Input: Month name (January / February / ... / December)
Output: Season (Winter / Spring / Summer / Autumn)
Winter: Dec - Feb
Spring: Mar - May
Summer: Jun - Aug
Autumn: Sept - Nov
I have tried:
Program Months;
var
Month:String;
begin
writeln('Insert month name:');
readln(Month);
if Month = 'March' or Month = 'April' or Month = 'May' then
begin
writeln(Month,' is Spring month');
end
...
etc
...
end.
But this program is not working.
Operator precedence - it's important. You need to write:
if (Month = 'March') or (Month = 'April') or (Month = 'May') then
This is because in Pascal, or has a higher priority than = so what is actually being evaluated is:
if ((Month = ('March' or Month)) = ('April' or Month)) = 'May' then
Which is obviously meaningless and will not compile (I might've made a mistake on the line above but it's the general idea). Please refer to this link to learn more about operator precedence in Pascal.
Related
I need to output to the console the calendar of the current month in Ruby. The result should be similar to ncal on UNIX-like systems. I found a solution for C ++ but can't adapt for Ruby. So far, I only realized that I need to use nested loops to output the height and width. Tell me in which direction to move?
require 'date'
days = %w[Mun Tue Wed Thu Fri Sat Sun]
puts " #{Date::MONTHNAMES[Date.today.month]} #{Date.today.year}"
i = 0
start_month = (Date.today - Date.today.mday + 1).strftime("%a")
while i < days.size
print days[i]
j = 1
while j <= 31
if days[i] == start_month
print " #{j}"
end
j += 7
end
i += 1
puts
end
I'll take your solution so far, and try to give some specific pointers for how to progress with it - but of course, there are many different ways to approach this problem in general, so this is by no means the only approach!
The first critical issue (as you're aware!) is that you're only printing things for the row starting on the 1st of the month, due to this line:
if days[i] == start_month
Sticking with the current overall design, we know we'll need to print something for every line, so clearly a conditional like this isn't going to work. Let's try removing it.
Firstly, it will be more convenient to know which day of the week the month started on as a number, not a string, so we can easily calculate offsets against another day. Let's do that with:
# e.g. for 1st July 2021 this was a Thursday, so we get `4`.
start_of_month_weekday = (Date.today - Date.today.mday + 1).cwday
Next (and this is the crucial step!), we can use the above information to find out "which day of the month is it, on this day of the week?"
Here a first version of that calculation, incorporated into your solution so far:
require 'date'
days = %w[Mon Tue Wed Thu Fri Sat Sun]
puts " #{Date::MONTHNAMES[Date.today.month]} #{Date.today.year}"
i = 0
# e.g. for 1st July 2021 this was a Thursday, so we get `4`.
start_of_month_weekday = (Date.today - Date.today.mday + 1).cwday
while i < days.size
print days[i]
day_of_month = i - start_of_month_weekday + 2 # !!!
while day_of_month <= 31
print " #{day_of_month}"
day_of_month += 7
end
i += 1
puts
end
This outputs:
July 2021
Mon -2 5 12 19 26
Tue -1 6 13 20 27
Wed 0 7 14 21 28
Thu 1 8 15 22 29
Fri 2 9 16 23 30
Sat 3 10 17 24 31
Sun 4 11 18 25
Not bad! Now we're getting somewhere!
I'll leave you to figure out the rest 😉 .... But here are some clues, for what I'd tackle next:
This code, print " #{day_of_month}", needs to print a "blank space" if the day number is less than 1. This could be done with a simple if statement.
Similarly, since you want this calendar to line up neatly in a grid, you need this code to always print a something two characters wide. sprintf is your friend here! Check out the "Examples of width", about halfway down the page.
You've hardcoded 31 for the number of days in the month. This should be fixed, of course. (Use the Date library!)
It's funny how you used strftime("%a") in one place, yet constructed the calendar title awkwardly in the line above! 😄 Take a look at the documentation for formatting dates; it's extremely flexible. I think you can use: Date.today.strftime("%B %Y").
If you'd like to add some colour (or background colour?) to the current day of the month, consider doing something like this, or use a library to assist.
Using while loops works OK, but is quite un-rubyish. In 99% of cases, ruby has even better tools for the job; it's a very expressive language - iterators are king! (I'm guessing you first learned another language, before ruby? Seeing while loops, and/or for loops, is a dead giveaway that you're more familiar with a different language.) Instead of the outer while loop (while i < days.size), you could use days.each_with_index. And instead of the inner while loop (while j < 31), you could use day_of_month.step(31, 7) (how cool is that!!).
This is one way:
Construct a one-dimensional array, beginning with the daynames (Mon Tue ...).
Figure out a way to determine with how many "blanks" the month starts (these are days from the previous month. wday might help). Attach that amount of empty strings to the array.
Determine how many days the month has (hint Date.new(2021,7,-1), and attach all these daynumbers to the array.
Attach empty strings to the array until the size of the array is divisible by 7 (or better, calculate). Skip this if you're skipping the last bullet.
Convert all elements of this array to right-adjusted strings of size 3 or some-such.
Use each_slice(7) to slice the array into weeks.
If desired, transpose this array of week-slices to mimic the ncal output.
Thank you for your help, literally 10 hours and I figured it out thanks to you. I apologize once again for the initially incorrectly posed question.
With the help of hints, I assembled such a solution.
require 'date'
days = %w[Mon Tue Wed Thu Fri Sat Sun]
p days
blanks = Date.new(2021,7,1).wday - 1
blanks.times do
days.push(' ')
end
days_in_month = Date.new(2021, 7, -1).day
days_in_month
day = 1
while day <= days_in_month
days.push(day)
day += 1
end
unless (days.size % 7) == 0
days.push(' ')
end
days.join(', ')
new_arr = days.each_slice(7).to_a
puts"Массив дней: #{new_arr}"
for i in 0...7
for j in 0...new_arr.size
print " #{new_arr[j][i]}"
end
puts
end
require 'date'
# init
DAYS_ORDER = ['Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun']
today = Date.today
month = today.month
year = today.year
first_day = Date.new(year, month, 1)
last_day = Date.new(year, month, -1)
hash_days = {}
# get all current months days and add to hash_days
first_day.upto(last_day) { |day| hash_days[day.day] = day.strftime('%a') }
# group by wday
grouped_hash = hash_days.group_by { |day| day.pop }.transform_values { |days| days.flatten }
# sort by wday from DAYS_ORDER
sorted_arr = grouped_hash.sort_by { |k, v| DAYS_ORDER.index(k) }
# rendering current month's calendar with mark current day
## title
print "\x1b[4m#{today.strftime("%B %Y")}\x1b[0m\n"
## calendar
indent = true
sorted_arr.each do |wday, days|
print wday
if days[0] != 1 && indent == true
print " "
else
indent = false
end
days.each do |value|
spaces = " " * (value > 9 ? 1 : 2)
str_day = spaces + value.to_s
current_day = "\x1b[1;31m#{str_day}\x1b[0m"
print value == today.day ? current_day : str_day
end
puts
end
view
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
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
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"]
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