根据Java 8中的不同案例字符串属性，从对象列表中删除重复项

Question

如何根据Java 8中的" String"属性列表从对象列表中删除重复项？我只能找到INT或长属性的代码，但在字符串为差异情况并获取唯一列表时无法找到字符串比较。

以下是我要实现的程序。

public class Diggu {
    class Employee{
        private int id;
        private String name;
        public Employee(int id, String name){
            this.id = id;
            this.name = name;
        }
        public int getId() {
            return id;
        }
        public void setId(int id) {
            this.id = id;
        }
        public String getName() {
            return name;
        }
        public void setName(String name) {
            this.name = name;
        }

    }
    public void ra(){
        List<Employee> employee = Arrays.asList(new Employee(1, "John"), new Employee(3, "JOHN"), new Employee(2, "BOB"));
        System.out.println(""+ employee.size());
        List<Employee> unique = employee.stream()
                                .collect(collectingAndThen(toCollection(() -> new TreeSet<>(comparing(Employee::getName))),
                                                           ArrayList::new));
        System.out.println(""+ unique.size());
    }
    public static void main(String[] args) {
        new Diggu().ra();
    }
}

结果：

run:
3
3
BUILD SUCCESSFUL (total time: 2 seconds)

在哪里应该是3和2

Answer 1

John和 JOHN是不同的字符串，请使用case不敏感的Comparator

 Comparator.comparing(Employee::getName, String.CASE_INSENSITIVE_ORDER)