I want to get the weeks as columns between two dates in an MDX query.
For example if I input these dates: 2015-01-01 and 2015-02-01 I want to get this:
Week ending January 1,Week ending January 8,Week ending January 15,Week ending January 22
I have created this MDX query with "Filter" method but it doesn't work:
WITH SET [WeeksBetweenDates] AS
Filter([Date].[Year - Week - Date Hierarchy].[Week],
[Date].[Year - Week - Date Hierarchy].CurrentMember.Member_Value >= [Date].[Year - Week - Date Hierarchy].[Week].&[2015-01-01T00:00:00]
AND [Date].[Year - Week - Date Hierarchy].CurrentMember.Member_Value <= [Date].[Year - Week - Date Hierarchy].[Week].&[2015-02-01T00:00:00])
SELECT [WeeksBetweenDates] ON COLUMNS
FROM [Team System]
I get "Unknown" column if I run this query.
You should define dates as subcube slice ({[Date1] : [Date2]}), this will greatly increase query performance + make code more transparent (tested against our TFS):
select
{
[Date].[Week].members
} on 0
from ( select {
[Date].[Date].&[2015-01-01T00:00:00]
:
[Date].[Date].&[2015-02-01T00:00:00]
} on 0
from [Team System])
To make it work through using hierarchy, you should correct
[Date].[Year - Week - Date Hierarchy].[Week].&[2015-01-01T00:00:00] to [Date].[Year - Week - Date Hierarchy].[Date].&[2015-01-01T00:00:00]
Related
How can I get the number of days passed in the current quarter?
For example, if today is 1/2/2021, it will return 2.
If today is 2/5, it will return 36.
If today is 4/2, it will return 2.
Use date_trunc() to get the start of the quarter and subtract dates:
WITH cte(day) AS (
VALUES
(date '2021-01-02')
, (date '2021-02-05')
, (date '2021-04-02')
)
SELECT day
, day - date_trunc('quarter', day)::date + 1 AS days_passed_in_quarter
FROM cte;
day | days_passed_in_quarter
------------+------------------------
2021-01-02 | 2
2021-02-05 | 36
2021-04-02 | 2
+ 1 to fix off-by-one error as you clearly want to include the current day as "passed".
Always use unambiguous ISO 8601 date format (YYYY-MM-DD - 2021-02-05), which is the default in Postgres and always unambiguous, or you depend on the current datestyle setting (and may be in for surprises). Also avoids misunderstandings general communication.
Related:
PostgreSQL: between with datetime
i need to calculate difference between two date excluding sunday. I have table with dates and i need to calculate number of dates of repeated days from last date.
if i have dates like that
27-05-2017
29-05-2017
30-05-2017
I use this code in script
date(max(Date)) as dateMax,
date(min(Date)) as dateMin
And i get min date = 27-05-2017 and max date = 30-05-2017 then i use in expressions
=floor(((dateMax - dateMin)+1)/7)*6 + mod((dateMax - dateMin)+1,7)
+ if(Weekday(dateMin) + mod((dateMax - dateMin)+1,7) < 7, 0, -1)
And get result 3 days. Thats OK, but the problem is if I have next dates:
10-05-2017
11-05-2017
27-05-2017
29-05-2017
30-05-2017
When use previously code I get min date = 10-05-2017 and max date = 30-05-2017 and result 18, but this is not OK.
I need to count only dates from
27-05-2017
29-05-2017
30-05-2017
I need to get max date and go throw loop repeated dates and if have brake to see is that date sunday if yes then step that date and continue to count repeated dates and if i again have break and if not sunday than close loop and remember number of days.
In my case instead of 18 days i need to get 3 days.
Any idea?
I'd recommend you creating a master calendar in the script where you can apply weights or any other rule to your days. Then in your table or app you can just loop through the dates or perform operations and sum their weights (0: if sunday, 1: if not). Let's see an example:
// In this case I'll do a master calendar of the present year
LET vMinDate = Num(MakeDate(year(today()),1,1));
LET vMaxDate = Num(MakeDate(year(today()),12,31));
Calendar_tmp:
LOAD
$(vMinDate) + Iterno() - 1 as Num,
Date($(vMinDate) + Iterno() - 1) as Date_tmp
AUTOGENERATE 1 WHILE $(vMinDate) + Iterno() - 1 <= $(vMaxDate);
Master_Calendar:
LOAD
Date_tmp AS Date,
Week(Date_tmp) as Week,
Year(Date_tmp) as Year,
Capitalize(Month(Date_tmp)) as Month,
Day(Date_tmp) as Day,
WeekDay(Date_tmp) as WeekDay,
if(WeekDay = '7',0,1) as DayWeight //HERE IS WHERE YOU COULD DEFINE A VARIABLE TO DIRECTLY COUNT THE DAY IF IT IS NOT SUNDAY
'T' & ceil(num(Month(Date_tmp))/3) as Quarter,
'T' & ceil(num(Month(Date_tmp))/3) & '-' & right(year(Date_tmp),2) as QuarterYear,
date(monthStart(Date_tmp),'MMMM-YYYY') as MonthYear,
date(monthstart(Date_tmp),'MMM-YY') as MonthYear2
RESIDENT Calendar_tmp
ORDER BY Date_tmp ASC;
DROP Table Calendar_tmp;
I need to make my Access query always return the Monday of the current week. I have seen a few solutions on Google/StackOverflow but they are written in SQL and I am a beginner in creating Access queries (I am using the Design view to make them).
Goal: The week should be considered as M T W T F S S. Then, the query should always return the Monday of the current week. Therefore, if it is Sunday, it should still return the Monday before, NOT the next week's Monday. Can anyone explain how to do this using the Design View in Access 2010?
Keep in mind that in this context we are working with dates, so if we do Date() - 1, we will get 1 day prior to today.
Date() ~ Today's date
DatePart(
"w" - Weekday
Date() - Today's date
2 - vBMonday (Access assumes Sunday is the first day of the week, which is why this is necessary.)
1 - vbFirstJan1 - This gets into using the first week of the year. We could have omitted this, as 1 is the default.
)
-1 - Subtract 1 from the DatePart value.
Values
Date() = 4/27/2015 (at time of this writing)
DatePart("w",Date(),2,1) = 1
DatePart("w",Date(),2,1)-1 = 0
So we have Date()-0... Okay, what's so great about that? Well, let's look at a more useful scenario where today's date is a day other than Monday.
Let's act like today is 4/28/2015 (Tuesday)
Date() = 4/28/2015
DatePart("w",Date(),2,1) = 2
DatePart("w",Date(),2,1)-1 = 1
So, from the outside, in; give me the current weekday value. (1 = Monday, 2 = Tuesday, etc.), and subtract 1 from that -> that's how many days we need to subtract from the current date to get back to the weekday value of 1 (Monday).
Here's a function that will do this:
Public Function DatePrevWeekday( _
ByVal datDate As Date, _
Optional ByVal bytWeekday As VbDayOfWeek = vbMonday) _
As Date
' Returns the date of the previous weekday, as spelled in vbXxxxday, prior to datDate.
' 2000-09-06. Cactus Data ApS.
' No special error handling.
On Error Resume Next
DatePrevWeekday = DateAdd("d", 1 - Weekday(datDate, bytWeekday), datDate)
End Function
As vbMonday is 2 and your date is today, you can use the core expression in a query:
PreviousMonday: DateAdd("d",1-Weekday(Date(),2),Date())
I've got the following problem:
i need to convert a datetime field in a where select to a date field.
scope :after_start_date, lambda { |value| where('end_time <= (?)', value) if value}
I need something like where(end_time.to_date + " >= (?)", value)
Is there a nice way to do this?
Update/Example
08.09.2014 14:40 as DB_END_TIME
08.09.2014 AS PARAM_END_TIME
I wanna get all results which until the date 08.09.2014 (including the 8th),
but the comparison where(DB_END_TIME <= PARAM_END_TIME) does not include 08.09.2014
because the PARAM_END_TIME has 00:00:00 as time value.
And i think its not nice to do something like this:
PARAM_END_TIME + 1.day - 1.second - this would work.
So i'd like to convert the database value to the date format in the where statement.
you can use ruby strftime method to create ur own date/time display.
time.strftime( string ) => string
%a - The abbreviated weekday name (``Sun'')
%A - The full weekday name (``Sunday'')
%b - The abbreviated month name (``Jan'')
%B - The full month name (``January'')
%c - The preferred local date and time representation
%d - Day of the month (01..31)
%H - Hour of the day, 24-hour clock (00..23)
%I - Hour of the day, 12-hour clock (01..12)
%j - Day of the year (001..366)
%m - Month of the year (01..12)
%M - Minute of the hour (00..59)
%p - Meridian indicator (``AM'' or ``PM'')
%S - Second of the minute (00..60)
%U - Week number of the current year,
starting with the first Sunday as the first
day of the first week (00..53)
%W - Week number of the current year,
starting with the first Monday as the first
day of the first week (00..53)
%w - Day of the week (Sunday is 0, 0..6)
%x - Preferred representation for the date alone, no time
%X - Preferred representation for the time alone, no date
%y - Year without a century (00..99)
%Y - Year with century
%Z - Time zone name
%% - Literal ``%'' character
t = Time.now
t.strftime("Printed on %m/%d/%Y") #=> "Printed on 04/09/2003"
t.strftime("at %I:%M%p") #=> "at 08:56AM"
Example:
date = Date.parse('2014-09-08 00:00:00 UTC')
# => Mon, 08 Sep 2014
date.strftime('%d-%m-%Y')
# => "08-09-2014"
And search in database accordingly.. I hope this may helps you
Edit:
If you want to fetch data from your SQL Server and convert datetime to date then have a look at this
When using LastFullWeek i get the last Sunday - Saturday week, but for my reports i want the last Monday - Sunday week.
Is there any simple way to get this behaviour, or do I have to write my own function for it (which isn't that hard, but unconvenient for such a common date span.)
custom function:
//LastFullWeekEx
Function (DateVar date, Optional NumberVar firstDayOfWeek := crSunday)
(date - DayOfWeek(date, firstDayOfWeek)) - 6 TO (date - DayOfWeek(date, firstDayOfWeek))
usage:
// use with non-volatile DataDate and Sunday
{TABLE.DATE} IN LastFullWeekEx(DataDate)
// use with non-volatile DataDate and Monday
{TABLE.DATE} IN LastFullWeekEx(DataDate, crMonday)
testing:
// should return True
( Minimum(LastFullWeek) = Minimum(LastFullWeekEx(DataDate, crSunday)) ) AND
( Maximum(LastFullWeek) = Maximum(LastFullWeekEx(DataDate, crSunday)) )
Instead of using
{DATE} in LastFullWeek
use the form
{DATE}-1 in LastFullWeek
The code, if no better answer comes along, for others who find this question through the search engine:
{DATE} >= currentdate - dayofweek(currentdate, crMonday) - 6 AND
{DATE} < currentdate - dayofweek(currentdate, crMonday) + 1
You have no other choice than to use a custom formule as Crystal Reports uses fixed week settings (US). So a week is Sunday to Saturday and DayOfWeek starts with Sunday (value/index 1).
Add +1 to LastFullWeek min and max:
DateVar Start := Minimum(LastFullWeek)+1; //Monday
DateVar End := Maximum(LastFullWeek)+1; //Sunday
Edit: Since LastFullWeek always takes Sunday as the first day of the week, we need to check if the current day is Sunday. If so, we need to subtract a week:
// Check if the current day is Sunday
IF (DAYOFWEEK(CurrentDate) = 1)
THEN (
Start := Start - 7;
End := End - 7
);
Example: Date range to string
StringVar LastWeekRange := ToText(Start) + " - " + ToText(End);
LastWeekRange;
Return example: "MM/DD/YYYY - MM/DD/YYYY"
Example: Selecting all dates within date range
{datefield} >= Start and {datefield} <= End;