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
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
Hello fellow Stack Overflowers,
I have a situation, where I need some help choosing the best way to make an algorithm work, the objective is to manage the occupation of a resource (Lets consider the resource A) to have multiple tasks, and where each task takes a specified amount of time to complete. At this first stage I don't want to involve multiple variables, so lets keep it the simple way, lets consider he only has a schedule of the working days.
For example:
1 - We have 1 resource, resource A
2 - Resource A works from 8 am to 4 pm, monday to friday, to keep it simple by now, he doesn't have lunch for now, so, 8 hours of work a day.
3 - Resource A has 5 tasks to complete, to avoid complexity at this level, lets supose each one will take exactly 10 hours to complete.
4 - Resource A will start working on this tasks at 2018-05-16 exactly at 2 pm.
Problem:
Now, all I need to know is the correct finish date for all the 5 tasks, but considering all the previous limitations.
In this case, he has 6 working days and additionaly 2 hours of the 7th day.
The expected result that I want would be: 2018-05-24 (at 4 pm).
Implementation:
I thought about 2 options, and would like to have feedback on this options, or other options that I might not be considering.
Algorithm 1
1 - Create a list of "slots", where each "slot" would represent 1 hour, for x days.
2 - Cross this list of slots with the hour schedule of the resource, to remove all the slots where the resource isn't here. This would return a list with the slots that he can actually work.
3 - Occupy the remaining slots with the tasks that I have for him.
4 - Finnaly, check the date/hour of the last occupied slot.
Disadvantage: I think this might be an overkill solution, considering that I don't want to consider his occupation for the future, all I want is to know when will the tasks be completed.
Algorithm 2
1 - Add the task hours (50 hours) to the starting date, getting the expectedFinishDate. (Would get expectedFinishDate = 2018-05-18 (at 4 pm))
2 - Cross the hours, between starting date and expectedFinishDate with the schedule, to get the quantity of hours that he won't work. (would basically get the unavailable hours, 16 hours a day, would result in remainingHoursForCalc = 32 hours).
3 - calculate new expectedFinishDate with the unavailable hours, would add this 32 hours to the previous 2018-05-18 (at 4 pm).
4 - Repeat point 2 and 3 with new expectedFinishDate untill remainingHoursForCalc = 0.
Disadvantage: This would result in a recursive method or in a very weird while loop, again, I think this might be overkill for calculation of a simple date.
What would you suggest? Is there any other option that I might not be considering that would make this simpler? Or you think there is a way to improve any of this 2 algorithms to make it work?
Improved version:
import java.util.Calendar;
import java.util.Date;
public class Main {
public static void main(String args[]) throws Exception
{
Date d=new Date();
System.out.println(d);
d.setMinutes(0);
d.setSeconds(0);
d.setHours(13);
Calendar c=Calendar.getInstance();
c.setTime(d);
c.set(Calendar.YEAR, 2018);
c.set(Calendar.MONTH, Calendar.MAY);
c.set(Calendar.DAY_OF_MONTH, 17);
//c.add(Calendar.HOUR, -24-5);
d=c.getTime();
//int workHours=11;
int hoursArray[] = {1,2,3,4,5, 10,11,12, 19,20, 40};
for(int workHours : hoursArray)
{
try
{
Date end=getEndOfTask(d, workHours);
System.out.println("a task starting at "+d+" and lasting "+workHours
+ " hours will end at " +end);
}
catch(Exception e)
{
System.out.println(e.getMessage());
}
}
}
public static Date getEndOfTask(Date startOfTask, int workingHours) throws Exception
{
int totalHours=0;//including non-working hours
//startOfTask +totalHours =endOfTask
int startHour=startOfTask.getHours();
if(startHour<8 || startHour>16)
throw new Exception("a task cannot start outside the working hours interval");
System.out.println("startHour="+startHour);
int startDayOfWeek=startOfTask.getDay();//start date's day of week; Wednesday=3
System.out.println("startDayOfWeek="+startDayOfWeek);
if(startDayOfWeek==6 || startDayOfWeek==0)
throw new Exception("a task cannot start on Saturdays on Sundays");
int remainingHoursUntilDayEnd=16-startHour;
System.out.println("remainingHoursUntilDayEnd="+remainingHoursUntilDayEnd);
/*some discussion here: if task starts at 12:30, we have 3h30min
* until the end of the program; however, getHours() will return 12, which
* substracted from 16 will give 4h. It will work fine if task starts at 12:00,
* or, generally, at the begining of the hour; let's assume a task will start at HH:00*/
int remainingDaysUntilWeekEnd=5-startDayOfWeek;
System.out.println("remainingDaysUntilWeekEnd="+remainingDaysUntilWeekEnd);
int completeWorkDays = (workingHours-remainingHoursUntilDayEnd)/8;
System.out.println("completeWorkDays="+completeWorkDays);
//excluding both the start day, and the end day, if they are not fully occupied by the task
int workingHoursLastDay=(workingHours-remainingHoursUntilDayEnd)%8;
System.out.println("workingHoursLastDay="+workingHoursLastDay);
/* workingHours=remainingHoursUntilDayEnd+(8*completeWorkDays)+workingHoursLastDay */
int numberOfWeekends=(int)Math.ceil( (completeWorkDays-remainingDaysUntilWeekEnd)/5.0 );
if((completeWorkDays-remainingDaysUntilWeekEnd)%5==0)
{
if(workingHoursLastDay>0)
{
numberOfWeekends++;
}
}
System.out.println("numberOfWeekends="+numberOfWeekends);
totalHours+=(int)Math.min(remainingHoursUntilDayEnd, workingHours);//covers the case
//when task lasts 1 or 2 hours, and we have maybe 4h until end of day; that's why i use Math.min
if(completeWorkDays>0 || workingHoursLastDay>0)
{
totalHours+=8;//the hours of the current day between 16:00 and 24:00
//it might be the case that completeWorkDays is 0, yet the task spans up to tommorrow
//so we still have to add these 8h
}
if(completeWorkDays>0)//redundant if, because 24*0=0
{
totalHours+=24*completeWorkDays;//for every 8 working h, we have a total of 24 h that have
//to be added to the date
}
if(workingHoursLastDay>0)
{
totalHours+=8;//the hours between 00.00 AM and 8 AM
totalHours+=workingHoursLastDay;
}
if(numberOfWeekends>0)
{
totalHours+=48*numberOfWeekends;//every weekend between start and end dates means two days
}
System.out.println("totalHours="+totalHours);
Calendar calendar=Calendar.getInstance();
calendar.setTime(startOfTask);
calendar.add(Calendar.HOUR, totalHours);
return calendar.getTime();
}
}
You may adjust the hoursArray[], or d.setHours along with c.set(Calendar.DAY_OF_MONTH, to test various start dates along with various task lengths.
There is still a bug , due to the addition of the 8 hours between 16:00 and 24:00:
a task starting at Thu May 17 13:00:00 EEST 2018 and lasting 11 hours will end at Sat May 19 00:00:00 EEST 2018.
I've kept a lot of print statements, they are useful for debugging purposes.
Here is the terminology explained:
I agree that algorithm 1 is overkill.
I think I would make sure I had the conditions right: hours per day (8), working days (Mon, Tue, Wed, Thu, Fri). Would then divide the hours required (5 * 10 = 50) by the hours per day so I know a minimum of how many working days are needed (50 / 8 = 6). Slightly more advanced, divide by hours per week first (50 / 40 = 1 week). Count working days from the start date to get a first shot at the end date. There was probably a remainder from the division, so use this to determine whether the tasks can end on this day or run into the next working day.
Trying to convert strings like 9 weeks ago, 1 year, 6 months ago, 20 hours ago into a ruby time object like Tue, 10 Mar 2015 12:06:15 PDT -07:00.
I've been doing this:
eval("10 days ago".gsub(' ', '.'))
This works fine, but for strings like 1 year, 6 months ago blows up.
I just need to do comparisons like:
eval("10 days ago".gsub(' ', '.')) < (Time.now - 7.days)
I'm using sinatra so no fancy rails helpers.
Please never use eval in production code..
Converting from timeago notation would be quite complex and resource intensive.
However, this way seems the least error prone: It will convert a string like "5 seconds ago" to "5S" and use mapping to find what it means in time, after which it will subtract that time from the current time.
The parse string is dynamically built so it can accomodate most every timeago notation.
require('date')
mapping = {"D"=> "%d","W"=>"%U","H"=>"%T","Y"=>"%Y","M"=>"%m","S"=>"%S"}
timerel = "1 year, 6 months ago".split(",").map { |n| n.gsub(/\s+/, "").upcase()[0,2].split('')}
Date.strptime(
timerel.map {|n| n[0]}.join(" "),
timerel.map {|n| mapping[n[1]]}.join(" ")
)
date = Date.new(0) + (Date.today - Date.strptime(timerel.map {|n| n[0]}.join(" "), timerel.map {|n| mapping[n[1]]}.join(" ")))
=> #<Date: 2014-10-10 ((2456941j,0s,0n),+0s,2299161j)>
It goes without saying that is very error prone. Use at your own risk:
def parse(date:)
eval(date.gsub(/ ?(,|and) ?/, '+').tr(' ', '.').gsub(/^(.*)(\.ago)$/, '(\1)\2'))
end
parse(date: '1 year, 6 months ago') # => Wed, 10 Sep 2014 21:29:11 BST +01:00
parse(date: '1 year, 6 months, 3 weeks, 6 days, 9 hours and 12 seconds ago')
# => Thu, 14 Aug 2014 12:33:07 BST +01:00
The idea is to convert the original string to:
'(1.year+6.months).ago'
I made new object Date.new with args (year, month). After create ruby added 01 number of day to this object by default. Is there any way to add not first day, but last day of month that i passed as arg(e.g. 28 if it will be 02month or 31 if it will be 01month) ?
use Date.civil
With Date.civil(y, m, d) or its alias .new(y, m, d), you can create a new Date object. The values for day (d) and month (m) can be negative in which case they count backwards from the end of the year and the end of the month respectively.
=> Date.civil(2010, 02, -1)
=> Sun, 28 Feb 2010
>> Date.civil(2010, -1, -5)
=> Mon, 27 Dec 2010
To get the end of the month you can also use ActiveSupport's helper end_of_month.
# Require extensions explicitly if you are not in a Rails environment
require 'active_support/core_ext'
p Time.now.utc.end_of_month # => 2013-01-31 23:59:59 UTC
p Date.today.end_of_month # => Thu, 31 Jan 2013
You can find out more on end_of_month in the Rails API Docs.
So I was searching in Google for the same thing here...
I wasn't happy with above so my solution after reading documentation
in RUBY-DOC was:
Example to get 10/31/2014
Date.new(2014,10,1).next_month.prev_day
require "date"
def find_last_day_of_month(_date)
if(_date.instance_of? String)
#end_of_the_month = Date.parse(_date.next_month.strftime("%Y-%m-01")) - 1
else if(_date.instance_of? Date)
#end_of_the_month = _date.next_month.strftime("%Y-%m-01") - 1
end
return #end_of_the_month
end
find_last_day_of_month("2018-01-01")
This is another way to find
You can do something like that:
def last_day_of_month?
(Time.zone.now.month + 1.day) > Time.zone.now.month
end
Time.zone.now.day if last_day-of_month?
This is my Time based solution. I have a personal preference to it compared to Date although the Date solutions proposed above read somehow better.
reference_time ||= Time.now
return (Time.new(reference_time.year, (reference_time.month % 12) + 1) - 1).day
btw for December you can see that year is not flipped. But this is irrelevant for the question because december always has 31 day. And for February year does not need flipping. So if you have another use case that needs year to be correct, then make sure to also change year.
Here is taking the first and third answers to find the last day of the previous month.
today_c = Date.civil(Date.today.prev_month.year, -1, -1)
p today_c
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