Bonfire Friendly Date Ranges

Author

Created by Rafase282

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Details

• Difficulty: 4/5

Implement a way of converting two dates into a more friendly date range that could be presented to a user.

It must not show any redundant information in the date range.

For example, if the year and month are the same then only the day range should be displayed.

Secondly, if the starting year is the current year, and the ending year can be inferred by the reader, the year should be omitted.

Input date is formatted as YYYY-MM-DD

Remember to use Read-Search-Ask if you get stuck. Try to pair program. Write your own code.

Useful Links

• String.split()
• String.substr()
• parseInt()

Problem Script:

function friendly(str) {
  return str;
}

friendly(['2015-07-01', '2015-07-04']);

Problem Explanation:

• Create a program that will take two dates and convert them into a more easy to understand date such as January 1st, 2017. It will also check the difference between them, and handles cases with no difference, more than a day, more than a month, more than a year, and more than a month and less than a year respectively.

Hint: 1

• Split the string into an array where you get the "YYYY", "MM", "DD"

Hint: 2

• You need to handle the case for "st", "nd", and "th". Note that 13 is "th" not "rd".

Hint: 3

• If you are using Date() to create instances of dates to work with, then use UTC time to avoid errors due to time zone difference between servers.

Spoiler Alert!

Solution ahead!

Code Solution:

function friendly(str) {

  var months = ['January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December'];

  // COnverst a YYYY-MM-DD string into a date object.
  function convertDate(str) {
    // Split the dates to work independently.
    var dateStr = str.split('-');

    // Force the dates into Universal time to avoid issues due to timezones.
    return (new Date(Date.UTC(dateStr[0], dateStr[1] - 1, dateStr[2])));

  }

  // Handles the case of the day's endings.
  function dateEnding(val) {
    switch (val) {
      case 1:
      case 21:
      case 31:
        return val + 'st';
      case 2:
      case 22:
        return val + 'nd';
      case 3:
      case 23:
        return val + 'rd';
      default:
        return val + 'th';
    }
  }

  // Checks for the real difference in months to avoid errors
  function monthDiff(date1, date2) {
    var month2 = date2.getUTCFullYear() * 12 + date2.getUTCMonth();
    var month1 = date1.getUTCFullYear() * 12 + date1.getUTCMonth();
    return month2 - month1;
  }

  // Get's the right month string.
  function getMonth(date) {
    return months[date.getUTCMonth()];
  }

  function displayDate() {

    // Handles same day
    if (date2.getTime() - date1.getTime() === 0) {
      return [getMonth(date1) + ' ' + dateEnding(date1.getUTCDate()) + ', ' + date1.getUTCFullYear()];
    }

    // Handles same month
    if (date1.getUTCMonth() === date2.getUTCMonth() && date1.getUTCFullYear() === date2.getUTCFullYear()) {
      return [getMonth(date1) + ' ' + dateEnding(date1.getUTCDate()), dateEnding(date2.getUTCDate())];
    }

    // Handles more than a month of difference, but less than 12 months
    if (monthDiff(date1, date2) < 12) {
      return [getMonth(date1) + ' ' + dateEnding(date1.getUTCDate()), getMonth(date2) + ' ' + dateEnding(date2.getUTCDate())];
    }

    // Handles cases with more than 12 months apaprt.
    return [getMonth(date1) + ' ' + dateEnding(date1.getUTCDate()) + ', ' + date1.getUTCFullYear(), getMonth(date2) + ' ' + dateEnding(date2.getUTCDate()) + ', ' + date2.getUTCFullYear()];
  }

  var date1 = convertDate(str[0]);
  var date2 = convertDate(str[1]);

  return displayDate();

}

Code Explanation:

• Explain your code here

Credits:

If you found this page useful, you can give thanks by copying and pasting this on the main chat: thanks @Rafase282

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