Hi all I am using kendo library in my page. I am parsing a date to get the month using the below code
kendo.toString(new Date(2000, 1, 1), "M")
and '
kendo.toString(new Date(2000, 1, 1), "MM")
but the result shows me "01 February" and "02" respectivly.
what documentation says is
"M" - The month, from 1 through 12.
"MM" - The month, from 01 through 12.
But I am not sure why the month is going for feb instead of Jan. Any help appreciated.
It is correct! Check Date documentation here
Months in JavaScript are 0 based which means that month 0 is January. In KendoUI "MM" format is the number of the month in 1 base (1 for January).
So the problem is that when you do Date(2000, 1, 1) you are saying February 1st, 2000.
Related
Is there anything like the Oracle "NEXT_DAY" function available in the syntax that Crystal Reports uses?
I'm trying to write a formula to output the following Monday # 9:00am if the datetime tested falls between Friday # 9:00pm and Monday # 9:00am.
So far I have
IF DAYOFWEEK ({DATETIMEFROMMYDB}) IN [7,1]
OR (DAYOFWEEK({DATETIMEFROMMYDB}) = 6 AND TIME({DATETIMEFROMMYDB}) in time(21,00,00) to time(23,59,59))
OR (DAYOFWEEK({DATETIMEFROMMYDB}) = 2 AND TIME({DATETIMEFROMMYDB}) in time(00,00,00) to time(08,59,59))
THEN ...
I know I can write seperate IF statements to do a different amount of DateAdd for each of Fri, Sat, Sun, Mon, but if I can keep it concise by lumping all of these into one I would much prefer it. I'm already going to be adding additional rules for if the datetime falls outside of business hours on the other weekdays so I want to do as much as possible to prevent this from becoming a very overgrown and ugly formula.
Since there is no CR equivalent that I know of, you can just cheat and borrow the NEXT_DAY() function from the Oracle database. You can do this by creating a SQL Expression and then entering something like:
-- SQL Expression {%NextDay}
(SELECT NEXT_DAY("MYTABLE"."MYDATETIME", 'MONDAY')
FROM DUAL)
then you could either use that directly in your formula:
IF DAYOFWEEK ({MYTABLE.MYDATETIME}) IN [7,1]
OR (DAYOFWEEK({MYTABLE.MYDATETIME}) = 6 AND TIME({MYTABLE.MYDATETIME}) in time(21,00,00) to time(23,59,59))
OR (DAYOFWEEK({MYTABLE.MYDATETIME}) = 2 AND TIME({MYTABLE.MYDATETIME) in time(00,00,00) to time(08,59,59))
THEN DateTime(date({%NextDay}),time(09,00,00))
Or, the even better way would be to just stuff ALL of the logic into the SQL Expression and do away with the formula altogether.
Considering Sunday is 1
And the first 7 is the week we want to back
7 = 1 week
14 = 2 weeks
The last Number (1) is 1 for Sunday, 2 for Monday, 3 for Tuestday
Last Sunday 1 week ago
Today - 7 + ( 7 - WEEKDAY(TODAY) )+1
Last Monday 2 weeks ago
Today - 14 + ( 7 - WEEKDAY(TODAY) )+2
So this 2 formulas give me MONDAY LAST WEEK and SUNDAY LAST WEEK.
EvaluateAfter({DATETIMEFROMMYDB}) ;
If DayOfWeek ({DATETIMEFROMMYDB}) In [crFriday,crSaturday,crSunday,crMonday]
then
IF DayOfWeek ({DATETIMEFROMMYDB}) In [crFriday]
AND TIME({DATETIMEFROMMYDB}) >= time(21,00,00)
then //your code here
Else if Not(DayOfWeek ({DATETIMEFROMMYDB}) In [crFriday] )
AND (TIME({DATETIMEFROMMYDB}) >= time(00,00,00) AND TIME({DATETIMEFROMMYDB}) <= time(23,59,59))
then //your code here
Else if DayOfWeek ({DATETIMEFROMMYDB})In [crMonday]
AND TIME({DATETIMEFROMMYDB}) < time(09,00,00)
then //your code here
My customer has an event each second Monday of each month.
I need to mark them with red in calendar.
How do i "cleanly" find out the date of that Mondays?
Here's my version.
If the eighth of the month is a Monday, then it is the second Monday. If it is not a Monday, then how many days until the next Monday?
oct_2012 = Date.new 2012, 10, 8
oct_2012.wday # => 1, We're done!
nov_2012 = Date.new 2012, 11, 8
nov_2012.wday # => 4
nov_2012 + (8 - nov_2012.wday) # => 2012-11-12
Does that help?
Edit
Easier version: Just add and be done. This algorithm works even if the month starts on a Monday.
oct_2012 = Date.new 2012, 10, 1
oct_2012 + (8 - oct_2012.wday) # => 2012-10-08
nov_2012 = Date.new 2012, 11, 1
nov_2012 + (8 - nov_2012.wday) # => 2012-11-12
One rule and done!
You second Monday will always fall within the 8th and 14th of each month.
I would like to know how to get the current week number from Rails and how do I manipulate it:
Translate the week number into date.
Make an interval based on week number.
Thanks.
Use strftime:
%U - Week number of the year. The week starts with Sunday. (00..53)
%W - Week number of the year. The week starts with Monday. (00..53)
Time.now.strftime("%U").to_i # 43
# Or...
Date.today.strftime("%U").to_i # 43
If you want to add 43 weeks (or days,years,minutes, etc...) to a date, you can use 43.weeks, provided by ActiveSupport:
irb(main):001:0> 43.weeks
=> 301 days
irb(main):002:0> Date.today + 43.weeks
=> Thu, 22 Aug 2013
irb(main):003:0> Date.today + 10.days
=> Sun, 04 Nov 2012
irb(main):004:0> Date.today + 1.years # or 1.year
=> Fri, 25 Oct 2013
irb(main):005:0> Date.today + 5.months
=> Mon, 25 Mar 2013
You are going to want to stay away from strftime("%U") and "%W".
Instead, use Date.cweek.
The problem is, if you ever want to take a week number and convert it to a date, strftime won't give you a value that you can pass back to Date.commercial.
Date.commercial expects a range of values that are 1 based.
Date.strftime("%U|%W") returns a value that is 0 based. You would think you could just +1 it and it would be fine. The problem will hit you at the end of a year when there are 53 weeks. (Like what just happened...)
For example, let's look at the end of Dec 2015 and the results from your two options for getting a week number:
Date.parse("2015-12-31").strftime("%W") = 52
Date.parse("2015-12-31").cweek = 53
Now, let's look at converting that week number to a date...
Date.commercial(2015, 52, 1) = Mon, 21 Dec 2015
Date.commercial(2015, 53, 1) = Mon, 28 Dec 2015
If you blindly just +1 the value you pass to Date.commercial, you'll end up with an invalid date in other situations:
For example, December 2014:
Date.commercial(2014, 53, 1) = ArgumentError: invalid date
If you ever have to convert that week number back to a date, the only surefire way is to use Date.cweek.
date.commercial([cwyear=-4712[, cweek=1[, cwday=1[, start=Date::ITALY]]]]) → date
Creates a date object denoting the given week date.
The week and the day of week should be a negative
or a positive number (as a relative week/day from the end of year/week when negative).
They should not be zero.
For the interval
require 'date'
def week_dates( week_num )
year = Time.now.year
week_start = Date.commercial( year, week_num, 1 )
week_end = Date.commercial( year, week_num, 7 )
week_start.strftime( "%m/%d/%y" ) + ' - ' + week_end.strftime("%m/%d/%y" )
end
puts week_dates(22)
EG: Input (Week Number): 22
Output: 06/12/08 - 06/19/08
credit: Siep Korteling http://www.ruby-forum.com/topic/125140
Date#cweek seems to get the ISO-8601 week number (a Monday-based week) like %V in strftime (mentioned by #Robban in a comment).
For example, the Monday and the Sunday of the week I'm writing this:
[ Date.new(2015, 7, 13), Date.new(2015, 7, 19) ].map { |date|
date.strftime("U: %U - W: %W - V: %V - cweek: #{date.cweek}")
}
# => ["U: 28 - W: 28 - V: 29 - cweek: 29", "U: 29 - W: 28 - V: 29 - cweek: 29"]
I can only find algorithm for getting ISO 8601 week (week starts on a Monday).
However, the iCal spec says
A week is defined as a seven day period, starting on the day of the
week defined to be the week start (see WKST). Week number one of the
calendar year is the first week that contains at least four (4) days
in that calendar year.
Therefore, it is more complex than ISO 8601 since the start of week can be any day of the week.
Is there an algorithm to determine what is the week number of a date, with a custom start day of week?
or... is there a function in iCal4j that does this? Determine a weekno from a date?
Thanks!
p.s. Limitation: I'm using a JVM language that cannot extend a Java class, but I can invoke Java methods or instantiate Java classes.
if (input_date < firstDateOfTheYear(WKST, year))
{
return ((isLeapYear(year-1))?53:52);
}
else
{
return ((dayOfYear(input_date) - firstDateOfTheYear(WKST, year).day)/7 + 1);
}
firstDateOfTheYear returns the first calendar date given a start of week(WKST) and the year, e.g. if WKST = Thursday, year = 2012, then it returns Jan 5th.
dayOfYear returns sequencial numerical day of the year, e.g. Feb 1st = 32
Example #1: Jan 18th, 2012, start of week is Monday
dayOfYear(Jan 18th, 2012) = 18
firstDateOfTheYear(Monday, 2012) = Jan 2nd, 2012
(18 - 2)/7 + 1 = 3
Answer Week no. 3
Example #2: Jan 18th, 2012, start of week is Thursday
dayOfYear(Jan 18th, 2012) = 18
firstDateOfTheYear(Thursday, 2012) = Jan 5th, 2012
(18 - 5)/7 + 1 = 2
Answer Week no. 2
Example #3: Jan 1st, 2012, start of week is Monday
firstDateOfTheYear(Monday, 2012) = Jan 2nd, 2012
IsLeapYear(2012-1) = false
Jan 1st, 2012 < Jan 2nd, 2012
Answer Week no. 52
Let daysInFirstWeek be the number of days on the first week of the year that are in January. Week starts on a WKST day. (e.g. if Jan 1st is a WKST day, return 7)
Set dayOfYear to the n-th days of the input date's year (e.g. Feb 1st = 32)
If dayOfYear is less than or equal to daysInFirstWeek
3.1. if daysInFirstWeek is greater than or equal to 4, weekNo is 1, skip to step 5.
3.2. Let daysInFirstWeekOfLastYear be the number of days on the first week of the previous year that are in January. Week starts on a WKST day.
3.3. if daysInFirstWeekOfLastYear is 4 or last year is Leap year and daysInFirstWeekOfLastYear is 5, weekNo is 53, otherwise weekNo is 52, skip to step 5.
Set weekNo to ceiling((dayOfYear - daysInFirstWeek) / 7)
4.1. if daysInFirstWeek greater than or equal to 4, increment weekNo by 1
4.2. if daysInFirstWeek equal 53 and count of days on the first week (starting from WKST) of January in the year of inputDate's year + 1 is greater than or equal to 4, set weekNo to 1
return weekNo
Would anyone know why the following code works correctly on Windows and not on Mac??
Today (24/11/2010) should return 47 not 48 as per MacOS
def fm_date = '24/11/2010'
import java.text.SimpleDateFormat
def lPad = {it ->
st = '00' + it.toString()
return st.substring(st.length()-2, st.length())
}
dfm = new SimpleDateFormat("dd/MM/yyyy")
cal=Calendar.getInstance()
cal.setTime( dfm.parse(fm_date) )
now = cal.get(Calendar.WEEK_OF_YEAR)
cal.add(Calendar.DAY_OF_MONTH,-7)
prev = cal.get(Calendar.WEEK_OF_YEAR)
cal.add(Calendar.DAY_OF_MONTH,14)
next = cal.get(Calendar.WEEK_OF_YEAR)
prev = 'diary' + lPad(prev) + '.shtml'
next = 'diary' + lPad(next) + '.shtml'
return 'diary' + lPad(now) + '.shtml'
I believe it's an ISO week number issue...
If I use this code adapted (and groovyfied) from yours:
import java.text.SimpleDateFormat
def fm_date = '24/11/2010'
Calendar.getInstance().with { cal ->
// We want ISO Week numbers
cal.firstDayOfWeek = MONDAY
cal.minimalDaysInFirstWeek = 4
setTime new SimpleDateFormat( 'dd/MM/yyyy' ).parse( fm_date )
now = cal[ WEEK_OF_YEAR ]
}
"diary${"$now".padLeft( 2, '0' )}.shtml"
I get diary47.shtml returned
As the documentation for GregorianCalendar explains, if you want ISO Month numbers:
Values calculated for the WEEK_OF_YEAR
field range from 1 to 53. Week 1 for a
year is the earliest seven day period
starting on getFirstDayOfWeek() that
contains at least
getMinimalDaysInFirstWeek() days from
that year. It thus depends on the
values of getMinimalDaysInFirstWeek(),
getFirstDayOfWeek(), and the day of
the week of January 1. Weeks between
week 1 of one year and week 1 of the
following year are numbered
sequentially from 2 to 52 or 53 (as
needed).
For example, January 1, 1998 was a
Thursday. If getFirstDayOfWeek() is
MONDAY and getMinimalDaysInFirstWeek()
is 4 (these are the values reflecting
ISO 8601 and many national standards),
then week 1 of 1998 starts on December
29, 1997, and ends on January 4, 1998.
If, however, getFirstDayOfWeek() is
SUNDAY, then week 1 of 1998 starts on
January 4, 1998, and ends on January
10, 1998; the first three days of 1998
then are part of week 53 of 1997.
Edit
Even Groovier (from John's comment)
def fm_date = '24/11/2010'
Calendar.getInstance().with { cal ->
// We want ISO Week numbers
cal.firstDayOfWeek = MONDAY
cal.minimalDaysInFirstWeek = 4
cal.time = Date.parse( 'dd/MM/yyyy', fm_date )
now = cal[ WEEK_OF_YEAR ]
}
"diary${"$now".padLeft( 2, '0' )}.shtml"
Edit2
Just ran this on Windows using VirtualBox, and got the same result