I want to display the short name for each month, 12 months back from the previous month, but with the below I get an error on mon1 and mon2 and I guess since that is last year?
mon1=MonthName(Month(Now())-11,1)
mon2=MonthName(Month(Now())-10,1)
mon3=MonthName(Month(Now())-9,1)
mon4=MonthName(Month(Now())-8,1)
mon5=MonthName(Month(Now())-7,1)
mon6=MonthName(Month(Now())-6,1)
mon7=MonthName(Month(Now())-5,1)
mon8=MonthName(Month(Now())-4,1)
mon9=MonthName(Month(Now())-3,1)
mon10=MonthName(Month(Now())-2,1)
mon11=MonthName(Month(Now())-1,1)
mon12=MonthName(Month(Now()),1)
So how can I display now,dec,jan,feb,mar,apr,may,jun,jul,aug,sep,oct
Thanks.
The problem here is the Month() function returns an integer between 1 and 12 to represent each month. Instead you want to minus the number of months from the Now() value before wrapping it with Month().
Below is an example that does this using a For loop and a single dimension Array.
Dim dt: dt = Now()
Dim i, mon(12)
Const numOfMonths = 12
For i = 1 To numOfMonths
mon(i) = MonthName(Month(DateAdd("m", i - numOfMonths, dt)), True)
Next
Call Response.Write(Join(mon, vbCrLf))
Output:
Sep
Aug
Jul
Jun
May
Apr
Mar
Feb
Jan
Dec
Nov
Oct
Related
I'm trying to compare one date value (ie. base value) with all other date values on a page and based on the difference between these days, I want to execute other commands.
Well, in the above UI, the base value is 11 Jul 2021 (Departure date in the first list) and other date values are 12 Jul 2021, 20 Jul 2021, 27 Jul 2021, 3 Aug 2021 and so on (Arrival dates from 2nd list onwards).
Now, I had to delete the all list(s) where the date difference between the base value and particular list is less than 15 days.
In this case, 12 Jul 2021, 20 Jul 2021 had to be deleted and all lists from 27 Jul 2021, 3 Aug 2021 and so on should be untouched as in the below picture.
So far, I have captured the value of the base value and came up with logic to compare it with another date value but I am not sure how I can save the 2nd and further date value(s) to a variable in order to compare with the base value.
{
cy.get("[data-test='departureTime']")
.eq(0)
.then((date) => {
const depDate_FirstPort = new Date(date.text());
cy.log(depDate_FirstPort.toISOString()); //2021-07-11T19:00:00.000Z
// const arrDate_SecondPort = new Date(cy.get('[data-test="arrivalTime"]').eq(1).invoke('text'));
// Since the above approach does not work, hard coding now.
const arrDate_SecondPort = new Date("22 Jul 2021 12:01")
cy.log(arrDate_SecondPort.toISOString()); //2021-07-22T10:01:00.000Z
cy.getDifferenceBetweenDates(depDate_FirstPort,arrDate_SecondPort).then((dif)=>{
if(dif < 16) {
cy.log("delete the port entry");
//do something
}
});
});
}
Cypress Command:
Cypress.Commands.add("getDifferenceBetweenDates", (Date1, Date2) => {
var diff_times = Math.abs(Date1.getTime() - Date2.getTime());
var diff_days = Math.ceil(diff_times / (1000 * 3600 * 24));
cy.log(diff_days) //11
})
Also, curious to know a possible approach to iterate all list falls under the 'to be deleted list' (12 Jul 2021, 20 Jul 2021) based on the condition mentioned above.
The iterative approach you have is ok, but you need to repeat the code you have for the first date to get the subsequent dates.
So, this bit but changing the index
cy.get("[data-test='departureTime']")
.eq(0) // 1,2,3 etc
.then((date) => {
A different approach is to filter the whole set,
const dayjs = require('dayjs') // replaces Cypress.moment
// first install with
// yarn add -D dayjs
it('finds the skipped ports', () => {
// helper func with format specific to this website
const toDate = (el) => dayjs(el.innerText, 'D MMM YYYY HH:mm')
cy.get("[data-test='departureTime']")
.then($departures => {
const departures = [...$departures] // convert jQuery object to an array
const first = toDate(departures[0]);
const cutoff = first.add(15, 'day')
const nextPorts = departures.slice(1) // all but the first
const skipPorts = nextPorts.filter(port => toDate(port).isBefore(cutoff))
expect(skipPorts.length).to.eq(2)
expect(skipPorts[0].innerText).to.eq('12 Jul 2021 14:02')
expect(skipPorts[1].innerText).to.eq('21 Jul 2021 04:00')
})
})
I'm not clear about your goal, but if you are going to actually delete the skipPorts from the page instead of just testing them, you should be wary of the DOM list changing as you do so.
Deleting from the list you have recently queried with cy.get("[data-test='departureTime']") would cause the internal subject to become invalid, and you might get "detached from DOM" errors or delete the wrong item.
my data looks like
JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC
22.60 24.60 30.60 34.60 36.20 35.70 32.10 30.20 31.40 31.60 28.00 24.80
25.40 27.60 32.40 34.60 36.50 38.10 31.70 31.40 30.30 30.20 27.00 23.90
and there are like hundreds of rows! I want to find a maximum value in each row and write it in different column next to data along with month
so my out put will be
36.20 MAY
38.10 JUN
.
.
I want to use maxloc function, but i have no idea how to use it!
Try
index = maxloc(myTable(3,:))
print *, myTable((/1,3/), index)
It should select the highest value from the third row and display the first and third value at this index.
I create a hash with months as keys and timelaps as values
biens_delai[bien_date.mon] = b.delai
I get this result without month parsing
{Wed, 18 Jan 2017=>3.0, Sat, 25 Feb 2017=>2.0, Fri, 17 Mar 2017=>3.0, Sat, 25 Mar 2017=>5.0, Tue, 18 Apr 2017=>2.0, Thu, 29 Jun 2017=>2.0}
In March i have 2 values but when i parse by month i get the most high value and i want a addition of 2 values for March not the most high
{1=>3.0, 2=>2.0, 3=>5.0, 4=>2.0, 6=>2.0}
That's not the high value which you are getting, the values are getting overwritten, try the following
biens_delai[bien_date.mon] = biens_delai[bien_date.mon].to_f + b.delai
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 hope this is obvious to someone. I have only had a vanilla use of GLPK/MathProg.
I am having trouble figuring out the syntax in GNU MathProg (within GLPK) to do the following, for example:
set PartsOfWeek;
set WeekDays;
data;
set PartsOfWeek := WorkWeek WeekEnd;
set WorkWeek := Mon Tue Wed Thu Fri;
set WeekEnd := Sat Sun;
set WeekDays := setof{d in (WorkWeek union WeekEnd)}(d);
The problem is that this is rejected by MathProg.
In general, I just want to be able to:
- declare a Partition (here PartsOfWeek) and a set (here Weekdays)
- build the partition from data
- populate the set with the elements of the of the sets from the partition.
A better example might be with seasons and months.
with #ALi's literature reference help:
set seasons;
set months;
set monthsOfseason {seasons} within months;
data;
set seasons := winter spring summer fall;
set months := jan feb mar apr may jun jul aug sep oct nov dec;
set monthsOfseason[winter] := dec jan feb;
set monthsOfseason[spring] := mar apr may;
set monthsOfseason[summer] := jun jul aug;
set monthsOfseason[fall] := sep oct nov;