I have requirement where i will need to get the number of days a role an employee was on.
Scenario 1
EmployeeId role effectiveFrom
1 A 1-Jan-2021
1 B 15-Jan-2021
No further roles are available for the month of Jan for role A therefore the number of days for role A would be 14.
Scenario 2
EmployeeId role effectiveFrom
1 A 1-Jan-2021
No further roles are available for the month of Jan therefore the number of days for role A would be 31 i.e the entire month of January. For the month of February i would expect to get 28 as the role would be effective for the entire month of february as well.
Scenario 3
EmployeeId role effectiveFrom
1 A 1-Jan-2021
1 B 15-Jan-2021
1 A 25-Jan-2021
To get the number of days for role A the logic would be
1 to 15th is 14 days.
25th to 31st(31st of Jan) would be 6 days.
14 + 6 = 20 days
The query i have come up with so far is this,
SELECT
DATEDIFF(MAX(effectiveFrom),
IF(MIN(effectiveFrom) = MAX(effectiveFrom),
MIN(effectiveFrom),
MIN(effectiveFrom))) + 1 daysWorked
FROM
EmployeeRoles
WHERE grade = 'A'
GROUP BY `employeeId`,effectiveFrom;
which would only give the result as 1 day for Scenario 1. Could someone guide me on the practical way of handling the scenarios. I have looked at loops, window functions but i am at a loss on the best way to proceed.
dbfiddle
When scenario2 has 31 days from 1-jan, until the end of the month, I would suspect that from 25-jan, until the end of the month, is 7 days, and not 6, as you write in scenario3.
The number of days, using above calculation:
SELECT
employeeID,
grade,
effectiveFrom,
DATEDIFF(COALESCE(LEAD(effectiveFrom)
OVER (PARTITION BY employeeID, grade ORDER By effectiveFrom),
DATE_ADD(LAST_DAY(effectiveFrom),INTERVAL 1 DAY)),
effectiveFrom) as '#Days'
FROM EmployeeRole;
This can be grouped, and summed giving:
SELECT
employeeID,
grade,
SUM(`#Days`)
FROM (
SELECT
employeeID,
grade,
effectiveFrom,
DATEDIFF(COALESCE(LEAD(effectiveFrom)
OVER (PARTITION BY employeeID, grade ORDER By effectiveFrom),
DATE_ADD(LAST_DAY(effectiveFrom),INTERVAL 1 DAY)),
effectiveFrom) as '#Days'
FROM EmployeeRole
) x
GROUP BY
employeeID,
grade;
output:
employeeID
grade
SUM(#Days)
1
A
14
1
B
17
2
A
31
3
A
21
3
B
10
see: DBFIDDLE
EDIT: The results were incorrect because the next effectiveFrom date was determined using OVER (PARTITION BY employeeID ORDER By effectiveFrom). this is not correct, because the grade should be taken into account too.
I corrected it to OVER (PARTITION BY employeeID, grade ORDER By effectiveFrom)
P.S. I also corrected this in the piece above the EDIT!
see: DBFIDDLE
Related
I have the following table called employees:
employee
name
101
John
102
Alexandra
103
Ruth
And the table called records:
employee
assistance
101
2022-02-01
101
2022-02-02
101
2022-02-07
Let's suppose that I want to display the employee number, name and the days of the month in which there were absences between 2022-02-01 and 2022-02-07 (taking into account that days 05 and 06 are weekends). In that case, the result would be the following:
employee
name
absence
101
John
4,5
How do I get that result?
So far I have developed a query where the days of the month in which there are attendances are displayed. Said query is as follows:
SELECT e.employee,
e.name,
r.assistance AS assistance,
OF employees and
JOIN LEFT(SELECT employee, GROUP_CONCAT(DIFFERENT EXTRACT(DAY SINCE assistance)
ORDER BY STATEMENT(DAY FROM assistance)) AS assistance FROM records
WHERE assistance BETWEEN '2022-02-01' AND '2022-02-07' GROUP BY employee) r ON e.employee = employee
WHERE (r.no_employee IS NOT NULL) ORDER BY name ASC
I would like to know how to implement the days in which there were absences and not consider the weekends. I've done several tests but I'm still stuck. I'm working with MariaDB 10.4.11
You use a recursive common table expression (requires mariadb 10.2+ or mysql 8) to get the list of dates in the date range, and join against that:
with recursive date_range as (
select '2021-12-01' dt
union all
select dt + interval 1 day from date_range where dt < '2021-12-07'
)
select employee.employee, group_concat(day(date_range.dt) order by date_range.dt) faults
from date_range
cross join employee
left join records on records.employee=employee.employee and records.assistance=date_range.dt
where weekday(date_range.dt) < 5 and records.employee is null
group by employee.employee
fiddle
If you are just looking for one employee, add that as a where condition.
I have an orders table
Order_id User_id Order_date
1 32 2020-07-19
2 24 2020-07-21
3 27 2020-07-27
4 24 2020-08-14
5 32 2020-08-18
6 32 2020-08-19
7 58 2020-08-20
Now I want to find how many of the users ordered in 1st month also ordered in the next month. In this case, user_id's 32,24,27 ordered in 7th month but only 24 and 32 ordered in the next month.
I want the result to be like :
Date Retained_Users Total_users
2020-07 Null 3
2020-08 2 3
I'm lost here. Can someone please help me with this?
In MySQL 8.0, you can do this with window functions:
select
order_month,
count(distinct case when cnt_orders_last_month > 0 then user_id end) retained_users,
count(distinct user_id) total_users
from (
select
user_id,
date_format(order_date, '%Y-%m-01') as order_month,
count(*) over(
partition by user_id
order by date(date_format(order_date, '%Y-%m-01'))
range between interval 1 month preceding and interval 1 day preceding
) cnt_orders_last_month
from mytable
) t
group by order_month
The logic lies in the range specification of the window function; it orders record by month, and counts how many orders the customer placed last month. Then all that is left to do is aggregate and count distinct users.
Demo on DB Fiddle
So, there is an account number and we have daily information about their payments. Suppose we have information of 1 year leading up to today which is 08/March/2019, I would want to calculate the number of times he/she overpaid in last 1 week. I have used mysql window function but for some reason it does not seem to work
#GMB A sample data would look like this:Suppose for this account we have info from last march 2018. I just want the number of times paid_status = overpaid from the last date that I have on my file which is of today - 08/March/2019 and previous 7 days, 14 days, 1 month or any duration of my choosing. Your query will hardcode it only for 7 days.
ACCOUNT_ID paid_status amt dte
-----------------------
1234 overpaid 100 01/March/2018
.
.
.
1234 overpaid 120 01/March/2019
1234 not paid 0 02/March/2019
1234 overpaid 110 03/March/2019
1234 overpaid 120 04/March/2019
1234 overpaid 130 05/March/2019
1234 overpaid 120 06/March/2019
1234 overpaid 120 07/March/2019
1234 overpaid 121 08/March/2019
Query:
,COUNT(CASE WHEN paid_status = 'OVERPAID' THEN 1 END)
over (PARTITION BY ACCOUNT_ID
ORDER BY DTE ROWS BETWEEN 7 PRECEDING AND UNBOUNDED FOLLOWING
) AS num_times_overpaid_week1
The output should be like this(not including today's info):
account_id num_times_overpaid_week1
1234 6
While I am getting multiple rows for the same account_id and it is not exactly calulating the field correctly
From your sample data it seems like you are looking for a simple aggregated query (no need for window functions):
SELECT account_id, SUM(paid_status = 'OVERPAID') AS num_times_overpaid_week1
FROM mytable
WHERE dte >= CURRENT_DATE - INTERVAL 7 DAY
GROUP BY account_id
Expression SUM(paid_status = 'OVERPAID') uses a nice MySQL feature where conditions return 1 when satisfied and 0 when not.
NB: if, for some reason, you do want to use window functions (maybe to perform other computation), then you would need to use ROW_NUMBER() to rank records by date, and the filter out only the most recent record per account in an outer query. I think that the definition of the window can be largely simplified:
SELECT *
FROM (
SELECT
account_id,
SUM(paid_status = 'OVERPAID') OVER(PARTITION BY account_id) AS num_times_overpaid_week1,
-- possibly other columns
ROW_NUMBER() OVER(PARTITION BY account_id ORDER BY dte DESC) rn
FROM mytable
WHERE dte >= CURRENT_DATE - INTERVAL 7 DAY
) x WHERE rn = 1
In MySQL I'm tasked with a big dataset, with data from 1970 to 2010.
I want to check for consistency: check if each instance occurs minimum one time per year. I took a snippet from 1970-1972 as example to demonstrate my problem.
input:
id year counts
-- ---- ---------
1 1970 1
1 1971 1
2 1970 3
2 1971 8
2 1972 1
3 1970 4
expected:
id 1970-1972
-- ----------
1 no
2 yes
3 no
I though about counting within the date range and then taking those out who had 3 counts: 1970, 1971, 1972. The following query doesn't force the check on each point in the range though.
select id, count(*)
from table1
WHERE (year BETWEEN '1970' AND '1972') AND `no_counts` >= 1
group by id
What to do?
You can use GROUP BY with CASE / inline if.
Using CASE. SQL Fiddle
select id,CASE WHEN COUNT(distinct year) = 3 THEN 'yes'ELSE 'No' END "1970-72"
from abc
WHERE year between 1970 and 1972
GROUP BY id
Using inline IF. SQL Fiddle
select id,IF( COUNT(distinct year) = 3,'yes','No') "1970-72"
from abc
WHERE year between 1970 and 1972
GROUP BY id
You can use a having clause with distinct count:
select `id`
from `table1`
where `year` between '1970' and '1972'
group by id
having count(distinct `year`) = 3
Do you expect this?
select id, count(*)
from table1
WHERE (year BETWEEN '1970' AND '1972')
group by id
having count(distinct year) = 3
I am having an issue with a SELECT command in MySQL. I have a database of securities exchanged daily with maturity from 1 to 1000 days (>1 mio rows). I would like to get the outstanding amount per day (and possibly per category). To give an example, suppose this is my initial dataset:
DATE VALUE MATURITY
1 10 3
1 15 2
2 10 1
3 5 1
I would like to get the following output
DATE OUTSTANDING_AMOUNT
1 25
2 35
3 15
Outstanding amount is calculated as the total of securities exchanged still 'alive'. That means, in day 2 there is a new exchange for 10 and two old exchanges (10 and 15) still outstanding as their maturity is longer than one day, for a total outstanding amount of 35 on day 2. In day 3 instead there is a new exchange for 5 and an old exchange from day 1 of 10. That is, 15 of outstanding amount.
Here's a more visual explanation:
Monday Tuesday Wednesday
10 10 10 (Day 1, Value 10, matures in 3 days)
15 15 (Day 1, 15, 2 days)
10 (Day 2, 10, 1 day)
5 (Day 3, 5, 3 days with remainder not shown)
-------------------------------------
25 35 15 (Outstanding amount on each day)
Is there a simple way to get this result?
First of all in the main subquery we find SUM of all Values for current date. Then add to them values from previous dates according their MATURITY (the second subquery).
SQLFiddle demo
select T1.Date,T1.SumValue+
IFNULL((select SUM(VALUE)
from T
where
T1.Date between
T.Date+1 and T.Date+Maturity-1 )
,0)
FROM
(
select Date,
sum(Value) as SumValue
from T
group by Date
) T1
order by DATE
I'm not sure if this is what you are looking for, perhaps if you give more detail
select
DATE
,sum(VALUE) as OUTSTANDING_AMOUNT
from
NameOfYourTable
group by
DATE
Order by
DATE
I hope this helps
Each date considers each row for inclusion in the summation of value
SELECT d.DATE, SUM(m.VALUE) AS OUTSTANDING_AMOUNT
FROM yourTable AS d JOIN yourtable AS m ON d.DATE >= m.MATURITY
GROUP BY d.DATE
ORDER BY d.DATE
A possible solution with a tally (numbers) table
SELECT date, SUM(value) outstanding_amount
FROM
(
SELECT date + maturity - n.n date, value, maturity
FROM table1 t JOIN
(
SELECT 1 n UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5
) n ON n.n <= maturity
) q
GROUP BY date
Output:
| DATE | OUTSTANDING_AMOUNT |
-----------------------------
| 1 | 25 |
| 2 | 35 |
| 3 | 15 |
Here is SQLFiddle demo