Question

我在Mainactivity中调用prog()函数，该函数使用进度条执行加载页面，并在程序函数LoginActivity之后立即调用。但是，它会在激活程序功能之前调用登录活动，我是android studio的新手，需要您的帮助。这是我的代码

public class MainActivity extends AppCompatActivity {

ProgressBar pb;
int counter = 0;
TextView textView;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    textView = findViewById(R.id.textView);

    prog();//loading page
     Intent intentLoginPage = new Intent(MainActivity.this,LoginActivity.class);
     startActivity(intentLoginPage);

}

public boolean prog(){// to dispay progress bar as loading bar in loading page

    pb = findViewById(R.id.progressBar);
    final Timer t = new Timer();
    TimerTask tt = new TimerTask() {
        @Override
        public void run(){
            counter++;
            textView.setText(counter + "%");//0 to 100%
            pb.setProgress(counter);
            if(counter == 100)
                t.cancel();

        }
    };
    t.schedule(tt,0,60);
    return true;
}
}

Answer 1

也许您需要在run（）中调用startActivity吗？：

import React, { Component } from 'react'

export default class Addroom extends Component {

    constructor(props) {
        
        super(props)
        
        }

        addRoomName=(e)=> {
            this.setState({room:e.target.value})

        }
        addType=(e)=> {
            this.setState({type:e.target.value})
        }


    createRoom=()=> {
        this.props.add(this.state.room,this.state.type);
    }


    render() {
        return (
            <div>

                <input onChange={this.addRoomName} placeholder='Name Your Room'/><br/>
                <input onChange={this.addType} placeholder='Whats The Room Type?'/><br/>
                <button onClick={this.createRoom}>Create</button>
                
            </div>
        )
    }
}

android studio活动在完成上一个之前开始

1 个答案: