I'm trying to find a result with two conditions in google bigquery.
Employees who worked from Monday through Friday consecutively will get an additional pay of 8 hour amount of wage.
Condition above is valid for workers who worked more than 15 hours (15 hrs <) per week.
id
date
hours
abc123
2020-01-05
12
abc123
2020-01-06
5
abc123
2020-01-07
14
abc123
2020-01-08
7
abc123
2020-01-09
6
abc123
2020-01-10
12
Thanks in advance.
Assuming you have one row per employee per day, then you can handle this by using window functions. The focus will be on Mondays, but the idea is to count the hours and days for a given row and the four days following.
So, to get the Mondays where a given id matches the conditions and is eligible for a bonus:
select id, date
from (select t.*,
count(*) over (partition by id
order by unix_date(date)
range between current row and 4 following
) as day_count,
sum(hours) over (partition by id
order by unix_date(date)
range between current row and 4 following
) as hours_count
from t
) t
where extract(dayofweek from date) = 2 and
day_count = 5 and
hours_count >= 15;
Related
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 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 query where I am counting the total number of new users signed up to a particular service each day since the service started.
So far I have:
SELECT DISTINCT CONVERT(DATE, Account_Created) AS Date_Created,
COUNT(ID) OVER (PARTITION BY CONVERT(DATE, Account_Created)) AS New_Users
FROM My_Table.dbo.NewAccts
ORDER BY Date_Created
This returns:
Date_Created | New_Users
--------------------------
2020-01-01 1
2020-01-03 3
2020-01-04 2
2020-01-06 5
2020-01-07 9
What I would like is to return a third column with a cumulative total for each day starting from the beginning until the present. So the first day there was only one new user. On January 3rd, three new users signed up for a total of four since the beginning--so on and so forth.
Date_Created | New_Users | Cumulative_Tot
------------------------------------------
2020-01-01 1 1
2020-01-03 3 4
2020-01-04 2 6
2020-01-06 5 11
2020-01-07 9 20
My thought process was to involve the ROW_NUMBER() function so that I can separate and add each consecutive row together, though I am not sure if that is correct. My feeling is that I am probably thinking about this too hard and the logic is simply just escaping me at the moment. Thank you for any help.
As a starter: I would recommend aggregation rather than DISTINCT and a window count. This makes the intent clearer, and is likely more efficient.
Then, you can make use of a window sum to compute the cumulative count.
SELECT
CONVERT(DATE, Account_Created) AS Date_Created,
COUNT(*) AS New_Users
SUM(COUNT(*)) OVER(ORDER BY CONVERT(DATE, Account_Created)) Cumulative_New_Users
FROM My_Table.dbo.NewAccts
GROUP BY CONVERT(DATE, Account_Created)
ORDER BY Date_Created
Following on from a previous question, in which i have a table called orders with information regarding the time an order was placed and who made that order.
order_timestamp user_id
-------------------- ---------
1-JUN-20 02.56.12 123
3-JUN-20 12.01.01 533
23-JUN-20 08.42.18 123
12-JUN-20 02.53.59 238
19-JUN-20 02.33.72 34
I would like to calculate a daily rolling count of the number of days a user made an order in a past 10 days.
For example, in the last 10 days from the 20th June, user 34 made an order on 5 of those days. Then in the last 10 days from the 21st June, user 34 made an order on 6 of those days
In the end the table should be like this:
date user_id no_of_days
----------- --------- ------------
20-JUN-20 34 5
20-JUN-20 123 10
20-JUN-20 533 2
20-JUN-20 238 3
21-JUN-20 34 6
21-JUN-20 123 10
How would the query be written for this kind of analysis?
Please let me know if my question is unclear/more infor is required.
Thanks to you in advancement.
You can use window functions for this. Start by getting one row per user per day. And then use a rolling sum:
select day, user_id,
count(*) over (partition by user_id range between interval '10' day preceding and current row)
from (select distinct trunc(order_timestamp) as day, user_id
from t
) t
Assuming that a user places one order a day maximum, you can use window functions as follows:
select
t.*,
count(*) over(partition by user_id order by trunc(order_timestamp) range 10 preceding) no_of_days
from mytable t
Otherwise, you can get the distinct orders per day first:
select
order_day,
user_id,
count(*) over(partition by user_id order by order_day range 10 preceding) no_of_days
from (select distinct trunc(order_timestamp) order_day, user_id from mytable) t
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.