I made something to test the time. This is the Code:
time1 = Time.new
puts time1.strftime("%H:%M:%S")
# |
# V Change here
t = Time.new(2017, 9, 8, 14, 30, 0)
dist = ((t - time1) /60 ).round
dist1 = dist/60
dist2 = dist
while dist2 > 60
dist2 = dist2 -60
end
puts "just #{dist1} hours and #{dist2} min left."
but I don't know how I create a fixed time for every day. I need to change it every day. Got some tips for me?
t = DateTime.parse("14:30").to_time
The above would construct a time instance for today, 2:30PM.
Related
What I have tried so far ...
start_hour = 7
start_minute = 0 * 0.01
end_hour = 17
end_minute = 45 * 0.01
step_time = 25
start_time = start_hour + start_minute
end_time = end_hour + end_minute
if step_time > 59
step_time = 1 if step_time == 60
step_time = 1.3 if step_time == 90
step_time = 2 if step_time == 120
else
step_time *= 0.01
end
hours = []
(start_time..end_time).step(step_time).map do |x|
next if (x-x.to_i) > 0.55
hours << '%0.2f' % x.round(2).to_s
end
puts hours
If I enter the step interval 0, 5, 10, 20, I can get the time interval I want. But if I enter 15, 25, 90, I can't get the right range.
You currently have:
end_hour = 17
end_minute = 45 * 0.01
end_time = end_hour + end_minute
#=> 17.45
Although 17.45 looks like the correct value, it isn't. 45 minutes is 3 quarters (or 75%) of an hour, so the correct decimal value is 17.75.
You could change your code accordingly, but working with decimal hours is a bit strange. It's much easier to just work with minutes. Instead of turning the minutes into hours, you turn the hours into minutes:
start_hour = 7
start_minute = 0
start_time = start_hour * 60 + start_minute
#=> 420
end_hour = 17
end_minute = 45
end_time = end_hour * 60 + end_minute
#=> 1065
The total amount of minutes can easily be converted back to hour-minute pairs via divmod:
420.divmod(60) #=> [7, 0]
1065.divmod(60) #=> [17, 45]
Using the above, we can traverse the range without having to convert the step interval:
def hours(start_time, end_time, step_time)
(start_time..end_time).step(step_time).map do |x|
'%02d:%02d' % x.divmod(60)
end
end
hours(start_time, end_time, 25)
#=> ["07:00", "07:25", "07:50", "08:15", "08:40", "09:05", "09:30", "09:55",
# "10:20", "10:45", "11:10", "11:35", "12:00", "12:25", "12:50", "13:15",
# "13:40", "14:05", "14:30", "14:55", "15:20", "15:45", "16:10", "16:35",
# "17:00", "17:25"]
hours(start_time, end_time, 90)
#=> ["07:00", "08:30", "10:00", "11:30", "13:00", "14:30", "16:00", "17:30"]
I'm looking for a concise way to get a Ruby Time object representing the top of the next minute (and hour/day/month/year, if possible). I want this to work in a pure Ruby environment, so the Rails function Time.change or similar doesn't fit the bill.
At first this seems simple - just add 1 to Time.now, but there are edge cases where if, for example, you try to instantiate a Time object with Time.now.min + 1 when the current minute is 59, you get an ArgumentError: min out of range. This goes for hour, day, and month as well.
I have some lengthy code that does the job. It's ugly, but I'm just experimenting:
def add_minute
now = Time.local
year = now.year
month = now.month
day = now.day
hour = now.hour
min = now.min
if now.min == 59
if now.hour == 23
if now.day == Date.civil(now.year, now.month, -1).day
if month == 12
year = year + 1
month = 1
day = 1
hour = 0
min = 0
else
month = now.month + 1
day = 1
hour = 0
min = 0
end
else
day = now.day + 1
hour = 0
min = 0
end
else
hour = now.hour + 1
min = 0
end
else
min = now.min + 1
end
Time.local year, month, day, hour, min, 0
end
This seems absurdly verbose for what seems like it should be a simple or built-in task, but I haven't found a native Ruby solution. Does one exist?
You could convert the Time object to UNIX epoch time (seconds since 1970) using #to_i, add 60 s, and then convert back to a Time object.
time_unix = Time.now.to_i
time_unix_one_min_later = time_unix + 60
time_one_min_later = t = Time.at(time_unix_one_min_later)
time_one_min_later_rounded_down = Time.new(t.year, t.month, t.day, t.hour, t.min)
EDIT: Even shorter - you can just add integer seconds to Time.now directly:
time_one_min_later = t = Time.now + 60
time_one_min_later_rounded_down = Time.new(t.year, t.month, t.day, t.hour, t.min)
EDIT 2: One-liner - just subtract Time.now.sec:
time_one_min_later_rounded_down = Time.now + 60 - Time.now.sec
Other option, given one second to midnight:
require 'time'
now = Time.strptime('2018-12-31 23:59:59', '%Y-%m-%d %H:%M:%S')
Within one minute:
Time.at(now + 60) #=> 2019-01-01 00:00:59 +0100
Time.at(now + 60 - now.sec) #=> 2019-01-01 00:00:00 +0100
You get: # HAPPY NEW YEAR!
Ruby has built in methods for adding months (>>) and days (+). A year is 12 months, and an hour is 1/24th of a day.
require 'date'
def add_time(time, year: 0 ,month: 0, day: 0, hour: 0, minute: 0)
time >>= 12*year
time >>= month
time += day
time += Rational(hour,24) # or (hour/24.0) if you dislike rationals
time += Rational(minute, 24*60) # (minute/24.0*60) if you dislike rationals
end
p t = DateTime.now
p add_time(t, year: 1, minute: 30)
Not that clean without ActiveSupport:
new_date = (DateTime.now + 1.to_f / (60*24))
DateTime.new(new_date.year, new_date.month, new_date.day, new_date.hour, new_date.minute)
We can make this calculation easier to understand by getting the current number of seconds we are through the day. (optional)
DateTime.current.to_i gives us the number of seconds since 1970
DateTime.current.to_i - DateTime.current.beginning_of_day.to_i gives us the number of seconds since the start of the day.
(((number_of_seconds_through_the_day + 60)/60) * 60) gives us the number of seconds we will be at when the next minute starts
Then we subtract the two to give us the number of seconds until the top of the next minute.
If we want the exact time at start of the next minute then we can do:
DateTime.current + seconds_until_start_of_the_next_minute.seconds
def seconds_until_start_of_the_next_minute
number_of_seconds_through_the_day = DateTime.current.to_i - DateTime.current.beginning_of_day.to_i
number_of_seconds_through_the_day_at_next_minute = (((number_of_seconds_through_the_day + 60)/60) * 60)
seconds_until_next_minute_starts = number_of_seconds_through_the_day_at_next_minute - number_of_seconds_through_the_day
return seconds_until_next_minute_starts
end
I have a string as follows:
"00:48:22"
From right to left, I am working with a power of 60, because I want to get the total number of seconds from hours, minutes, seconds.
This is what I have tried:
clock
=> "00:48:22"
i = 0
result = clock.split(":").reverse.reduce(0) do |acc, segment|
acc += segment.to_i + (60 ** i)
i += 1
acc
end
=> 3731
The result is off. It should be 2902. Any idea what I am doing wrong?
Your algorithm is a little messy and error prone.
This will give you the right answer:
clock = '00:48:22'
clock.split(':').map(&:to_i).reduce(0) do |acc, segment|
acc * 60 + segment
end
You are adding where you should be multiplying
acc += segment.to_i * (60 ** i)
require 'date'
d = Date.today
(Time.new(d.year,d.month,d.day,*str.split(':').map(&:to_i))-d.to_time).to_i
#=> 2902
#scores_raw.each do |score_raw|
# below is code if time was being sent in milliseconds
hh = ((score_raw.score.to_i)/100)/3600
mm = (hh-hh.to_i)*60
ss = (mm-mm.to_i)*60
crumbs = [hh,mm,ss]
sum = crumbs.first.to_i*3600+crumbs[1].to_i*60+crumbs.last.to_i
#scores << {:secs => sum, :hms => hh.round.to_s+":"+mm.round.to_s+":"+ss.round.to_s}
#scores_hash << {:secs => sum, :hms => hh.round.to_s+":"+mm.round.to_s+":"+ss.round.to_s}
# milliseconds case end
end
That's my current code but I hate it. It's looks messy. It doesn't just look great at all. Maybe someone whose an expert in ruby could tell how to do this by chaining collects, reduces etc and making it look good?
Time class ruby provides provides at function to get time from seconds. Use this it will cure.
miliseconds = 32290928
seconds = miliseconds/1000
Time.at(seconds).strftime("%H:%M:%S")
OR to get utc time
#Get UTC Time
Time.at(seconds).utc.strftime("%H:%M:%S")
You can wrap this in a helper method:
def format_milisecs(m)
secs, milisecs = m.divmod(1000) # divmod returns [quotient, modulus]
mins, secs = secs.divmod(60)
hours, mins = mins.divmod(60)
[secs,mins,hours].map { |e| e.to_s.rjust(2,'0') }.join ':'
end
format_milisecs 10_600_00
=> "03:13:20"
Nice solution given by #Mike Woodhouse :
Use divmod :
t = 270921000
ss, ms = t.divmod(1000) #=> [270921, 0]
mm, ss = ss.divmod(60) #=> [4515, 21]
hh, mm = mm.divmod(60) #=> [75, 15]
dd, hh = hh.divmod(24) #=> [3, 3]
puts "%d days, %d hours, %d minutes and %d seconds" % [dd, hh, mm, ss]
#=> 3 days, 3 hours, 15 minutes and 21 seconds
Answer is how to convert 270921sec into days + hours + minutes + sec ? (ruby)
I have an hour selection drop down 0-23 and minutes selection drop down 0-59 for Start time and End time respectively (so four controls).
I'm looking for an algorithm to calculate time difference using these four values.
Since they're not stored in fancy date/time selection controls, I don't think I can use any standard date/time manipulation functions.
How do I calculate the difference between the two times?
This pseudo-code gives you the algorithm to work out the difference in minutes. It assumes that, if the start time is after the end time, the start time was actually on the previous day.
const MINS_PER_HR = 60, MINS_PER_DAY = 1440
startx = starthour * MINS_PER_HR + startminute
endx = endhour * MINS_PER_HR + endminute
duration = endx - startx
if duration < 0:
duration = duration + MINS_PER_DAY
The startx and endx values are the number of minutes since midnight.
This is basically doing:
Get number of minutes from start of day for start time.
Get number of minutes from start of day for end time.
Subtract the former from the latter.
If result is negative, add number of minutes in a day.
Don't be so sure though that you can't use date/time manipulation functions. You may find that you could easily construct a date/time and calculate differences with something like:
DateTime startx = new DateTime (1, 1, 2010, starthour, startminute, 0);
DateTime endx = new DateTime (1, 1, 2010, endhour , endminute , 0);
Integer duration = DateTime.DiffSecs(endx, startx) / 60;
if (duration < 0)
duration = duration + 1440;
although it's probably not needed for your simple scenario. I'd stick with the pseudo-code I gave above unless you find yourself doing some trickier date/time manipulation.
If you then want to turn the duration (in minutes) into hours and minutes:
durHours = int(duration / 60)
durMinutes = duration % 60 // could also use duration - (durHours * 60)
This will compute duration in minutes including the year as factor
//* Assumptions:
Date is in Julian Format
startx = starthour * 60 + startminute
endx = endhour * 60 + endminute
duration = endx - startx
if duration <= 0:
duration = duration + 1440
end-if
if currday > prevday
duration = duration + ((currday-preday) - 1 * 1440)
end-if
First you need to check to see if the end time is greater than or equal to the start time to prevent any problems. To do this you first check to see if the End_Time_Hour is greater than Start_Time_Hour. If they're equal you would instead check to see if End_Time_Min is greater than or equal to Start_Time_Min.
Next you would subtract Start_Time_Hour from End_Time_Hour. Then you would subtract Start_Time_Min from End_Time_Min. If the difference of the minutes is less than 0 you would decrement the hour difference by one and add the minute difference to 60 (or 59, test that). Concat these two together and you should be all set.
$start_time_hr = 5;
$start_time_mi = 50;
$end_time_hr = 8;
$end_time_mi = 30;
$diff = (($end_time_hr*60)+$end_time_mi) - (($start_time_hr*60)+$start_time_mi);
$diff_hr = (int)($diff / 60);
$diff_mi = (int)($diff) - ($diff_hr*60);
echo $diff_hr . ':' . $diff_mi;
simple equation should help:
mindiff = 60 + endtime.min - starttime.min
hrdiff = ((mindiff/60) - 1) + endtime.hr - starttime.hr
This gives you the duration in hours and minutes
h1 = "hora1"
m1 "min1"
h2 "hora2"
m2 = "min2"
if ( m1 > m2)
{
h3 = (h2 - h1) - 1;
}
else
{
h3 = h2 - h1;
}
m1 = 60 - m1;
if (m1 + m2 >= 60)
{
m3 = 60 - (m1 + m2);
} else if (m3 < 0)
{
m3 = m3 * -1;
}
else
{
m3 = m1 + m2;
}
System.out.println("duration:" + h3 + "h" + m3 + "min");
If you have a function that returns the number of days since some start date (e.g. dayssince1900) you can just convert both dates to seconds since that start date, do the ABS(d1-d2) then convert the seconds back to whatever format you want e.g. HHHH:MM:SS
Simple e.g.
SecondsSince1900(d)
{
return dayssince1900(d)*86400
+hours(d)*3600
+minutes(d)*60
+seconds(d);
}
diff = ABS(SecondsSince1900(d1)-SecondsSince1900(d2))
return format(diff DIV 3600)+':'+format((diff DIV 60) MOD 60)+':'+format(diff MOD 60);
Hum: Not that simple if you have to take into account the leap seconds astronomers are keen to put in from time to time.