Sorting string property using comparable

Question

I have an employee class & a client class. I am able to sort using compareTo() via Employee's id & age as they are of integer type. But how do I sort by employee's name or salary? compareTo is not accepting any data type other than int, throws a compile time exception.

public class Employee implements Comparable<Employee> {

    private int id;
    private String name;
    private int age;
    private long salary;

    public int getId() {
        return id;
    }

    public String getName() {
        return name;
    }

    public int getAge() {
        return age;
    }

    public long getSalary() {
        return salary;
    }

    public Employee(int id, String name, int age, int salary) {
        this.id = id;
        this.name = name;
        this.age = age;
        this.salary = salary;
    }

    @Override
    public int compareTo(Employee emp) {
        //let's sort the employee based on id in ascending order
        //returns a negative integer, zero, or a positive integer as this employee id
        //is less than, equal to, or greater than the specified object.
        return (this.age - emp.age);
    }

    @Override
    //this is required to print the user friendly information about the Employee
    public String toString() {
        return "[id=" + this.id + ", name=" + this.name + ", age=" + this.age + ", salary=" +
                this.salary + "]";
    }

}

Client class

import java.util.Arrays;

public class Test {

    public static void main(String a[]){

        //sorting custom object array
        Employee[] empArr = new Employee[4];
        empArr[0] = new Employee(10, "Mikey", 25, 10000);
        empArr[1] = new Employee(20, "Arun", 29, 20000);
        empArr[2] = new Employee(5, "Lisa", 35, 5000);
        empArr[3] = new Employee(1, "Pankaj", 32, 50000);

        //sorting employees array using Comparable interface implementation
        Arrays.sort(empArr);
        System.out.println("Default Sorting of Employees list:\n"+Arrays.toString(empArr));

    }
}

I just googled seems like I can also achieve it by implementing comparator

public static Comparator<Employee> SalaryComparator = new Comparator<Employee>() {

        @Override
        public int compare(Employee e1, Employee e2) {
            return (int) (e1.getSalary() - e2.getSalary());
        }
    };

Which one is advisable using the comparator or

@Override
public int compareTo(Employee emp) {
// compare salaries, using the builtin Long.compare:
    return Long.compare (salary, emp.salary);
}

compareTo wants you, according to the documentation of Comparable, to return an int describing the ternary comparison result (less, equal, or greater than), don't confuse that. — Smutje
– Smutje, Commented May 7, 2014 at 6:15
As your name is a String and String implements Comparable as well, just reuse it: return this.name.compareTo(emp.name); — Smutje
– Smutje, Commented May 7, 2014 at 6:19
Use index to get fields by name and use it in the comparator model. — Roman C
– Roman C, Commented May 7, 2014 at 6:32

Suren Raju · Accepted Answer · 2014-05-07 06:14:12Z

4

If you want to sort based on name, you can try

@Override
    public int compareTo(Employee emp) {
        return this.name.compareTO(emp.name);
    }

answered May 7, 2014 at 6:14

Suren Raju

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1 Comment

AgilePro Over a year ago

this will not work if you are using a primitive type like int. Age is declared as an int.

Mureinik · Accepted Answer · 2014-05-07 06:18:56Z

Basically, you want to sort your class based on several properties. The trick here is to decide the order in which they should be evaluated, and treat one as a Comparable on its own right. If the comparison isn't 0 - you found which instance should come before the other, and you can just return the value. If not, you need to evaluate a different property.

E.g., assuming the properties you want to use are id, age, name and salary:

@Override
public int compareTo(Employee emp) {
    // compare IDs:
    int cmp = Integer.compare(id, emp.id);
    if (cmp != 0) {
        return cmp;
    }

    // compare ages:
    cmp = Integer.compare(age, emp.age);
    if (cmp != 0) {
        return cmp;
    }

    // compare names. Luckily, Strings are comparable:
    cmp = name.compareTo(emp.name);
    if (cmp != 0) {
        return cmp;
    }

    // compare salaries, using the builtin Long.compare:
    return Long.compare (salary, emp.salary);
}

AgilePro · Accepted Answer · 2014-05-07 06:11:44Z

1

I would implement compareTo this way:

public int compareTo(Employee emp) {
    return Integer.compare(this.age, emp.age);
}

answered May 7, 2014 at 6:11

AgilePro

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