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
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;
}
});