How to check the age is between 18 and 75 years from today

Question

I have written the code like below, but it it is not working in below case.

LocalDate today = LocalDate.now(ZoneId.systemDefault());
DateTimrFormatter dobFormat = DateTimeFormatter.ofPattern("dd-MM-uuuu");
int age = Period.between(inputdate, today).getYears();
if (age >= 18 && age <= 75) {
    return true;
} else {
    return false
}

It is not checking age including days i.e. if we give input date as 24-06-1948, then age is 75 years + 2 days (greater than 75), but it returns true.

Answer 1

You can create new Period objects, and utilize the isZero and isNegative methods, to determine comparison.

Period minus18 = period.minusYears(18);
boolean gte18 = minus18.isZero() || !minus18.isNegative();
Period minus75 = period.minusYears(75);
boolean lte75 = minus75.isZero() || minus75.isNegative();
if (gte18 && lte75) {

} else {

}

Here is an example.

LocalDate today = LocalDate.now(ZoneId.systemDefault());
LocalDate[] dates = {
    LocalDate.of(1948, 6, 25), LocalDate.of(1948, 6, 26), LocalDate.of(1948, 6, 27),
    LocalDate.of(2005, 6, 25), LocalDate.of(2005, 6, 26), LocalDate.of(2005, 6, 27) };
Period period, minus18, minus75;
boolean gte18, lte75;
for (LocalDate inputdate : dates) {
    period = Period.between(inputdate, today);
    minus18 = period.minusYears(18);
    gte18 = minus18.isZero() || !minus18.isNegative();
    System.out.printf("%-12s - 18 = %-12s >= 18 = %b%n", period, minus18, gte18);
    minus75 = period.minusYears(75);
    lte75 = minus75.isZero() || minus75.isNegative();
    System.out.printf("%-12s - 75 = %-12s <= 75 = %b%n", period, minus75, lte75);
}

Output

P75Y1D       - 18 = P57Y1D       >= 18 = true
P75Y1D       - 75 = P1D          <= 75 = false
P75Y         - 18 = P57Y         >= 18 = true
P75Y         - 75 = P0D          <= 75 = true
P74Y11M30D   - 18 = P56Y11M30D   >= 18 = true
P74Y11M30D   - 75 = P-1Y11M30D   <= 75 = true
P18Y1D       - 18 = P1D          >= 18 = true
P18Y1D       - 75 = P-57Y1D      <= 75 = true
P18Y         - 18 = P0D          >= 18 = true
P18Y         - 75 = P-57Y        <= 75 = true
P17Y11M30D   - 18 = P-1Y11M30D   >= 18 = false
P17Y11M30D   - 75 = P-58Y11M30D  <= 75 = true

Answer 2

I will recommend to use LocalDate, you can create two local date objects one with minimum date and another with maximum date

 LocalDate minDate = LocalDate.now().minusYears(18).minusDays(1);
 LocalDate maxDate = LocalDate.now().plusYears(57).plusDays(1);

Considering today's date as center point, you can create minDate with subtracting 18 years and 1 day, and the same way for maxDate of 75 years

Now you can use isBefore and isAfter methods to check in between condition

if(minDate.isBefore(toDayDate) && maxDate.isAfter(toDayDate))

Note : If you are looking for DOB check, always take DOB as center point

Answer 3

Former library Joda-time had the solution. See How do I calculate someone's age in Java?.

With the new Java-Time API we have two options:

• Duration with compareTo, not suitable for months and years
• Period with arithmetic, see Reilas 's answer
• compare different LocalDates

Duration for short and precise (weeks, days, milli-/nano-seconds)

Usually we can use Duration when we need to compare objects to express temporal distance like age. The Duration class implements the Comparable interface and has a method compareTo(otherDuration).

This would work for approximate relative events like:

• updated 5 minutes ago
• next run in 2 weeks and 4 days

So, in theory we could compare the actual-age to a boundary like 18 years like actualAge.compareTo(AGE_18_YEARS) and evaluate the returned -1, 0 or 1. In practice the age-reference of AGE_18_YEARS must be specified as Duration which only allows ChronoUnit.Days as largest unit. How many days would be 18 years ?

Period based on calendar for long (especially months and years)

Unfortunately distant events like the age of a person take months and years. Both chrono-units depend on a calendar and their relative position. Months can be 28, 29, 30 or 31 days. Years can be 365, even 366 days for a leap-year.

LocalDate orientation within a calendar of birthdays

The age is special for any person, it has a special day and comprises a special number of days (depending on the birthday and leap-years).

Let's use the birthday to evaluate if someone is 18, in between or 75 years old.

Using a single concept like calendar, birthdays and date-comparison leads to a meaningful and readable solution.

Test the demo on IDEone:

// DOB: 2000-12-24
var dob =  LocalDate.of(2000, 12, 24);


// Minimum age: 18 years
var birthdayAge18years = dob.plusYears(18);
// Maximum age: 75 years
var birthdayAge75years = dob.plusYears(75);

var today = LocalDate.now();

// is today
//      on or after  18th birthday 
//  and on or before 75th birthday
var olderThanOr18 = today.equals(birthdayAge18years) || today.isAfter(birthdayAge18years);
var youngerThanOr75 = today.equals(birthdayAge75years) || today.isBefore(birthdayAge75years);

// test
System.out.println(birthdayAge18years + " > older than or exactly 18 years: " + olderThanOr18);
System.out.println(birthdayAge75years + " > younger than or exactly 75 years: " + youngerThanOr75);

Answer 4

The issue is that getYears() returns the rounded number of full years, which is 75 in this case. So what you should really do is if (age >= 18 && age < 75).

That said, I have a few more remarks:

1. Your question should not have syntax errors.
2. You should really, really, learn to use a debugger if you want to pursue any form of professional software development career. It is a fundamental and essential skill.
3. For future questions, please have a look at the StackOverflow code of conduct