1,首先创建一个类
class Person3{ private String name; private int age; private double score; public Person3(String name, int age, double score) { super(); this.name = name; this.age = age; this.score = score; } @Override public String toString() { return "Person3 [name=" + name + ", age=" + age + ", score=" + score + "]"; } public String getName() { return name; } public void setName(String name) { this.name = name; } public int getAge() { return age; } public void setAge(int age) { this.age = age; } public double getScore() { return score; } public void setScore(double score) { this.score = score; } }
这个时候再产生这个类的对象,将对象存入数组里,并对这个数组进行排序
public class Test17 { public static void main(String[] args) { Person3[] array = new Person3[3]; array[0] = new Person3("caocao",18,29); array[1] = new Person3("liubei",20,49); array[2] = new Person3("guanyu",21,77); //对这个数组排序 Arrays.sort(array); System.out.println(Arrays.toString(array)); } }
我们会发现运行结果出现了异常
因为你并不知道应该是依照这个对象的什么属性进行排序
因此我们需要去重写一个compareTo方法,重写这个方法的时候应该要继承一个接口Comparable。
查看这个接口的源码
public interface Comparable<T> { /** * Compares this object with the specified object for order. Returns a * negative integer, zero, or a positive integer as this object is less * than, equal to, or greater than the specified object. * * <p>The implementor must ensure <tt>sgn(x.compareTo(y)) == * -sgn(y.compareTo(x))</tt> for all <tt>x</tt> and <tt>y</tt>. (This * implies that <tt>x.compareTo(y)</tt> must throw an exception iff * <tt>y.compareTo(x)</tt> throws an exception.) * * <p>The implementor must also ensure that the relation is transitive: * <tt>(x.compareTo(y)>0 && y.compareTo(z)>0)</tt> implies * <tt>x.compareTo(z)>0</tt>. * * <p>Finally, the implementor must ensure that <tt>x.compareTo(y)==0</tt> * implies that <tt>sgn(x.compareTo(z)) == sgn(y.compareTo(z))</tt>, for * all <tt>z</tt>. * * <p>It is strongly recommended, but <i>not</i> strictly required that * <tt>(x.compareTo(y)==0) == (x.equals(y))</tt>. Generally speaking, any * class that implements the <tt>Comparable</tt> interface and violates * this condition should clearly indicate this fact. The recommended * language is "Note: this class has a natural ordering that is * inconsistent with equals." * * <p>In the foregoing description, the notation * <tt>sgn(</tt><i>expression</i><tt>)</tt> designates the mathematical * <i>signum</i> function, which is defined to return one of <tt>-1</tt>, * <tt>0</tt>, or <tt>1</tt> according to whether the value of * <i>expression</i> is negative, zero or positive. * * @param o the object to be compared. * @return a negative integer, zero, or a positive integer as this object * is less than, equal to, or greater than the specified object. * * @throws NullPointerException if the specified object is null * @throws ClassCastException if the specified object's type prevents it * from being compared to this object. */ public int compareTo(T o); }
接下来我们来重写这个方法,并使通过年龄来排序
class Person3 implements Comparable<Person3>{ private String name; private int age; private double score; public Person3(String name, int age, double score) { super(); this.name = name; this.age = age; this.score = score; } @Override public String toString() { return "Person3 [name=" + name + ", age=" + age + ", score=" + score + "]"; } public String getName() { return name; } public void setName(String name) { this.name = name; } public int getAge() { return age; } public void setAge(int age) { this.age = age; } public double getScore() { return score; } public void setScore(double score) { this.score = score; } @Override public int compareTo(Person3 o) { // TODO Auto-generated method stub return age-o.age; } }
分别通过姓名和成绩
但是这样写的话我们会发现每次想要改一下比较什么比较麻烦
于是我们改用这个方法来实现
Arrays.sort(array, new Comparator<Person3>(){ //传入的时候由用户自己选 @Override public int compare(Person3 o1, Person3 o2) { // TODO Auto-generated method stub int age1 = o1.getAge(); int age2 = o2.getAge(); return age1 > age2 ? age1:(age1==age2) ? 0:-1; } });Comparator是一个接口,这个是使用这个接口实现了一个匿名内部类,在里面重写compare方法,来用来比较这个对象