zoukankan      html  css  js  c++  java
  • 双击Back退出应用 android中弹出吐司

    第一种方法:
    public void onBackPressed() {
    
    
    		if (isState) { //isState初始值为true
    			isState = false;
    			Toast.makeText(this, "请双击back退出程序", Toast.LENGTH_SHORT).show();
    			new Timer().schedule(new TimerTask() {
    
    				@Override
    				public void run() {
    					// TODO Auto-generated method stub
    					isState = true;
    				}
    			}, 2000);
    		} else {
    			finish();
    		}
    	}

    第二种方法:


    public class MainActivity extends Activity {
    
    
        private Toast toast;
    
        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.activity_main);
            toast = Toast.makeText(getApplicationContext(), "确定退出?", 0);
    
        }
        public void onBackPressed() {
            quitToast();
        }
        /*
        public boolean onKeyDown(int keyCode, KeyEvent event) {
            System.out.println(keyCode + "...." + event.getKeyCode());
            if(keyCode == KeyEvent.KEYCODE_BACK){
                quitToast();
            }
            return super.onKeyDown(keyCode, event);
        }
        */
        private void quitToast() {
            if(null == toast.getView().getParent()){
                toast.show();
            }else{
                System.exit(0);
            }
        }
    }

    程序员小冰博客:http://blog.csdn.net/qq_21376985 技术交流QQ986945193 微博:http://weibo.com/mcxiaobing
  • 相关阅读:
    VUE 路由
    JS小表达式
    JS正则表达式
    react案例—实现组件之间的通信
    React脚手架搭建
    react属性三ref
    react属性二props
    java多线程2
    jvm小结2
    spark小结3
  • 原文地址:https://www.cnblogs.com/mcxiaobing/p/5472127.html
Copyright © 2011-2022 走看看