I have a query like the following:
select
table.date,
table.shift,
sum(table.value)
from
db.table
where
table.date >= date '2020-01-01' and
table.filter = 'type'
group by
table.date,
table.shift
order by
table.date,
table.shift;
That returns data this way:
date | shift | sum(value)
-----------|-------|------------
2020-01-06 | 1 | 15
2020-01-06 | 3 | 12
2020-01-07 | 1 | 20
2020-01-07 | 2 | 38
2020-01-09 | 1 | 6
2020-01-09 | 2 | 22
2020-01-09 | 3 | 14
2020-01-10 | 1 | 17
2020-01-10 | 2 | 3
2020-01-10 | 3 | 10
I'm trying to get it like this but I don't know how:
date | 1 | 2 | 3
-----------|----|----|----
2020-01-06 | 15 | | 12
2020-01-07 | 20 | 38 |
2020-01-09 | 6 | 22 | 14
2020-01-10 | 17 | 3 | 10
No need for an addition subquery or CTE. You can pivot your dataset using conditional aggregation with slight modifications of your query: just remove shift from the group by clause, and then implement conditional logic in the sum()s:
select
date,
sum(case when shift = 1 then value end) shift1,
sum(case when shift = 2 then value end) shift2,
sum(case when shift = 3 then value end) shift3
from
db.table
where
date >= date '2020-01-01'
and filter = 'type'
group by date
order by date
Note:
there is no need to prefix the column names since a single table comes into play. I removed those
date is the name of datatype in Oracle, hence not a good choice for a column name
You can do conditional aggregation :
select t.date,
sum(case when t.shift = 1 then t.value else 0 end),
sum(case when t.shift = 2 then t.value else 0 end),
sum(case when t.shift = 3 then t.value else 0 end)
from db.table as t
where t.date >= date '2020-01-01' and
t.filter = 'type'
group by t.date;
You can use conditional aggregation
with cte as
(
select
table.date,
table.shift,
sum(table.value) as val
from
db.table
where
table.date >= date '2020-01-01' and
table.filter = 'type'
group by
table.date,
table.shift
order by
table.date,
table.shift
)
select date, max(case when shift=1 then val end) as 1,
max(case when shift=1 then val end) as 2,
max(case when shift=1 then val end) as 3
from cte
group by date
You can use PIVOT for this as follows:
SELECT
*
FROM ( SELECT
table.date,
table.shift,
table.value
from
db.table
where
table.date >= date '2020-01-01' and
table.FILTER = 'type' )
PIVOT
( SUM ( VALUE ) FOR SHIFT IN ( 1,2,3 ))
ORDER BY date;
Cheers!!
Related
I need some help with SQL.
I have
Table1 with columns Id, Date1 and Date2
Table2 with columns Table1Id and Table2Id
Table3 with columns Id and Name
Here is my try:
with tmp_tab as (
select
v."Name" as name
, date_part('month', cv."OfferAcceptedDate") as MonthAcceptedName
, date_part('month', cv."OfferSentDate") as MonthSentName
, 1 as cntAcc
, 1 as cntSent
from hr_metrics."CvInfo" as cv
join hr_metrics."CvInfoVacancy" as civ
on civ."CvInfosId" = cv."Id"
join hr_metrics."Vacancy" as v
on civ."VacanciesId" = v."Id"
where cv."OfferSentDate" is not null
and date_part('year', cv."OfferSentDate") = date_part('year', CURRENT_DATE)
group by v."Name" , date_part('month', cv."OfferAcceptedDate"),
date_part('month', cv."OfferSentDate")
)
select distinct
tmp_tab."name" as name,
tmp_tab.MonthSentName as mSent,
tmp_tab.MonthAcceptedName as mAcc,
Sum(tmp_tab.cntSent) as sented,
Sum(tmp_tab.cntacc) as accepted
from tmp_tab as tmp_tab
group by tmp_tab.name, tmp_tab.MonthSentName, tmp_tab.MonthAcceptedName;
I need to take Count(date2)/Count(date1) grouped by monthes and name.
I have no idea how to do that, as there is no table with monthes.
DB - Postgres
sample data from comment:
t1
1 | 01/01/2021 | 31/03/2021
2 | 05/01/2021 | 18/01/2021
3 | 12/01/2021 | 31/01/2021
4 | 13/03/2021 | 22/03/2021
t2
1 | 1
2 | 1
3 | 2
4 | 1
t3
1 | SomeName1
2 | someName2
Desired result:
Name | month | value
SomeName1 | 1 | 1\2
SomeName1 | 3 | 2
SomeName2 | 1 | 1
Update: if count(date2) == 0, than count(date2) = -1
Source answer
Here code for my question thats work. And yeah, i've asked it on ru too.
select name, month, sum((SRC=1)::int) as AcceptedCount, sum((SRC=2)::int) as SentCount,
case when sum((SRC=1)::int) = 0 then -1
else sum((SRC=2)::int)::float / sum((SRC=1)::int) end as Result
from (
select v.name, SRC,
extract('month' from case SRC when 1 then OfferAcceptedDate else OfferSentDate end) as month
from (select (date_part('year', CURRENT_DATE)::char(4) || '-01-01')::timestamptz as from_date) x
cross join (select 1 as SRC union all select 2) s
join CvInfo as cv on (SRC=1 and cv.OfferAcceptedDate >= from_date and cv.OfferAcceptedDate < from_date + interval '1 year')
or (SRC=2 and cv.OfferSentDate >= from_date and cv.OfferSentDate < from_date + interval '1 year')
join CvInfoVacancy as civ on civ.CvInfosId = cv.Id
join Vacancy as v on civ.VacanciesId = v.Id
where case SRC when 1 then OfferAcceptedDate else OfferSentDate end is not null
) x
group by name, month
I have a table that stores per day if a user worked or if he was on vacation based on a value.
Example table, Value = 1 -> WorkDay, Value = 2 -> Vacation:
User | Day | Value
--------|------------|-------
user-1 | 2021-01-01 | 1
user-1 | 2021-01-02 | 1
user-1 | 2021-01-03 | 1
user-1 | 2021-01-04 | 1
user-1 | 2021-01-05 | 2
user-1 | 2021-01-06 | 2
...
I'll like to convert this table to this (Using the simple example above):
User | Year | Month | WorkDay | Vacation
--------|------|-------|---------|---------
user-1 | 2021 | 01 | 4 | 2
...
I tried using group by, subqueries and case, but the whole thing become a mess.
SELECT
YEAR(Day),
MONTH(DAY),
User,
...
From Table1
Group By YEAR(Day), MONTH(DAY), User
Just use conditional aggregation:
select year(day), month(day),
sum(case when value = 1 then 1 else 0 end) as workday,
sum(case when value = 2 then 1 else 0 end) as vacation
from table1
group by year(day), month(day)
You can use conditional aggregation as below:
select User,
year(day),
month(day),
sum(case when value = 1 then 1 else 0 end) as WorkDay,
sum(case when value = 2 then 1 else 0 end) as Vacation
from table1
group by User, year(day), month(day)
DB-Fiddle:
Schema and insert statements:
create table table1([User] varchar(50), Day date, Value int);
insert into table1 values('user-1' , '2021-01-01' , 1);
insert into table1 values('user-1' , '2021-01-02' , 1);
insert into table1 values('user-1' , '2021-01-03' , 1);
insert into table1 values('user-1' , '2021-01-04' , 1);
insert into table1 values('user-1' , '2021-01-05' , 2);
insert into table1 values('user-1' , '2021-01-06' , 2);
Query:
select [User] as "user",
year(day) as "year",
month(day) as "month",
sum(case when value = 1 then 1 else 0 end) as WorkDay,
sum(case when value = 2 then 1 else 0 end) as Vacation
from table1
group by [User], year(day), month(day)
GO
Output:
user
year
month
WorkDay
Vacation
user-1
2021
1
4
2
db<fiddle here
I am not sure if this scenario can be achieved using TSQL. I have a table called WorkingDays, which have this info
ID | EmployeeId | Monday | Tuesday | Wednesday | Thursday | Friday
----------------------------------------------------------------------
1 | 1 | 2 | 2 | 3 | 6 | 5
2 | 2 | 1 | 7 | 5 | 2 | 3
The days columns store Ids of WorkingSchedule table, which has this columns:
ID int Primary Key
StartTime time
EndTime time
So what I need id get the StartTime and EndTime of an employee depending on the current date.
What I need to get from query is the start and end time depending on the day. The day I want to filter is de current date (using getdate() function)
So need to select the correct day column name to make the join.
How can I achieve this scenario?
The dynamic sql version:
declare #sql nvarchar(max) ='
select
t.EmployeeId
, StarTime = max(case when t.rn=1 then '+quotename(datename(weekday,getdate()))+' end)
, EndTime = max(case when t.rn=2 then '+quotename(datename(weekday,getdate()))+' end)
from (
select *
, rn = row_number() over (partition by t.EmployeeId order by t.Id)
from t
) t
group by t.EmployeeId;'
exec sp_executesql #sql;
rextester demo: http://rextester.com/WNH34961
returns:
+------------+----------+---------+
| EmployeeId | StarTime | EndTime |
+------------+----------+---------+
| 1 | 5 | 3 |
+------------+----------+---------+
Depending on how you want the output, here are two other ways that do not use dynamic sql:
Both use cross apply() to unpivot the data, and WorkDay = datename(weekday,getdate()) to get the current WorkDay column.
For one row output we add some conditional aggregation:
/* one row per employeeId */
select
t.EmployeeId
, x.WorkDay
, StarTime = max(case when t.rn=1 then x.Time end)
, EndTime = max(case when t.rn=2 then x.Time end)
from (
select *
, rn = row_number() over (partition by t.EmployeeId order by t.Id)
from t
) t
cross apply (values
('Monday',Monday),('Tuesday',Tuesday),('Wednesday',Wednesday)
,('Thursday',Thursday),('Friday',Friday)
) x (WorkDay,Time)
where WorkDay = datename(weekday,getdate())
group by t.EmployeeId, x.WorkDay
returns:
+------------+---------+----------+---------+
| EmployeeId | WorkDay | StarTime | EndTime |
+------------+---------+----------+---------+
| 1 | Friday | 5 | 3 |
+------------+---------+----------+---------+
If you want the output on two rows, like your current output:
/* two rows per employeeId */
select
t.Id
, t.EmployeeId
, x.WorkDay
, t.StartEnd
, x.Time
from (
select *
, StartEnd = case
when row_number() over (partition by t.EmployeeId order by t.Id) = 1
then 'StartTime'
else 'EndTime'
end
from t
) t
cross apply (values
('Monday',Monday),('Tuesday',Tuesday),('Wednesday',Wednesday)
,('Thursday',Thursday),('Friday',Friday)
) x (WorkDay,Time)
where WorkDay = datename(weekday,getdate());
returns:
+----+------------+---------+-----------+------+
| Id | EmployeeId | WorkDay | StartEnd | Time |
+----+------------+---------+-----------+------+
| 1 | 1 | Friday | StartTime | 5 |
| 2 | 1 | Friday | EndTime | 3 |
+----+------------+---------+-----------+------+
select wd.Employee, ws.StartTime, ws.EndTime
from WorkingDays wd
join WorkingSchedule ws on ws.Id = case datename(weekday, getdate())
when 'Monday' then ws.Monday
when 'Tuesday' then ws.Tuesday
when 'Wednesday' then ws.Wednesday
when 'Thursday' then ws.Thursday
when 'Friday' then ws.Friday
else 0
end
Hint: datename(weekday, getdate()) returns you the weekday name in your current locale! This might be better:
select wd.Employee, ws.StartTime, ws.EndTime
from WorkingDays wd
join WorkingSchedule ws on ws.Id = case datepart(weekday, getdate())
when 1 then wd.Monday
when 2 then wd.Tuesday
when 3 then wd.Wednesday
when 4 then wd.Thursday
when 5 then wd.Friday
else 0
end
But then you have to check which day is the first of week (0, 1), depending on your settings.
I would like to identify the returning customers from an Oracle(11g) table like this:
CustID | Date
-------|----------
XC321 | 2016-04-28
AV626 | 2016-05-18
DX970 | 2016-06-23
XC321 | 2016-05-28
XC321 | 2016-06-02
So I can see which customers returned within various windows, for example within 10, 20, 30, 40 or 50 days. For example:
CustID | 10_day | 20_day | 30_day | 40_day | 50_day
-------|--------|--------|--------|--------|--------
XC321 | | | 1 | |
XC321 | | | | 1 |
I would even accept a result like this:
CustID | Date | days_from_last_visit
-------|------------|---------------------
XC321 | 2016-05-28 | 30
XC321 | 2016-06-02 | 5
I guess it would use a partition by windowing clause with unbounded following and preceding clauses... but I cannot find any suitable examples.
Any ideas...?
Thanks
No need for window functions here, you can simply do it with conditional aggregation using CASE EXPRESSION :
SELECT t.custID,
COUNT(CASE WHEN (last_visit- t.date) <= 10 THEN 1 END) as 10_day,
COUNT(CASE WHEN (last_visit- t.date) between 11 and 20 THEN 1 END) as 20_day,
COUNT(CASE WHEN (last_visit- t.date) between 21 and 30 THEN 1 END) as 30_day,
.....
FROM (SELECT s.custID,
LEAD(s.date) OVER(PARTITION BY s.custID ORDER BY s.date DESC) as last_visit
FROM YourTable s) t
GROUP BY t.custID
Oracle Setup:
CREATE TABLE customers ( CustID, Activity_Date ) AS
SELECT 'XC321', DATE '2016-04-28' FROM DUAL UNION ALL
SELECT 'AV626', DATE '2016-05-18' FROM DUAL UNION ALL
SELECT 'DX970', DATE '2016-06-23' FROM DUAL UNION ALL
SELECT 'XC321', DATE '2016-05-28' FROM DUAL UNION ALL
SELECT 'XC321', DATE '2016-06-02' FROM DUAL;
Query:
SELECT *
FROM (
SELECT CustID,
Activity_Date AS First_Date,
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '10' DAY FOLLOWING )
- 1 AS "10_Day",
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '20' DAY FOLLOWING )
- 1 AS "20_Day",
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '30' DAY FOLLOWING )
- 1 AS "30_Day",
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '40' DAY FOLLOWING )
- 1 AS "40_Day",
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '50' DAY FOLLOWING )
- 1 AS "50_Day",
ROW_NUMBER() OVER ( PARTITION BY CustID ORDER BY Activity_Date ) AS rn
FROM Customers
)
WHERE rn = 1;
Output
USTID FIRST_DATE 10_Day 20_Day 30_Day 40_Day 50_Day RN
------ ------------------- ---------- ---------- ---------- ---------- ---------- ----------
AV626 2016-05-18 00:00:00 0 0 0 0 0 1
DX970 2016-06-23 00:00:00 0 0 0 0 0 1
XC321 2016-04-28 00:00:00 0 0 1 2 2 1
Here is an answer that works for me, I have based it on your answers above, thanks for contributions from MT0 and Sagi:
SELECT CustID,
visit_date,
Prev_Visit ,
COUNT( CASE WHEN (Days_between_visits) <=10 THEN 1 END) AS "0-10_day" ,
COUNT( CASE WHEN (Days_between_visits) BETWEEN 11 AND 20 THEN 1 END) AS "11-20_day" ,
COUNT( CASE WHEN (Days_between_visits) BETWEEN 21 AND 30 THEN 1 END) AS "21-30_day" ,
COUNT( CASE WHEN (Days_between_visits) BETWEEN 31 AND 40 THEN 1 END) AS "31-40_day" ,
COUNT( CASE WHEN (Days_between_visits) BETWEEN 41 AND 50 THEN 1 END) AS "41-50_day" ,
COUNT( CASE WHEN (Days_between_visits) >50 THEN 1 END) AS "51+_day"
FROM
(SELECT CustID,
visit_date,
Lead(T1.visit_date) over (partition BY T1.CustID order by T1.visit_date DESC) AS Prev_visit,
visit_date - Lead(T1.visit_date) over (
partition BY T1.CustID order by T1.visit_date DESC) AS Days_between_visits
FROM T1
) T2
WHERE Days_between_visits >0
GROUP BY T2.CustID ,
T2.visit_date ,
T2.Prev_visit ,
T2.Days_between_visits;
This returns:
CUSTID | VISIT_DATE | PREV_VISIT | DAYS_BETWEEN_VISIT | 0-10_DAY | 11-20_DAY | 21-30_DAY | 31-40_DAY | 41-50_DAY | 51+DAY
XC321 | 2016-05-28 | 2016-04-28 | 30 | | | 1 | | |
XC321 | 2016-06-02 | 2016-05-28 | 5 | 1 | | | | |
Which would be the best way to get the number of people hired on each day of the week for 7 years from a table People that has their entry_date with a day-month-year as 01-Jun-91.
For example:
2000 2001 2002 etc..
SUN 2 0 1
MON 0 0 2
Do I have to create a counter for each day of each year? Like Sun2000, Sun2001 etc?
You need to join each day of the week with your entry_date and pivot the results.
SQL Fiddle
Query:
with x(days) as (
select 'sunday' from dual union all
select 'monday' from dual union all
select 'tuesday' from dual union all
select 'wednesday' from dual union all
select 'thursday' from dual union all
select 'friday' from dual union all
select 'saturday' from dual
)
select * from (
select x.days,
extract(year from emp.entry_date) entry_year
from x left outer join emp
on x.days = to_char(emp.entry_date,'fmday')
)
pivot(count(entry_year)
for entry_year in (
2007,
2008,
2009,
2010,
2011,
2012
)
)
order by
case days when 'sunday' then 1
when'monday' then 2
when'tuesday' then 3
when'wednesday' then 4
when'thursday' then 5
when'friday' then 6
when'saturday' then 7
end
Results:
| DAYS | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 |
|-----------|------|------|------|------|------|------|
| sunday | 0 | 0 | 0 | 0 | 0 | 0 |
| monday | 0 | 0 | 0 | 2 | 0 | 0 |
| tuesday | 0 | 0 | 0 | 0 | 1 | 0 |
| wednesday | 0 | 0 | 0 | 1 | 2 | 1 |
| thursday | 0 | 0 | 0 | 0 | 0 | 3 |
| friday | 0 | 0 | 0 | 0 | 0 | 0 |
| saturday | 0 | 0 | 0 | 0 | 0 | 0 |
You need to use group by on the year and entry_date to get the count of employees joined for each date.
For example:
Rem -- Assuming following table structure
create table People(id number, name varchar2(20), entry_date date);
Rem -- Following groups the results
select extract(year from entry_date) "Year", entry_date, count(id)
from People
where extract(year from entry_date) between 2008 and 2015
group by extract(year from entry_date), entry_date
order by extract(year from entry_date), entry_date;
Check this sqlfiddle to explore more.
Depending on the version of Oracle you're using (10g doesn't have the PIVOT function, for example), you might try something like the following conditional aggregation:
SELECT day_abbrev
, SUM(CASE WHEN year_num = 2000 THEN person_cnt ELSE 0 END) AS "2000"
, SUM(CASE WHEN year_num = 2001 THEN person_cnt ELSE 0 END) AS "2001"
, SUM(CASE WHEN year_num = 2002 THEN person_cnt ELSE 0 END) AS "2002"
, SUM(CASE WHEN year_num = 2003 THEN person_cnt ELSE 0 END) AS "2003"
, SUM(CASE WHEN year_num = 2004 THEN person_cnt ELSE 0 END) AS "2004"
, SUM(CASE WHEN year_num = 2005 THEN person_cnt ELSE 0 END) AS "2005"
, SUM(CASE WHEN year_num = 2006 THEN person_cnt ELSE 0 END) AS "2006"
FROM (
SELECT TO_CHAR(entry_date, 'DY') AS day_abbrev
, EXTRACT(YEAR FROM entry_date) AS year_num
, COUNT(*) AS person_cnt
FROM people
GROUP BY TO_CHAR(entry_date, 'DY'), EXTRACT(YEAR FROM entry_date)
) GROUP BY day_abbrev
ORDER BY TO_CHAR(NEXT_DAY(SYSDATE, day_abbrev), 'D');