Yes 100
1 201803 20 3 Yes 100
1 201802 10 4 No
2 201803 20 5 Yes 100
Summary:
Partition by ID & Month
Aggregate Days & Hours
My SQL: (my work)
SELECT (CASE
WHEN AUDIT IN ('YES')
THEN HOURS * DAYS
END) OVER (PARTITION BY ID ,c.month) AS TOTAL
FROM TABLEA
Use sum as the window function.
SELECT t.*,SUM(CASE WHEN AUDIT = 'YES' THEN HOURS * DAYS END)
OVER(PARTITION BY ID,month) AS TOTAL
FROM TABLEA t
Related
Date Amt ytd mtd
01-Jan-21 1 2 2
01-Jan-21 1 2 2
02-Jan-21 1 3 3
03-Jan-21 1 4 4
01-Feb-21 1 5 1
02-Feb-21 1 6 2
03-Feb-21 1 7 3
04-Feb-21 1 8 4
05-Feb-21 1 9 5
01-Mar-21 1 10 1
02-Mar-21 1 11 2
03-Mar-21 1 12 3
04-Mar-21 1 13 4
01-Apr-21 1 14 1
02-Apr-21 1 15 2
03-Apr-21 1 16 3
01-May-21 1 17 1
02-May-21 1 18 2
03-May-21 1 19 3
04-May-21 1 20 4
05-May-21 1 21 5
06-May-21 1 22 6
I have the first two columns (Date, Amt) and i need the YTD and MTD columns in MS SQL so that i can show the above table.
Seems like a rolling COUNT OVER was used to calculate the ytd & mtd in the Oracle source.
(Personally, I would prefere RANK or DENSE_RANK)
And since Oracle datestamps can be casted to a DATE as-is.
SELECT [Date], Amt
, ytd = COUNT(*) OVER (ORDER BY CAST([Date] AS DATE))
, mtd = COUNT(*) OVER (PARTITION BY EOMONTH(CAST([Date] AS DATE)) ORDER BY CAST([Date] AS DATE))
FROM your_table
ORDER BY CAST([Date] AS DATE)
Date
Amt
ytd
mtd
01-Jan-21
1
2
2
01-Jan-21
1
2
2
02-Jan-21
1
3
3
03-Jan-21
1
4
4
01-Feb-21
1
5
1
02-Feb-21
1
6
2
03-Feb-21
1
7
3
04-Feb-21
1
8
4
05-Feb-21
1
9
5
db<>fiddle here
I'm trying to get a running total as of a date. This is the data I have
Date
transaction Amount
End of Week Balance
jan 1
5
100
jan 2
3
100
jan 3
4
100
jan 4
3
100
jan 5
1
100
jan 6
3
100
I would like to find out what the daily end balance is. My thought is to get a running total from each day to the end of the week and subtract it from the end of week balance, like below
Date
transaction Amount
Running total
End of Week Balance
Balance - Running total
jan 1
5
19
100
86
jan 2
3
14
100
89
jan 3
4
11
100
93
jan 4
3
7
100
96
jan 5
1
4
100
97
jan 6
3
3
100
100
I can use
SUM(transactionAmount) OVER (Order by Date)
to get a running total, is there a way to specify that I only want the total of transactions that have taken place after the date?
You can use sum() as a window function, but accumulate in reverse:
select t.*,
(end_of_week_balance -
sum(transactionAmount) over (order by date desc)
)
from t;
If you have this example:
1> select i, sum(i) over (order by i) S from integers where i<10;
2> go
i S
----------- -----------
1 1
2 3
3 6
4 10
5 15
6 21
7 28
8 36
9 45
you can also do:
1> select i, sum(case when i>3 then i else 0 end) over (order by i) S from integers where i<10;
2> go
i S
----------- -----------
1 0
2 0
3 0
4 4
5 9
6 15
7 22
8 30
9 39
I have a table like below,
id
number
date
1
23
2020-01-01
2
12
2020-03-02
3
23
2020-09-02
4
11
2019-03-04
5
12
2019-03-23
6
23
2019-04-12
I want to know is that how many times each number appears per year, such as,
number
2019
2020
23
1
2
12
1
1
11
1
0
I'm kinda stuck.. tried with left join or just a single select, but still, cannot figure out how to make it, please help thank you!
SELECT C.NUMBER,
SUM
(
CASE
WHEN C.DATE BETWEEN '20190101'AND '20191231'
THEN 1 ELSE NULL
END
) AS A_2019,
SUM
(
CASE
WHEN C.DATE BETWEEN '20200101'AND '20201231'
THEN 1 ELSE NULL
END
) AS A_2020
FROM I_have_a_table_like_below AS C
GROUP BY C.NUMBER
I have DB2 table like below -
Date1 Item_code Amt
2018-06-01 1 2
2018-06-02 1 3
2018-06-03 2 4
2018-06-03 2 5
2018-06-04 3 6
2018-06-05 3 7
2018-06-06 4 8
I need the cumulative sum item_code wise per day. The result should look like -
Date1 Item_code Amt
2018-06-01 1 2
2018-06-02 1 5
2018-06-03 2 9
2018-06-04 3 6
2018-06-05 3 13
2018-06-06 4 8
I have tried a lot by myself and search also on SO but nothing is fulfilling my need. There are a lot of examples if I just need the cumulative sum day wise irrespective of item code.
Any help is greatly appreciated. Thanks in advance.
I think you want aggregation with a cumulative sum:
select item_code, date1,
sum(sum(amt)) over (partition by item_code order by date1) as running_amt
from t
group by item_code, date;
I have a set of data that looks like below
Name Time Perc Group Mode Control Cancelled
A 10:52 10.10 10 0 1 0
B 09:00 10.23 10 1 1 1
C 12:02 12.01 12 0 1 1
D 10:45 12.12 12 1 7 1
E 12:54 12.56 12 1 3 0
F 01:01 13.90 13 0 11 1
G 02:45 13.23 13 1 12 1
H 09:10 13.21 13 1 1 0
I need an output like below;
Group Perc Cancelled
10 20.33 1
12 36.69 2
13 40.34 2
What I'm getting was something like;
Group Perc Cancelled
10 20.33 5
12 36.69 5
13 40.34 5
I don't know what to call this, I have something in my mind to call it like CTE?, but I really can't figure it out.
Here's my source;
SELECT Group, SUM(Perc), Cancelled FROM
(SELECT Group, Perc, (SELECT COUNT(*) FROM tblName WHERE Cancelled=1) AS Cancelled FROM tblName WHERE 1=1 AND Group>=10)dt
GROUP BY Group, Cancelled
From your example, you don't need the nested query, any recursion, etc...
SELECT
Group,
SUM(Perc) AS total_perc,
SUM(cancelled) AS total_cancelled
FROM
tblName
WHERE
1=1
AND Group >= 10
GROUP BY
Group
If you did have some different data, then you might want to use something like...
SUM(CASE WHEN cancelled > 0 THEN 1 ELSE 0 END) AS total_cancelled