I i have a table [holidays] with the following structure :
id | start_date | end_date | user_id
How to get the number of holidays for a user in the previous year ?
I wanna something like that with correct syntax:
SELECT SUM(end_date - start_date)
FROM holidays
WHERE user_id = 342
AND YEAR(end_date) = YEAR(CURRENT) - 1
I think this should work in informix:
SELECT SUM(end_date-start_date)
from holidays
where user_id = 342 and
YEAR(end_date) = YEAR(TODAY)-1;
Note: it is not clear whether the end date is inclusive or not. You might want:
SELECT SUM((end_date-start_date) + 1)
from holidays
where user_id = 342 and
YEAR(end_date) = YEAR(TODAY)-1;
Your query is perfectly fine. You don't have any syntax improvements to be done. The only thing Gordon Linoff's approach (which will work) differs from yours is that he is using TODAY (that is YEAR TO DAY) instead of CURRENT (that is YEAR TO FRACTION), but YEAR(CURRENT) and YEAR(TODAY) would get the exact same result.
The only way you would get a different result was if you had CURRENT inside the SUM, like SUM(CURRENT-start_date). That way you would have more precision than only the days (You would get the days and then hh:mm:ss.fff) like you want, but if you used TODAY, you would get only the days.
Other than that, it's perfectly fine.
Related
I have two date fields, DATE_FIELD_ONE = 8/30/2018 and DATE_FIELD_TWO = DATE_FIELD_ONE + 20. I need to find what DATE_FIELD_TWO should be if I'm only added 20 business days . How would I accomplish this? I thought maybe trying 'DY' but not sure how to get it to work. Thanks.
CASE WHEN TO_CHAR(TO_DATE(DATE_FIELD_ONE),'DY')='SAT' THEN 1 ELSE 0 END
CASE WHEN TO_CHAR(TO_DATE(DATE_FIELD_ONE),'DY')='SUN' THEN 1 ELSE 0 END
You may try this :
select max(date_field_two) as date_field_two
from
(
select date'2018-08-30'+
cast(case when to_char(date'2018-08-30'+level,'D','NLS_DATE_LANGUAGE=ENGLISH')
in ('6','7') then
0
else
level
end as int) as date_field_two,
sum(cast(case when to_char(date'2018-08-30'+level,'D','NLS_DATE_LANGUAGE=ENGLISH')
in ('6','7') then
0
else
1
end as int)) over (order by level) as next_day
from dual
connect by level <= 20*1.5
-- 20 is the day to be added, every time 5(#of business days)*1.5 > 7(#of week days)
-- 7=5+2<5+(5/2)=5*(1+1/2)=5*1.5 [where 1.5 is just a coefficient might be replaced a greater one like 2]
-- so 4*5*1.5=20*1.5 > 4*7
)
where next_day = 20;
DATE_FIELD_TWO
-----------------
27.09.2018
by using connect by dual clause.
P.S. Ignored the case for public holidays, which differ from one culture to another , depending on the question being related with only weekends.
Rextester Demo
Edit : Assume you have a national holidays on '2018-09-25' and '2018-09-26' (in this set of days), then consider the following :
select max(date_field_two) as date_field_two
from
(
select date'2018-08-30'+
(case when to_char(date'2018-08-30'+level,'D','NLS_DATE_LANGUAGE=ENGLISH')
in ('6','7') then
0
when date'2018-08-30'+level in (date'2018-09-25',date'2018-09-26') then
0
else
level
end) as date_field_two,
sum(cast(case when to_char(date'2018-08-30'+level,'D','NLS_DATE_LANGUAGE=ENGLISH')
in ('6','7') then
0
when date'2018-08-30'+level in (date'2018-09-25',date'2018-09-26') then
0
else
1
end as int)) over (order by level) as next_day
from dual
connect by level <= 20*2
)
where next_day = 20;
DATE_FIELD_TWO
-----------------
01.10.2018
which iterates one day next, as in this case, unless this holiday coincides with weekend.
you can define workdays to be whatever you like if you use a PL/SQL function
Have a simple prototype here - without any holidays - but it could be adapted for that purpose using the same kind of logic.
create or replace function add_business_days (from_date IN date, bd IN integer) return date as
fd date := trunc(from_date,'iw');
cnt int := (from_date-fd)+bd-1;
ww int := ceil(cnt/5);
wd int := mod(cnt,5);
begin
return from_date + (ww*7)+wd;
end;
/
I realize you already have an answer, but for what it's worth this is something we deal with all the time and have what has turned out to be a very good solution.
In effect, we maintain a separate table called "work days" that has every conceivable date we would ever compare (and that definition will vary from application to application, of course -- but in any case it will never be "huge" by RDBMS standards). There is a boolean flag that dictates if the date is a work day or a weekend/holiday, but more importantly there is an index value that only increments on work days. The table looks like this:
The advantage to this is transparency and scalability. If you want the difference between two dates in work days:
select
h.entry_date, h.invoice_date, wd2.workday_index - wd1.workday_index as delta
from
sales_order_data h
join util.work_days wd1 on h.sales_order_entry_dte = wd1.cal_date
join util.work_days wd2 on h.invoice_dte = wd2.cal_date
If you need to take a date in a table and add 20 days (like your original problem statement):
select
h.date_field_1, wd2.cal_date as date_field_1_plus_20
from
my_table h
join util.work_days wd1 on h.date_field_1 = wd1.cal_date
join util.work_days wd2 on
wd1.workday_index + 20 = wd2.workday_index and
wd2.is_workday
(disclaimer, this is in PostgreSQL, which is why I have the boolean. In Oracle, I'm guessing you need to change that to an integer and say =1)
Also, for the bonus question, this also gives two different options for defining "work day," one that rolls forward and another that rolls backwards (hence the workday_index and workday_index_back). For example, if you need something on a Saturday, and Saturday is not a work day, that means you need it on Friday. Conversely, if something is to be delivered on Saturday, and Saturday is not on a work day, then that means it will be available on Monday. The context of how to handle non-workdays differs, and this method affords you the option of chosing the right one.
As a final selling point, this option allows you to define holidays as non-work days also... and you can do this or not do this; it's up to you. The solution permits either option. You could theoretically add two more columns for work day index weekend only that gave you both options.
the title would be too wordy if I actually tried to cram it all in there but here's what I need help with...
We are trying to calculate retention of users. Our users have assignment start dates and assignment end dates that may overlap. What I need to do is look at all candidate assignments and determine if they are retained (30 days or less between previous end and new start). The tricky part: I need to assign the retention credit to the previous assignment end date. Here's a preview of the data:
month | user_id | start_date | end_date | rank | days_btw_assignment
1 5 1-1-16 1-31-16 1 NULL
2 5 2-14-16 4-15-16 2 15
6 4 6-01-16 11-01-16 1 NULL
8 4 8-01-16 11-01-16 2 -81
Therefore for user 5, I would need to give credit of retention to the month of jan-16' because their assignment end date ends 1-31-16. For User 4, where there assignments overlap, I would give credit of retention to nov-16' because their previous assignment end date ends 11-01-16.
I've restricted this example to use cases where they only have 2 assignments, though, there could be more. I just need a step in the right direction and I can probably handle all other use cases by myself.
Here's the sample code I'm currently using:
with placement_facts as (select date_trunc('month',assignment_start_date) as month, user_id, assignment_start_date, assignment_end_date, rank () over (partition by user_id order by assignment_start_date asc), extract( day from assignment_start_date - lag(assignment_end_date, 1) over (partition by user_id order by assignment_start_date asc)) as time_btw_placement
from activations as ca
join offers on ca.offer_id = offers.id
where assignment_start_date != assignment_end_date
order by 2,4 asc)
select placement_facts.month, count(distinct case when time_btw_placement <=30 then user_id else null end) as retained_raw
from placement_facts
group by 1;
Appreciate the help and please lmk if I nee to clarify anything!
If I understand your question then I think you can achieve what you want by replacing your use of LAG() with LEAD(). It's basically the same function but it looks at a given number of rows ahead.
I'd like to show every row where date_added equals '2015-02-18' and every seven days after, so '2015-02-25' and '2015-03-04' etc..
here's what I have so far
select * from table
where ((to_char(date_added, 'j')) /
((select to_char(d,'j') from (select date '2015-02-18' d from dual)))) = 1
That gets me the first desired date, however I'm stuck as to how to express it to show the next 7 days as a step additive function.
Any help is appreciated. Thank you.
One way to do this is with mod():
select *
from table
where mod(date_added - date '2015-02-18', 7) = 0;
Note: this assumes that the dates have no time component. If they do, then use trunc() to get rid of it.
I like Gordon's "mod()" solution but he's missing part of the requested solution.
In this case, I have a "calendar" table that includes a series of dates:
http://www.perpendulum.com/2012/06/calendar-table-script-for-oracle/
select *
from calendar
where date_time_start >= to_date('01-Jan-2013')
and mod(trunc(date_time_start) - to_date('01-Jan-2013'), 7) = 0;
Per the original question, you want records where the dates are equal to a given date and every seven days thereafter.
I am trying to query my postgresql db to return results where a date is in certain month and year. In other words I would like all the values for a month-year.
The only way i've been able to do it so far is like this:
SELECT user_id
FROM user_logs
WHERE login_date BETWEEN '2014-02-01' AND '2014-02-28'
Problem with this is that I have to calculate the first date and last date before querying the table. Is there a simpler way to do this?
Thanks
With dates (and times) many things become simpler if you use >= start AND < end.
For example:
SELECT
user_id
FROM
user_logs
WHERE
login_date >= '2014-02-01'
AND login_date < '2014-03-01'
In this case you still need to calculate the start date of the month you need, but that should be straight forward in any number of ways.
The end date is also simplified; just add exactly one month. No messing about with 28th, 30th, 31st, etc.
This structure also has the advantage of being able to maintain use of indexes.
Many people may suggest a form such as the following, but they do not use indexes:
WHERE
DATEPART('year', login_date) = 2014
AND DATEPART('month', login_date) = 2
This involves calculating the conditions for every single row in the table (a scan) and not using index to find the range of rows that will match (a range-seek).
From PostreSQL 9.2 Range Types are supported. So you can write this like:
SELECT user_id
FROM user_logs
WHERE '[2014-02-01, 2014-03-01]'::daterange #> login_date
this should be more efficient than the string comparison
Just in case somebody land here... since 8.1 you can simply use:
SELECT user_id
FROM user_logs
WHERE login_date BETWEEN SYMMETRIC '2014-02-01' AND '2014-02-28'
From the docs:
BETWEEN SYMMETRIC is the same as BETWEEN except there is no
requirement that the argument to the left of AND be less than or equal
to the argument on the right. If it is not, those two arguments are
automatically swapped, so that a nonempty range is always implied.
SELECT user_id
FROM user_logs
WHERE login_date BETWEEN '2014-02-01' AND '2014-03-01'
Between keyword works exceptionally for a date. it assumes the time is at 00:00:00 (i.e. midnight) for dates.
Read the documentation.
http://www.postgresql.org/docs/9.1/static/functions-datetime.html
I used a query like that:
WHERE
(
date_trunc('day',table1.date_eval) = '2015-02-09'
)
or
WHERE(date_trunc('day',table1.date_eval) >='2015-02-09'AND date_trunc('day',table1.date_eval) <'2015-02-09')
I've hit a little snag with one of my queries. I'm throwing together a simple chart to plot a number of reports being submitted by day of week.
My query to start was :
SELECT Weekday(incidentdate) AS dayOfWeek
, Count(*) AS NumberOfIncidents
FROM Incident
GROUP BY Weekday(incidentdate);
This works fine and returns what I want, something like
1 200
2 323
3 32
4 322
5 272
6 282
7 190
The problem is, I want the number returned by the weekday function to read the corresponding day of week, like case when 1 then 'sunday' and so forth. Since Access doesn;t have the SQL server equivalent that returns it as the word for the weekday, I have to work around.
Problem is, it's not coming out the way I want. So I wrote it using iif since I can't use CASE. The problem is, since each iif statement is treated like a column selection (the way I'm writing it), my data comes out unusable, like this
SELECT
iif(weekday(incidentdate) =1,'Sunday'),
iif(weekday(incidentdate) =2,'Monday')
'so forth
, Count(*) AS NumberOfIncidents
FROM tblIncident
GROUP BY Weekday(incidentdate);
Expr1000 Expr1001 count
Sunday 20
Monday 106
120
186
182
164
24
Of course, I want my weekdays to be in the same column as the original query. Halp pls
Use the WeekdayName() function.
SELECT
WeekdayName(Weekday(incidentdate)) AS dayOfWeek,
Count(*) AS NumberOfIncidents
FROM Incident
GROUP BY WeekdayName(Weekday(incidentdate));
As BWS Suggested, Switch was what I wanted. Here's what I ended up writing
SELECT
switch(
Weekday(incidentdate) = 1, 'Sunday',
Weekday(incidentdate) = 2,'Monday',
Weekday(incidentdate) = 3,'Tuesday',
Weekday(incidentdate) = 4,'Wednesday',
Weekday(incidentdate) = 5,'Thursday',
Weekday(incidentdate) = 6,'Friday',
Weekday(incidentdate) = 7,'Saturday'
) as DayOfWeek
, Count(*) AS NumberOfIncidents
FROM tblIncident
GROUP BY Weekday(incidentdate);
Posting this here so there's actual code for future readers
Edit: WeekdayName(weekday(yourdate)) as HansUp said it probably a little easier :)
check this previous post:
What is the equivalent of Select Case in Access SQL?
Why not just create a 7 row table with day number & day name then just join to it?