Check Time Between 9.30 to 4 Ruby

check time between 9.30 to 4 ruby

t = Time.now

Range.new(
  Time.local(t.year, t.month, t.day, 9),
  Time.local(t.year, t.month, t.day, 16, 30)
) === t

Time range method not working in Rails 4/Ruby 2 app

I would take advantage of the fact that DateTime objects support comparison. E.g.

   Time.local(2014, 12, 31, 6, 30) > Time.local(2015, 1, 1, 6, 30)
   => false

Hence you could rewrite the method during_business_hours:

   # Check if time `t` is between 6:30 and 21:00 of `t`'s day, a.k.a. in between business hours.
   # If that's the case, return true, otherwise false.
   def during_business_hours(t)
     if t < Time.local(t.year, t.month, t.day, 6, 30) ||
        t > Time.local(t.year, t.month, t.day, 21)
        return false
     else
        return true
     end
   end

Personally, your current code for that method is very hard to read, the new version is much clearer.

...and BTW, I would make it a private method.

def validate_each [snip]

private

def during_business_hours(t)
  [snip]

Check if current datetime is between 08 PM and 08 AM and calculate total hours left until 08 AM

Here is one simple way to do it:

var date = DateTime.Now;
if(date.Hour >= 20 || date.Hour < 8)
{
    var DateTime8 = date.Date.AddHours((date.Hour > 8) ? 24 + 8 : 8);
    TimeSpan diff = DateTime8 - date;
}

Custom Validator to Prevent an appointment from being scheduled too early in day?

You can create your own, the same way active record does. This results in clean, portable, easy to test custom validations. An example might look like:

create app/validators/during_business_hours_validator.rb with:

class DuringBusinessHoursValidator < ActiveModel::EachValidator

  def validate_each(record, attribute, value)
    # example - do whatever you want here
    unless value.present? && during_business_hours(value)
      record.errors[attribute] << 'must be during business hours'
    end
  end

  def during_business_hours(time)
    # from http://stackoverflow.com/q/10090962/525478
    Range.new(
      Time.local(time.year, time.month, time.day, 6, 30),
      Time.local(time.year, time.month, time.day, 21)) === time
  end

end

then, in your model add:

validates :appointment_date,
          during_business_hours: true

PHP - compare variables to see if between specified start and end time

I'm not sure with that kind of time format.. maybe it is best if you first convert it on something that can be easily read by php.

try using strtotime() or do some manual conversion.

Generate array of times (as strings) for every X minutes in JavaScript

If the interval is only to be set in minutes[0-60], then evaluate the below solution w/o creating the date object and in single loop:

var x = 5; //minutes intervalvar times = []; // time arrayvar tt = 0; // start timevar ap = ['AM', 'PM']; // AM-PM
//loop to increment the time and push results in arrayfor (var i=0;tt<24*60; i++) {  var hh = Math.floor(tt/60); // getting hours of day in 0-24 format  var mm = (tt%60); // getting minutes of the hour in 0-55 format  times[i] = ("0" + (hh % 12)).slice(-2) + ':' + ("0" + mm).slice(-2) + ap[Math.floor(hh/12)]; // pushing data in array in [00:00 - 12:00 AM/PM format]  tt = tt + x;}
console.log(times);

Rails Rounding float with different options

Here's a generic way to do it for any precision:

class Float
  def round_currency(precision: 1, direction: :none)
    round_method = case direction
      when :none then :round
      when :up   then :ceil
      when :down then :floor
    end

    integer_value = (self * 100).round
    ((integer_value / precision.to_f).send(round_method) * precision / 100.0)
  end
end

# USAGE
9.37.round_currency(direction: :none, precision: 10)
# => 9.4

9.37.round_currency(direction: :up, precision: 25)
# => 9.5

9.37.round_currency(direction: :none)
# => 9.37

# Precision is defined in pennies: 10 dime, 25 quarter, 100 dollar. 1 penny is default

This code converts the float into an integer first to ensure accuracy. Be wary using ceil and floor with floating number arithmetic - due to accuracy errors you could get odd results e.g. 9.37 * 100 = 936.9999999999999. If you floor the result, you'll end up rounding to 9.36

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