I have a whole bunch of tariffs, some work on weekends, some work on weekdays some on both. Sometimes I'll be querying on NOW() but sometimes I'll be querying on datetime column.
id | Weekday | Weekend | Price
1 | 1 | 0 | 0.04
2 | 0 | 1 | 0.02
date
2020-04-15 00:00:00
2012-04-16 00:00:00
The date is from another table and is not related to the Price / days of week.
I know I can get the weekend dates by
SELECT * FROM tariff where EXTRACT(ISODOW FROM date) IN (6,7)
however I can't think of how I'd get rows that are either weekend / weekdays or both given a date.
** edit **
Updated the tables to show the dates are seperate. What I'm trying to get is the tariff that corresponds to the date in that table, whether it's on a week day or a weekend (or both but I can extrapolate that).
The weekend 1 is the tariff that is used for weekends, weekdays 1, all days is both.
Cannot give you a query, supply anything to query. Nor can we be sure that the columns Weekday and Weekend mean as you didn't tell us. But if we take them as boolean indicator where 1 means desired may some thing like will work for you.
select ...
from ...
where ...
and ( (weekday = 1 and weekend =1)
or (weekday = 1 and extract(isodow from date) not in (6,7))
or (weekend = 1 and extract(isodow from date) in (6,7))
) ;
Related
I am trying to get week numbers in a Year starting from a certain day
I've checked the stack but quite confused.
SELECT EXTRACT(WEEK FROM TIMESTAMP '2021-01-01'),
extract('year' from TIMESTAMP '2021-01-01')
The output is 53|2021
I want it to be 01|2021
I understand the principle of the isoweek but I want the year to start in 01-01-2021
The aim is to use intervals from this day to determine week numbers
Week N0| End Date
1 | 01-01-2021
2 | 01-08-2021
5 | 01-29-2021
...
This is really strange way to determine the week number, but in the end it's a simple math operation: the number of days since January first divided by 7.
You can create a function for this:
create function custom_week(p_input date)
returns int
as
$$
select (p_input - date_trunc('year', p_input)::date) / 7 + 1;
$$
language sql
immutable;
So this:
select date, custom_week(date)
from (
values
(date '2021-01-01'),
(date '2021-01-08'),
(date '2021-01-29')
) as v(date)
yields
date | custom_week
-----------+------------
2021-01-01 | 1
2021-01-08 | 2
2021-01-29 | 5
I am currently writing a Crystal Report that has a DB2 query as its backend. I have finished the query but am stuck on the date portion of it. I am going to be running it twice a month - once on the 16th, and once on the 1st of the next month. Here's how it should work:
If I run it on the 16th of the month, it will give me results from the 1st of that same month to the 15th of that month.
If I run it on the 1st of the next month, it will give me results from the 16th of the previous month to the last day of the previous month.
This comes down a basic bi-monthly report. I've found plenty of hints to do this in T-SQL, but no efficient ways on how to accomplish this in DB2. I'm having a hard time wrapping my head around the logic to get this to consistently work, taking into account differences in month lengths and such.
There are 2 expressions for start and end date of an interval depending on the report date passed, which you may use in your where clause.
The logic is as follows:
1) If the report date is the 1-st day of a month, then:
DATE_START is 16-th of the previous month
DATE_END is the last day of the previous month
2) Otherwise:
DATE_START is 1-st of the current month
DATE_END is 15-th of the current month
SELECT
REPORT_DATE
, CASE DAY(REPORT_DATE) WHEN 1 THEN REPORT_DATE - 1 MONTH + 15 ELSE REPORT_DATE - DAY(REPORT_DATE) + 1 END AS DATE_START
, CASE DAY(REPORT_DATE) WHEN 1 THEN REPORT_DATE - 1 ELSE REPORT_DATE - DAY(REPORT_DATE) + 15 END AS DATE_END
FROM
(
VALUES
DATE('2020-02-01')
, DATE('2020-02-05')
, DATE('2020-02-16')
) T (REPORT_DATE);
The result is:
|REPORT_DATE|DATE_START|DATE_END |
|-----------|----------|----------|
|2020-02-01 |2020-01-16|2020-01-31|
|2020-02-05 |2020-02-01|2020-02-15|
|2020-02-16 |2020-02-01|2020-02-15|
In Db2 (for Unix, Linux and Windows) it could be a WHERE Condition like
WHERE
(CASE WHEN date_part('days', CURRENT date) > 15 THEN yourdatecolum >= this_month(CURRENT date) AND yourdatecolum < this_month(CURRENT date) + 15 days
ELSE yourdatecolum > this_month(CURRENT date) - 1 month + 15 DAYS AND yourdatecolum < this_month(CURRENT date)
END)
Check out the THIS_MONTH function - there are multiple ways to do it. Also DAYS_TO_END_OF_MONTH might be helpful
To find the number of days between two dates we can use something like this:
SELECT date_part('day',age('2017-01-31','2017-01-01')) as total_days;
In the above query we got 30 as output instead of 31. Why is that?
And I also want to find the number of days except Sundays. Expected output for the interval ('2017-01-01', '2017-01-31'):
Total Days = 31
Total Days except Sundays = 26
You need to define "between two dates" more closely. Lower and upper bound included or excluded? A common definition would be to include the lower and exclude the upper bound of an interval. Plus, define the result as 0 when lower and upper bound are identical. This definition happens to coincide with date subtraction exactly.
SELECT date '2017-01-31' - date '2017-01-01' AS days_between
This exact definition is important for excluding Sundays. For the given definition an interval from Sun - Sun (1 week later) does not include the upper bound, so there is only 1 Sunday to subtract.
interval in days | sundays
0 | 0
1-6 | 0 or 1
7 | 1
8-13 | 1 or 2
14 | 2
...
An interval of 7 days always includes exactly one Sunday.
We can get the minimum result with a plain integer division (days / 7), which truncates the result.
The extra Sunday for the remainder of 1 - 6 days depends on the first day of the interval. If it's a Sunday, bingo; if it's a Monday, too bad. Etc. We can derive a simple formula from this:
SELECT days, sundays, days - sundays AS days_without_sundays
FROM (
SELECT z - a AS days
, ((z - a) + EXTRACT(isodow FROM a)::int - 1 ) / 7 AS sundays
FROM (SELECT date '2017-01-02' AS a -- your interval here
, date '2017-01-30' AS z) tbl
) sub;
Works for any given interval.
Note: isodow, not dow for EXTRACT().
To include the upper bound, just replace z - a with (z - a) + 1. (Would work without parentheses, due to operator precedence, but better be clear.)
Performance characteristic is O(1) (constant) as opposed to a conditional aggregate over a generated set with O(N).
Related:
How do I determine the last day of the previous month using PostgreSQL?
Calculate working hours between 2 dates in PostgreSQL
You could try using generate_series to generate all the dates between given date and then take count of required days.
SELECT
count(case when extract(dow from generate_series) <> 0 then 1 end) n
from generate_series('2017-01-01'::date,'2017-01-31'::date, '1 day');
Hello guys I need know the number of the week of a month
For example:
Date | WeekOfTheMonth
2015-04-15 | 3
2015-03-01 | 1
2015-01-08 | 2
Beacuse in docs only see Week of the year
Thanks
Test this SQL query
SELECT KURRENT - FIRSTY + 1 FROM ( -- AVOID CURRENT AND FIRST KEYWORD
SELECT
WEEK_ISO(DATE(1) + (YEAR(date_colum)-1) YEARS + (MONTH(date_colum)-1) MONTHS )AS FIRSTY ,
WEEK_ISO(date(date_column)) AS KURRENT
FROM TEST_DATE_TABLE
) AS T
Try this:
SELECT week_iso(current date) -
week_iso(last_day(current date) - 1 month + 1 day)
FROM sysibm.sysdummy1
You don't mention your platform, but VARCHAR_FORMAT has a W format string, which gives the week of the month:
Week of the month (1-5), where week 1 starts on the first day of the
month and ends on the seventh day.
SELECT VARCHAR_FORMAT(CURRENT TIMESTAMP, 'W') FROM SYSIBM.SYSDUMMY1
Requirements
I have data table that saves data in date ranges.
Each record is allowed to overlap previous record(s) (record has a CreatedOn datetime column).
New record can define it's own date range if it needs to hence can overlap several older records.
Each new overlapping record overrides settings of older records that it overlaps.
Result set
What I need to get is get per day data for any date range that uses record overlapping. It should return a record per day with corresponding data for that particular day.
To convert ranges to days I was thinking of numbers/dates table and user defined function (UDF) to get data for each day in the range but I wonder whether there's any other (as in better* or even faster) way of doing this since I'm using the latest SQL Server 2008 R2.
Stored data
Imagine my stored data looks like this
ID | RangeFrom | RangeTo | Starts | Ends | CreatedOn (not providing data)
---|-----------|----------|--------|-------|-----------
1 | 20110101 | 20110331 | 07:00 | 15:00
2 | 20110401 | 20110531 | 08:00 | 16:00
3 | 20110301 | 20110430 | 06:00 | 14:00 <- overrides both partially
Results
If I wanted to get data from 1st January 2011 to 31st May 2001 resulting table should look like the following (omitted obvious rows):
DayDate | Starts | Ends
--------|--------|------
20110101| 07:00 | 15:00 <- defined by record ID = 1
20110102| 07:00 | 15:00 <- defined by record ID = 1
... many rows omitted for obvious reasons
20110301| 06:00 | 14:00 <- defined by record ID = 3
20110302| 06:00 | 14:00 <- defined by record ID = 3
... many rows omitted for obvious reasons
20110501| 08:00 | 16:00 <- defined by record ID = 2
20110502| 08:00 | 16:00 <- defined by record ID = 2
... many rows omitted for obvious reasons
20110531| 08:00 | 16:00 <- defined by record ID = 2
Actually, since you are working with dates, a Calendar table would be more helpful.
Declare #StartDate date
Declare #EndDate date
;With Calendar As
(
Select #StartDate As [Date]
Union All
Select DateAdd(d,1,[Date])
From Calendar
Where [Date] < #EndDate
)
Select ...
From Calendar
Left Join MyTable
On Calendar.[Date] Between MyTable.Start And MyTable.End
Option ( Maxrecursion 0 );
Addition
Missed the part about the trumping rule in your original post:
Set DateFormat MDY;
Declare #StartDate date = '20110101';
Declare #EndDate date = '20110501';
-- This first CTE is obviously to represent
-- the source table
With SampleData As
(
Select 1 As Id
, Cast('20110101' As date) As RangeFrom
, Cast('20110331' As date) As RangeTo
, Cast('07:00' As time) As Starts
, Cast('15:00' As time) As Ends
, CURRENT_TIMESTAMP As CreatedOn
Union All Select 2, '20110401', '20110531', '08:00', '16:00', DateAdd(s,1,CURRENT_TIMESTAMP )
Union All Select 3, '20110301', '20110430', '06:00', '14:00', DateAdd(s,2,CURRENT_TIMESTAMP )
)
, Calendar As
(
Select #StartDate As [Date]
Union All
Select DateAdd(d,1,[Date])
From Calendar
Where [Date] < #EndDate
)
, RankedData As
(
Select C.[Date]
, S.Id
, S.RangeFrom, S.RangeTo, S.Starts, S.Ends
, Row_Number() Over( Partition By C.[Date] Order By S.CreatedOn Desc ) As Num
From Calendar As C
Join SampleData As S
On C.[Date] Between S.RangeFrom And S.RangeTo
)
Select [Date], Id, RangeFrom, RangeTo, Starts, Ends
From RankedData
Where Num = 1
Option ( Maxrecursion 0 );
In short, I rank all the sample data preferring the newer rows that overlap the same date.
Why do it all in DB when you can do it better in memory
This is the solution (I eventually used) that seemed most reasonable in terms of data transferred, speed and resources.
get actual range definitions from DB to mid tier (smaller amount of data)
generate in memory calendar of a certain date range (faster than in DB)
put those DB definitions in (much easier and faster than DB)
And that's it. I realised that complicating certain things in DB is not not worth it when you have executable in memory code that can do the same manipulation faster and more efficient.