Java 如何使用按钮单击意图打开已安装的 android 应用程序?

声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow 原文地址: http://stackoverflow.com/questions/18937677/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me): StackOverFlow

提示:将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间:2020-08-12 12:40:51  来源:igfitidea点击:

How to open a installed android app with a button click intent?

javaandroideclipse

提问by rapsong11

Hi i'm working on a android launcher for education and I need it to be able to when the user clicks the school tools button it launches the school tools app that is installed on the device

嗨,我正在开发用于教育的 android 启动器,我需要它能够在用户单击学校工具按钮时启动安装在设备上的学校工具应用程序

Here's the code

这是代码

package com.d4a.stzh;

import android.net.Uri;
import android.os.Bundle;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.content.Intent;

import com.actionbarsherlock.app.SherlockFragment;

public class FragmentTab1 extends SherlockFragment {
    private Button appbtn;
    private Button webbtn;
    private Button toolsbttn;



    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,
            Bundle savedInstanceState) {
        // Get the view from fragmenttab1.xml
        View view = inflater.inflate(R.layout.fragmenttab1, container, false);


        //Get the button from layout
        appbtn = (Button) view.findViewById(R.id.app);
        webbtn = (Button) view.findViewById(R.id.web);
        toolsbttn = (Button) view.findViewById(R.id.tools);

       //show all apps installed on the device 
        appbtn.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) {
                Intent intent = new Intent(FragmentTab1.this.getActivity(), MyLauncherActivity.class);
                startActivity(intent);

            }


            });


        //luanches google on the default web browser
        webbtn.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) {
                String url = "http://www.google.com";
                Intent i = new Intent(Intent.ACTION_VIEW);
                i.setData(Uri.parse(url));
                startActivity(i);
            }
        });
        //tools button i know ths code is wrong!I need help here!
        toolsbttn.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) {
                Intent intent = new Intent(FragmentTab1.this.getActivity(), MyLauncherActivity.class);
                startActivity(intent);

            }


            });

        return view;
    }


    @Override
    public void onSaveInstanceState(Bundle outState) {
        super.onSaveInstanceState(outState);
        setUserVisibleHint(true);
    }

}

I am still new to android coding so please don't judge me

我还是 android 编码的新手,所以请不要评判我

Any help would be amazing Thanks way in advance

任何帮助都会很棒提前谢谢

Regards

问候

Rapsong11

Rapsong11

采纳答案by Daniel Bo

This code snippet should do exactly what you are trying to achieve

此代码片段应该完全符合您要实现的目标

Intent i;
PackageManager manager = getPackageManager();
try {
   i = manager.getLaunchIntentForPackage("com.example.schoolToolApp");
if (i == null)
    throw new PackageManager.NameNotFoundException();
i.addCategory(Intent.CATEGORY_LAUNCHER);
startActivity(i);
} catch (PackageManager.NameNotFoundException e) {

}

It will just launch another app by its package name

它只会通过包名启动另一个应用程序

source - Open another application from your own (intent)

源 -从您自己的(意图)打开另一个应用程序