I have a table of months, and it has two columns of dates : start date, and end date. And I need your help to check the missing months in date ranges using a simple sql query.
The example below illustrates sample of data:
08/12/2018 - 06/01/2019
07/01/2019 - 05/02/2019
08/03/2019 - 05/04/2019
06/04/2019 - 05/05/2019
Expected result:
Missing months
06/02/2019 - 07/03/2019
06/05/2019 - 03/06/2019
Note that l’m using Hijri calendar not Gregorian calendar, so the first month in my example which is 08/12/2018 G refers to this date 01/04/1440 H in Hijri calendar , and its end date 06/01/2019 refers to 30/04/1440 H and so on.. Note that I used this date format DD/MM/YYYY and Oracle sql DB.
Any help is greatly appreciated.
Your expected result does not seem to match your sample data and explanation, but I understand that you want to exhibit the boundaries of intervals that are not adjacent. You could do this with lag():
select
end_date + interval 1 day start_gap,
lead_start_date - interval 1 day end_gap
from (
select
t.*,
lead(start_date) over(order by start_date) lead_start_date
from mytable t
) t
where lead_start_date > end_date + interval 1 day
This uses standard date arithmetic; the exact syntax may vary depending on your RDBMS.
Related
I have a date field in one of my tables and the column name is from_dt. Now I have to compare a month and year combination against this from_dt field and check whether the month has already passed. The current database function uses separate conditions for the month and the year, but this is wrong as it will compare month and year separately. The current code is like this
SELECT bill_rate, currency FROM table_name WHERE
emp_id = employee_id_param
AND EXTRACT(MONTH FROM from_dt) <= month_param
AND EXTRACT(YEAR FROM from_dt) <= year_param
Now the fromt_dt field has value 2021-10-11. If I give month_param as 01 and year_param as 2022, this condition will not work as the month 10 is greater than 1, which I have given. Basically, I need to check whether 01-2022 (Jan 2022) is greater than r equal to 2021-10-01(October 1st, 2021). It would be very much helpful if someone can shed some light here.
If you just want to check whether one date is >= then another:
# select '2022-01-01'::date >= '2021-10-11'::date;
?column?
----------
t
If you want to restrict to year/month then:
select date_trunc('month','2022-01-01'::date) >= date_trunc('month', '2021-10-11'::date);
?column?
----------
t
Where the date_trunc components are:
select date_trunc('month','2022-01-01'::date) ;
date_trunc
------------------------
2022-01-01 00:00:00-08
select date_trunc('month','2021-10-11'::date) ;
date_trunc
------------------------
2021-10-01 00:00:00-07
See Postgres date_trunc for more information.
Assuming the given year_param and month_param are integers you can use the make_date function to create the first of the year_month and date_trunc to get the first on the month from the table. Just compare those values. (See date functions) So:
select bill_rate, currency
from table_name
where emp_id = employee_id_param
and date_trunc('month',from_dt) =
make_date( year_param, month_param, 01);
I am new to Big Query. I am trying to do a where condition to only select yesterday's data and that of same day last year (in this case, 10/25/2021 data and 10/25/2020 data). I know how to select a range of data, but I couldn't figure out a way to only select those 2 days of data. Any help is appreciated.
I recommend using BigQuery functions to define dates. You can read about them here.
WHERE DATE(your_date_field) IN ((DATE_SUB(CURRENT_DATE(), INTERVAL 1 DAY),
DATE_SUB(DATE_SUB(CURRENT_DATE(), INTERVAL 1 DAY), INTERVAL 1 YEAR))
This is dynamic to any day that you run the query. It will take the current date, then subtract 1 day. For the other date, it will take the current date and subtract 1 day and then 1 year, making it yesterday's date 1 year prior.
WHERE date_my_field IN (DATE('2021-10-25'), DATE('2020-10-25'))
Use IN which is a short cut for OR operator
Consider below (less verbose approach - especially if you remove time zone)
select
current_date('America/Los_Angeles') - 1 as yesterday,
date(current_date('America/Los_Angeles') - 1 - interval 1 year) same_day_last_year
with output
So, now you can use it in your WHERE clause as in below example (with dummy data via CTE)
with data as (
select your_date_field
from unnest(generate_date_array(current_date() - 1000, current_date())) your_date_field
)
select *
from data
where your_date_field in (
current_date('America/Los_Angeles') - 1,
date(current_date('America/Los_Angeles') - 1 - interval 1 year)
)
with output
I am new to SQL, here is my problem.
I have a table with daily dates:
Date:
20190101
20190102
20190103
.
**20190131**
20190201
20190202
20190203
.
**20190228**
20190301
20190302
20190303
.
**20190331**
I want to select only the month-end dates, what would be the code to do that?
thanks
I am using MS SQL Studio.
One method in standard SQL would be:
select t.*
from t
where extract(month from date + interval '1' day) <> extract(month from date);
Date/time functions vary significantly by database, so the exact functions might not match your database. However, the idea is simple: add one day and see if the month changes.
In standard SQL, you could do:
select date
from mytable
where date = date_trunc('month', date) + interval '1' month - interval '1' day
Edit
In SQL Server, you can just use eomonth(). Given a date, this functions returns the corresponding end of month, which you can compare against the date. So:
select date
from mytable
where date = eomonth(date)
I have below scenario that Business want to calculate Week Number based on Given Start Date to End Date.
For Ex: Start Date = 8/24/2020 End Date = 12/31/2020 ( These Start date & end date are not constant they may change from year to year )
Expected Output below:
[Date 1 Date 2 Week Number
8/24/2020 8/30/2020 week1
8/31/2020 9/6/2020 week2
9/7/2020 9/14/2020 week3
9/15/2020 9/21/2020 week4
9/22/2020 9/28/2020 week5
9/29/2020 10/5/2020 week6
10/6/2020 10/12/2020 week7
10/13/2020 10/19/2020 week8
10/20/2020 10/26/2020 week9
10/27/2020 11/02/2020 week10
11/03/2020 11/09/2020 week11
11/10/2020 11/16/2020 week12
11/17/2020 11/23/2020 week13
11/24/2020 11/30/2020 week14
I need Oracle Query to calculate Week Number(s) like above .. Based on Start date for 7 days then week number will be calcuated.. But remember that crossing months some month have 30 days and some month 31 days etc.. How to calculate ? Appreciate your help!!
Seems your looking for custom week definition rather that built-ins. But not overly difficult. The first thing is to convert from strings to dates (if columns actually coming off table this conversion is not required), and from there let Oracle do all the calculations as you can apply arithmetic operations to dates, except adding 2 dates. Oracle will automatically handle differing number of days per month correctly.
Two methods for this request:
Use a recursive CTE (with)
with dates(start_date,end_date) as
( select date '2020-08-24' start_date
, date '2020-12-31' end_date
from dual
)
, weeks (wk, wk_start, wk_end, e_date) as
( select 1, start_date, start_date+6 ld, end_date from dates
union all
select wk+1, wk_end+1, wk_end+7, e_date
from weeks
where wk_end<e_date
)
select wk, wk_start, wk_end from weeks;
Use Oracle connect by
with dates(start_date,end_date) as
( select date '2020-08-24' start_date
, date '2020-12-31' end_date
from dual
)
select level wk
, start_date+7*(level-1) wk_start
, start_date+6+7*(level-1)
from dates
connect by level <= ceil( (end_date-start_date)/7.0);
Depend on how strict you need to be with the end date specified you may need to adjust the last row returned. Both queries do not make adjust for that. They just ensure no week begins after that date. But the last week contains the full 7 days, which may end after the specified end date.
If your date datatype is varchar then first convert it to date and then convert it back to varchar.
convert date to to_char(to_date('8/24/2020','MM/DD/YYYY'),'WW')
If you to keep week datatype as a number then you can do something like this
to_number(to_char(to_date('8/24/2020','MM/DD/YYYY'),'WW'))
Few options according to your need.
WW Week of year (1-53) where week 1 starts on the first day of the year and continues to the seventh day of the year.
W Week of month (1-5) where week 1 starts on the first day of the month and ends on the seventh.
IW Week of year (1-52 or 1-53) based on the ISO standard.
I'm having a hard time explaining this through writing, so please be patient.
I'm making this project in which I have to choose a month and a year to know all the active employees during that month of the year.. but in my database I'm storing the dates when they started and when they finished in dd/mm/yyyy format.
So if I have an employee who worked for 4 months eg. from 01/01/2013 to 01/05/2013 I'll have him in four months. I'd need to make him appear 4 tables(one for every active month) with the other employees that are active during those months. In this case those will be: January, February, March and April of 2013.
The problem is I have no idea how to make a query here or php processing to achieve this.
All I can think is something like (I'd run this query for every month, passing the year and month as argument)
pg_query= "SELECT employee_name FROM employees
WHERE month_and_year between start_date AND finish_date"
But that can't be done, mainly because month_and_year must be a column not a variable.
Ideas anyone?
UPDATE
Yes, I'm very sorry that I forgot to say I was using DATE as data type.
The easiest solution I found was to use EXTRACT
select * from employees where extract (year FROM start_date)>='2013'
AND extract (month FROM start_date)='06' AND extract (month FROM finish_date)<='07'
This gives me all records from june of 2013 you sure can substite the literal variables for any variable of your preference
There is no need to create a range to make an overlap:
select to_char(d, 'YYYY-MM') as "Month", e.name
from
(
select generate_series(
'2013-01-01'::date, '2013-05-01', '1 month'
)::date
) s(d)
inner join
employee e on
date_trunc('month', e.start_date)::date <= s.d
and coalesce(e.finish_date, 'infinity') > s.d
order by 1, 2
SQL Fiddle
If you want the months with no active employees to show then change the inner for a left join
Erwin, about your comment:
the second expression would have to be coalesce(e.finish_date, 'infinity') >= s.d
Notice the requirement:
So if I have an employee who worked for 4 months eg. from 01/01/2013 to 01/05/2013 I'll have him in four months
From that I understand that the last active day is indeed the previous day from finish.
If I use your "fix" I will include employee f in month 05 from my example. He finished in 2013-05-01:
('f', '2013-04-17', '2013-05-01'),
SQL Fiddle with your fix
Assuming that you really are not storing dates as character strings, but are only outputting them that way, then you can do:
SELECT employee_name
FROM employees
WHERE start_date <= <last date of month> and
(finish_date >= <first date of month> or finish_date is null)
If you are storing them in this format, then you can do some fiddling with years and months.
This version turns the "dates" into strings of the form "YYYYMM". Just express the month you want like this and you can do the comparison:
select employee_name
from employees e
where right(start_date, 4)||substr(start_date, 4, 2) <= 'YYYYMM' and
(right(finish_date, 4)||substr(finish_date, 4, 2) >= 'YYYYMM' or finish_date is null)
NOTE: the expression 'YYYYMM' is meant to be the month/year you are looking for.
First, you can generate multiple date intervals easily with generate_series(). To get lower and upper bound add an interval of 1 month to the start:
SELECT g::date AS d_lower
, (g + interval '1 month')::date AS d_upper
FROM generate_series('2013-01-01'::date, '2013-04-01', '1 month') g;
Produces:
d_lower | d_upper
------------+------------
2013-01-01 | 2013-02-01
2013-02-01 | 2013-03-01
2013-03-01 | 2013-04-01
2013-04-01 | 2013-05-01
The upper border of the time range is the first of the next month. This is on purpose, since we are going to use the standard SQL OVERLAPS operator further down. Quoting the manual at said location:
Each time period is considered to represent the half-open interval
start <= time < end [...]
Next, you use a LEFT [OUTER] JOIN to connect employees to these date ranges:
SELECT to_char(m.d_lower, 'YYYY-MM') AS month_and_year, e.*
FROM (
SELECT g::date AS d_lower
, (g + interval '1 month')::date AS d_upper
FROM generate_series('2013-01-01'::date, '2013-04-01', '1 month') g
) m
LEFT JOIN employees e ON (m.d_lower, m.d_upper)
OVERLAPS (e.start_date, COALESCE(e.finish_date, 'infinity'))
ORDER BY 1;
The LEFT JOIN includes date ranges even if no matching employees are found.
Use COALESCE(e.finish_date, 'infinity')) for employees without a finish_date. They are considered to be still employed. Or maybe use current_date in place of infinity.
Use to_char() to get a nicely formatted month_and_year value.
You can easily select any columns you need from employees. In my example I take all columns with e.*.
The 1 in ORDER BY 1 is a positional parameter to simplify the code. Orders by the first column month_and_year.
To make this fast, create an multi-column index on these expressions. Like
CREATE INDEX employees_start_finish_idx
ON employees (start_date, COALESCE(finish_date, 'infinity') DESC);
Note the descending order on the second index-column.
If you should have committed the folly of storing temporal data as string types (text or varchar) with the pattern 'DD/MM/YYYY' instead of date or timestamp or timestamptz, convert the string to date with to_date(). Example:
SELECT to_date('01/03/2013'::text, 'DD/MM/YYYY')
Change the last line of the query to:
...
OVERLAPS (to_date(e.start_date, 'DD/MM/YYYY')
,COALESCE(to_date(e.finish_date, 'DD/MM/YYYY'), 'infinity'))
You can even have a functional index like that. But really, you should use a date or timestamp column.