How to sort object argument if some arguments are equal

Question

I have an ArrayList with some objects that contains 3 parameters (id, name, salary). Unsorted list:

public class Employee {
        private int id;
        private String name;
        private double salary;

        public Employee(int _id, String _name, double _salary) {
            this.id = _id;
            this.name = _name;
            this.salary = _salary;
        }

        public int getEmployeeID () {
            return this.id;
        }


        public String getEmployeeName () {
            return this.name;
        }

        public double getMonthSalary() {
            return this.salary;
        }

        public static void main(String[] argc) {
            ArrayList<Employee> list = new ArrayList<Employee>();
            list.add(new Employee(1, "asd", 1300));
            list.add(new Employee(7, "asds", 14025)); // this
            list.add(new Employee(2, "nikan", 1230));
            list.add(new Employee(3, "nikalo", 12330));
            list.add(new Employee(8, "aaaa", 14025)); // and this are equal
            list.add(new Employee(4, "nikaq", 140210));
            list.add(new Employee(5, "nikas", 124000)); 
            list.add(new Employee(6, "nikab", 14020));
        }
}

And I want to sort list by salary, and if salaries are equal then sort their names alphabetically. I tried to use Collections.sort();, but it sorts by all of arguments.

e.g: 14025 and 14025 are equal, so sort this one by name: asds, aaaa --> aaaa, asds

Here's what I tried:

for (int i = 0; i < list.size() - 1; i++) {
    for (int j = 0; j < list.size() - 1; j++) {
        boolean isSorted = false;
        if (list.get(j + 1).getMonthSalary() > list.get(j).getMonthSalary()) {
            Employee temprary = list.get(j);
            list.set(j, list.get(j + 1));
            list.set(j + 1, temprary);
            isSorted = true;
        }

        if (isSorted) {
            if (list.get(j).getMonthSalary() == list.get(j + 1).getMonthSalary()) {
                Collections.sort(list, new Comparator < Employee > () {
                    @Override
                    public int compare(Employee s1, Employee s2) {
                        return s1.getEmployeeName().compareToIgnoreCase(s2.getEmployeeName());
                    }
                });
            }
        }
    }
}

@ΦXocę웃Пepeúpa thanks, Mureinik already answered my question — phen0menon
– phen0menon, Commented Jun 26, 2016 at 6:42

Andreas · Accepted Answer · 2016-06-25 20:50:22Z

6

Java has a built-in mechanism to handle such a problem - Comparators. You shouldn't reinvent the wheel:

list.sort(Comparator.comparing(Employee::getMonthSalary)
                    .thenComparing(Employee::getEmployeeName));

edited Jun 25, 2016 at 20:50

Andreas

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answered Jun 25, 2016 at 20:45

Mureinik

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4 Comments

phen0menon Over a year ago

So what if I have another constructor from another class that extends Employee.java? e.g Employee2(int id, String name, double salary) How to sort ArrayList<Employee> that contains Employee.add(new Employee2(1, "test", 1000);?

Mureinik Over a year ago

If Employee2 extends Employee, it's still a type of Employee and the same code will work for it too.

phen0menon Over a year ago

I get an error: Employee.sort(Comparator.comparing(Employee::getEmployeeSalary) and etc - Cannot resolve method GetMonthSalary, Cannot resolve method getEmployeeName. But if Employee.sort(Comparator.comparing(Employee2::getEmployeeSalary) and etc it works! What's wrong?

Mureinik Over a year ago

@phen0men works just fine for me, so there's obviously something missing here. However, it's extremely hard to read code in the comments like this. Could you please post a new question with the entire code and the exact error so it would be easier to answer?

Olav Trauschke · Accepted Answer · 2016-06-25 20:46:25Z

0

Collections.sort() sort by natural ordering. Try passing a comparator that compares the way you want to or implement compareTo.

answered Jun 25, 2016 at 20:46

Olav Trauschke

1368 bronze badges

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