运用面向对象思维对数组的增删改查

1. 创建animal类，添加动物属性，添加构造方法和方法

public class Animal { public String type; public String name; public int age; public int legnum; public Animal(String type,String name,int age,int legnum){ this.type=type; this.name=name; this.age= age; this.legnum = legnum; } public String getinfo(){ return "动物种类:"+type+" 名字:"+name+" 岁数:"+age+" 腿的数量:"+legnum; } 

2.数组的增加

 Scanner input = new Scanner(System.in); Animal[] ani = new Animal[3]; ani[0] = new Animal("袋鼠", "花花", 2, 2); ani[1] = new Animal("河马", "憨憨", 1, 3); ani[2] = new Animal("*", "草草", 3, 4); // 数组的遍历 for (int i = 0; i < ani.length; i++) { System.out.println(ani[i].getinfo()); } // 数组的增加 System.out.println("请输入您需要增加的动物信息："); String type1 = input.next(); String name1 = input.next(); int age1 = input.nextInt(); int legnum1 = input.nextInt(); Animal num = new Animal(type1, name1, age1, legnum1); ani = Arrays.copyOf(ani, ani.length + 1); ani[ani.length-1] = num; for (int i = 0; i < ani.length; i++) { System.out.println(ani[i].getinfo()); } 

3.数组的删除操作

int index = 0; System.out.println("请输入您需要删除的动物"); String type2 =input.next(); String name2 = input.next(); int age2 =input.nextInt(); int legnum2 = input.nextInt(); Animal num = new Animal(type2, name2, age2, legnum2); System.out.println(num.getinfo()); for (int i = 0; i < ani.length;i++) { if (ani[i].type.equals(num.type) && ani[i].name.equals(num.name) && ani[i].age==num.age &&ani[i].legnum==num.legnum) { index = i; break; } } System.out.println(index); if(index==-1){ System.out.println("动物园内不存在该动物"); } else{ for(int j=index;j<ani.length-1;j++){ ani[j]=ani[j+1]; } } ani[ani.length-1]=new Animal(null,null,0,0); Arrays.copyOf(ani,ani.length-1); for (int i = 0; i < ani.length; i++) { System.out.println(ani[i].getinfo()); } 

4.数组的查找操作

int index = 0; System.out.println("请输入您需要查找的动物"); String type2 =input.next(); String name2 = input.next(); int age2 =input.nextInt(); int legnum2 = input.nextInt(); Animal num = newAnimal(type2, name2, age2, legnum2); System.out.println(num.getinfo()); for (int i = 0; i < ani.length;i++) { if (ani[i].type.equals(num.type) &&ani[i].name.equals(num.name) && ani[i].age == num.age && ani[i].legnum == num.legnum) { index = i; break; } } System.out.println("动物园内存在该动物，他所在的位置是：" + index); if (index == -1) { System.out.println("动物园内不存在该动物"); } 

5.数组的改操作

System.out.println("请输入您需要改变的动物类型："); String type2 = input.next(); String name2 = input.next(); int age2 = input.nextInt(); int legnum2= input.nextInt(); Animal num = new Animal(type2, name2, age2,legnum2); System.out.println(num.getinfo()); for (int i = 0; i < ani.length; i++) { if (ani[i].type.equals(num.type) &&ani[i].name.equals(num.name) && ani[i].age == num.age && ani[i].legnum == num.legnum) { System.out.println("请输入新的动物："); Stringtype3 = input.next(); String name3 = input.next(); int age3 =input.nextInt(); int legnum3 = input.nextInt(); Animal num1 = newAnimal(type3, name3, age3, legnum3); ani[i]= num1; } } for (int i =0; i < ani.length; i++) { System.out.println(ani[i].getinfo()); } 

本文地址：https://blog.csdn.net/qq_45960200/article/details/107409915