Question

嗨，我正在开发用于教育的Android启动器，我需要它能够在用户点击学校工具按钮时启动设备上安装的学校工具应用程序

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);
    }

}

我还是Android编码的新手，所以请不要评判我

任何帮助都会很棒提前谢谢

问候

Rapsong11

Answer 1

此代码段应该完全符合您的目标

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) {

}

它将按包名称

启动另一个应用程序

来源 - Open another application from your own (intent)

如何使用按钮单击意图打开已安装的Android应用程序？

1 个答案: