i have table like this:
Date |Shop Code |Good code |Value
01.11 1001 1 1000.00
01.11 1001 2 799.00
01.11 1002 1 899.00
03.12 1003 2 500.00
03.12 1003 3 760.00
and i need to have
table with SUM for each shop code for each good code for each date
And then i need to have 10 day movement average on the column of sum(value) per each shop (5 days before and 5 days after)
You can express what you want in standard SQL as:
select shop_code, good_code, date, sum(value),
avg(sum(value)) over (partition by shop_code, good_code
order by date
range between interval '9 day' preceding and current row
) as avg10
from t
group by shop_code, good_code, date;
Although this is standard functionality, not all databases support it or have variations on the syntax.
Related
I have a table t with:
DATE
LOCATION
PRODUCT_ID
AMOUNT
2021-10-29
1
123
10
2021-10-30
1
123
9
2021-10-31
1
123
8
2021-10-29
1
456
100
2021-10-30
1
456
90
2021-10-31
1
456
80
2021-10-29
2
123
18
2021-10-30
2
123
17
2021-11-29
2
456
18
I need to find the AMOUNT of each PRODUCT_ID for each combination of LOCATION + PRODUCT_ID.
If a PRODUCT_ID has no entry for that day the AMOUNT is NULL.
So the result should look like:
DATE
LOCATION
PRODUCT_ID
AMOUNT
2021-10-31
1
123
8
2021-10-31
1
456
80
2021-10-31
2
123
NULL
2021-11-30
2
456
NULL
Sadly EXASOL has no LAST_DAY() or EOMONTH() function. How can I solve this?
You can get to the last day of the month using a date_trunc function in combination with date_add:
case
when t.date = date_add('day', -1, date_add('month', 1, date_trunc('month', t.date)))
then 'Y' else 'N' end as end_of_month
That being said, if you group your table for all combinations of locations and products, you will not get NULLs for products without sales on the last day of the month as shown in your output table.
When you group your data, any value that does not exist will simply not show up in your output table. If you want to force nulls to show up, you can create a new table that contains all combinations of products, locations, and hard-coded end of month dates.
Then, you can left join your old table with this new hard-coded table by date, location, and product. This method will give you the NULL values you expect.
I have a table of EMPLOYEES that contains information about the DATE and WORKTIME per that day. Fx:
ID | DATE | WORKTIME |
----------------------------------------
1 | 1-Sep-2014 | 4 |
2 | 2-Sep-2014 | 6 |
1 | 3-Sep-2014 | 5.5 |
1 | 4-Sep-2014 | 7 |
2 | 4-Sep-2014 | 4 |
1 | 9-Sep-2014 | 8 |
and so on.
Question: How can I create a query that would allow me to calculate amount of time worked per week (HOURS_PERWEEK). I understand that I need a summation of WORKTIME together with grouping considering both, ID and week, but so far my trials as well as googling didnt yield any results. Any ideas on this? Thank you in advance!
edit:
Got a solution of
select id, sum (worktime), trunc(date, 'IW') week
from employees
group by id, TRUNC(date, 'IW');
But will need somehow to connect that particular output with DATE table by updating a newly created column such as WEEKLY_TIME. Any hints on that?
You can find the start of the ISO week, which will always be a Monday, using TRUNC("DATE", 'IW').
So if, in the query, you GROUP BY the id and the start of the week TRUNC("DATE", 'IW') then you can SELECT the id and aggregate to find the SUM the WORKTIME column for each id.
Since this appears to be a homework question and you haven't attempted a query, I'll leave it at this to point you in the correct direction and you can complete the query.
Update
Now I need to create another column (lets call it WEEKLY_TIME) and populate it with values from the current output, so that Sep 1,3,4 (for ID=1) would all contain value 16.5, specifying that on that day (that is within the certain week) that person worked 16.5 in total. And for ID=2 it would then be a value of 10 for both Sep 2 and 4.
For this, if I understand correctly, you appear to not want to use aggregation functions and want to use the analytic version of the function:
select id,
"DATE",
trunc("DATE", 'IW') week,
worktime,
sum (worktime) OVER (PARTITION BY id, trunc("DATE", 'IW'))
AS weekly_time
from employees;
Which, for the sample data:
CREATE TABLE employees (ID, "DATE", WORKTIME) AS
SELECT 1, DATE '2014-09-01', 4 FROM DUAL UNION ALL
SELECT 2, DATE '2014-09-02', 6 FROM DUAL UNION ALL
SELECT 1, DATE '2014-09-03', 5.5 FROM DUAL UNION ALL
SELECT 1, DATE '2014-09-04', 7 FROM DUAL UNION ALL
SELECT 2, DATE '2014-09-04', 4 FROM DUAL UNION ALL
SELECT 1, DATE '2014-09-09', 8 FROM DUAL;
Outputs:
ID
DATE
WEEK
WORKTIME
WEEKLY_TIME
1
2014-09-01 00:00:00
2014-09-01 00:00:00
4
16.5
1
2014-09-03 00:00:00
2014-09-01 00:00:00
5.5
16.5
1
2014-09-04 00:00:00
2014-09-01 00:00:00
7
16.5
1
2014-09-09 00:00:00
2014-09-08 00:00:00
8
8
2
2014-09-04 00:00:00
2014-09-01 00:00:00
4
10
2
2014-09-02 00:00:00
2014-09-01 00:00:00
6
10
db<>fiddle here
edit: answer submitted without noticing "Oracle" tag. Otherwise, question answered here: Oracle SQL - Sum and group data by week
Select employee_Id,
DATEPART(week, workday) as [Week],
sum (worktime) as [Weekly Hours]
from WORK
group by employee_id, DATEPART(week, workday)
https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=238b229156a383fa3c466b6c3c2dee1e
I have table in Teradata SQL like below:
ID trans_date
------------------------
123 | 2021-01-01
887 | 2021-01-15
123 | 2021-02-10
45 | 2021-03-11
789 | 2021-10-01
45 | 2021-09-02
And I need to calculate average monthly number of transactions made by customers in a period between 2021-01-01 and 2021-09-01, so client with "ID" = 789 will not be calculated because he made transaction later.
In the first month (01) were 2 transactions
In the second month was 1 transaction
In the third month was 1 transaction
In the nineth month was 1 transactions
So the result should be (2+1+1+1) / 4 = 1.25, isn't is ?
How can I calculate it in Teradata SQL? Of course I showed you sample of my data.
SELECT ID, AVG(txns) FROM
(SELECT ID, TRUNC(trans_date,'MON') as mth, COUNT(*) as txns
FROM mytable
-- WHERE condition matches the question but likely want to
-- use end date 2021-09-30 or use mth instead of trans_date
WHERE trans_date BETWEEN date'2021-01-01' and date'2021-09-01'
GROUP BY id, mth) mth_txn
GROUP BY id;
Your logic translated to SQL:
--(2+1+1+1) / 4
SELECT id, COUNT(*) / COUNT(DISTINCT TRUNC(trans_date,'MON')) AS avg_tx
FROM mytable
WHERE trans_date BETWEEN date'2021-01-01' and date'2021-09-01'
GROUP BY id;
You should compare to Fred's answer to see which is more efficent on your data.
I have a dataset in bigquery which contains order_date: DATE and customer_id.
order_date | CustomerID
2019-01-01 | 111
2019-02-01 | 112
2020-01-01 | 111
2020-02-01 | 113
2021-01-01 | 115
2021-02-01 | 119
I try to count distinct customer_id between the months of the previous year and the same months of the current year. For example, from 2019-01-01 to 2020-01-01, then from 2019-02-01 to 2020-02-01, and then who not bought in the same period of next year 2020-01-01 to 2021-01-01, then 2020-02-01 to 2021-02-01.
The output I am expect
order_date| count distinct CustomerID|who not buy in the next period
2020-01-01| 5191 |250
2020-02-01| 4859 |500
2020-03-01| 3567 |349
..........| .... |......
and the next periods shouldn't include the previous.
I tried the code below but it works in another way
with customers as (
select distinct date_trunc(date(order_date),month) as dates,
CUSTOMER_WID
from t
where date(order_date) between '2018-01-01' and current_date()-1
)
select
dates,
customers_previous,
customers_next_period
from
(
select dates,
count(CUSTOMER_WID) as customers_previous,
count(case when customer_wid_next is null then 1 end) as customers_next_period,
from (
select prev.dates,
prev.CUSTOMER_WID,
next.dates as next_dates,
next.CUSTOMER_WID as customer_wid_next
from customers as prev
left join customers
as next on next.dates=date_add(prev.dates,interval 1 year)
and prev.CUSTOMER_WID=next.CUSTOMER_WID
) as t2
group by dates
)
order by 1,2
Thanks in advance.
If I understand correctly, you are trying to count values on a window of time, and for that I recommend using window functions - docs here and here a great article explaining how it works.
That said, my recommendation would be:
SELECT DISTINCT
periods,
COUNT(DISTINCT CustomerID) OVER 12mos AS count_customers_last_12_mos
FROM (
SELECT
order_date,
FORMAT_DATE('%Y%m', order_date) AS periods,
customer_id
FROM dataset
)
WINDOW 12mos AS ( # window of last 12 months without current month
PARTITION BY periods ORDER BY periods DESC
ROWS BETWEEN 12 PRECEEDING AND 1 PRECEEDING
)
I believe from this you can build some customizations to improve the aggregations you want.
You can generate the periods using unnest(generate_date_array()). Then use joins to bring in the customers from the previous 12 months and the next 12 months. Finally, aggregate and count the customers:
select period,
count(distinct c_prev.customer_wid),
count(distinct c_next.customer_wid)
from unnest(generate_date_array(date '2020-01-01', date '2021-01-01', interval '1 month')) period join
customers c_prev
on c_prev.order_date <= period and
c_prev.order_date > date_add(period, interval -12 month) left join
customers c_next
on c_next.customer_wid = c_prev.customer_wid and
c_next.order_date > period and
c_next.order_date <= date_add(period, interval 12 month)
group by period;
I have a database in SQL Server 2012 and there is a table with dates in D.M.YYYY format like below:
ID | Date(date type) | Amount(Numeric)
1 3.4.2013 16.00
1 12.4.2013 13.00
1 2.5.2013 9.50
1 18.5.2013 10.00
I need to sum the total amount for every month in a given year. For example:
ID | Month | TotalAmount
1 1 0.00
...
1 4 29.00
1 5 19.50
I thought what I needed was to determine the number of days in a month, so I created a function which is described in determine the number of days, and it worked. After that I tried to compare two dates(date type) and got stuck; there are some examples out there, but all of them about datetime.
Is this wrong? How can I accomplish this?
I think you just want an aggregation:
select id, month(date) as "month", sum(amount) as TotalAmount
from t
where year(date) = 2013
group by id, month(date)