java - 如何通过检查这 3 个条件对通用列表进行排序

Question

如果我们有如下列表

Rollno name age address
====== ==== === =======
01      abc  11   pqr
02      stu  21   pqr

我们应该根据名称对该列表进行排序。如果学生姓名相同，则根据年龄排序；如果年龄也相同，则根据数字排序。我对如何对名称进行排序感到困惑我尝试过的代码是

public void sortData(List<Student> studentList) {
     String a1=null;
     String a2[] = null;
     int i=0;
    for (Iterator<Student> iter = list.listIterator(); iter.hasNext();) {
        Student a = iter.next();
       System.out.println(a1 = a.getStudentName());

    }


    System.out.println(a2);
}

它正在打印卷号列表，但我现在不知道如何对它们进行排序。

Answer 1

使用Comparable和Collections.sort():

public class TestClass {

@Test
public void testSort() {
    List<Student> list = new ArrayList<Student>();
    list.add(  new Student( "two", "4", "street7" ) );
    list.add(  new Student( "two", "4", "street6" ) );
    list.add(  new Student( "one", "1", "street3" ) );
    list.add(  new Student( "two", "2", "street5" ) );



    Collections.sort( list );

    assertEquals( 
        "[Student [name=one, age=1, adress=street3], " +
        "Student [name=two, age=2, adress=street5], " +
        "Student [name=two, age=4, adress=street6], " +
        "Student [name=two, age=4, adress=street7]]", 
        list.toString() 
    );
}


public static class Student implements Comparable<Student> {

    String name;
    String age;
    String adress;



    public Student( String name, String age, String adress ) {
        super();
        this.name = name;
        this.age = age;
        this.adress = adress;
    }


    @Override
    public int compareTo( Student o ) {
        if ( o == null ) {
            return -1;
        }

        int compare = name.compareTo( o.name );
        if ( compare != 0 ) {
            return compare;
        }

        compare = age.compareTo( o.age );
        if ( compare != 0 ) {
            return compare;
        }

        return adress.compareTo( o.adress );
    }


    @Override
    public String toString() {
        StringBuilder builder = new StringBuilder();
        builder.append( "Student [name=" );
        builder.append( name );
        builder.append( ", age=" );
        builder.append( age );
        builder.append( ", adress=" );
        builder.append( adress );
        builder.append( "]" );
        return builder.toString();
    }

}

}