打开浏览器到 web 页面 Android App

Question

Trying to grasp Java and Android would like help with a simple task of opening a users browser after they click a button.

I have been doing tutorials for the last two days though it might help if I just took a stab at it and got feedback. thanks in advance for any help.

main.xml:

<AbsoluteLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:orientation="vertical"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"
    android:background="@drawable/bgimage2"> 
    >

<Button
android:id="@+id/goButton"
android:layout_width="150px"
android:layout_height="wrap_content"
android:text="@string/start"
android:layout_x="80px"
android:layout_y="21px"
>
</AbsoluteLayout>

GetURL.java:

package com.patriotsar;

import android.app.Activity;
import android.content.Intent;
import android.view.View.OnClickListener;

String url = "http://www.yahoo.com";
Intent i = new Intent(Intent.ACTION_VIEW);
Uri u = Uri.parse(url);
i.setData(u);


public class patriosar extends Activity {

     private Button goButton;

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);


        goButton.setOnClickListener(new OnClickListener() {         
            @Override
            public void onClick(View v){
                try {
                      // Start the activity
                      startActivity(i);
                    } catch (ActivityNotFoundException e) {
                      // Raise on activity not found
                      Toast toast = Toast.makeText(context, "Browser not found.", Toast.LENGTH_SHORT);
                    }
                  } 
        });

    }
}

Answer 1

It's close, but a few things are in the wrong place or missing. The below code works -- I tried to make the minimum necessary alterations. You could load both versions into something like WinMerge to see exactly what changed.

main.xml:

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:orientation="vertical"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"
    android:background="@drawable/bgimage2"
    >

    <Button
        android:id="@+id/goButton"
        android:layout_width="150px"
        android:layout_height="wrap_content"
        android:text="@string/start"
        android:layout_x="80px"
        android:layout_y="21px"
    ></Button>
</LinearLayout>

GetURL.java:

import android.app.Activity;
import android.content.ActivityNotFoundException;
import android.content.Context;
import android.content.Intent;
import android.net.Uri;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.Toast;

public class GetURL extends Activity {
    private Button goButton;
    String url = "http://www.yahoo.com";
    Intent i = new Intent(Intent.ACTION_VIEW);
    Uri u = Uri.parse(url);
    Context context = this;

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        goButton = (Button)findViewById(R.id.goButton);
        goButton.setOnClickListener(new OnClickListener() {         
            @Override
            public void onClick(View v){
                try {
                      // Start the activity
                        i.setData(u);
                      startActivity(i);
                    } catch (ActivityNotFoundException e) {
                      // Raise on activity not found
                      Toast.makeText(context, "Browser not found.", Toast.LENGTH_SHORT);
                    }
                  } 
        });
    }
}

(You also need a bgimage2.png file in /res/drawable/ and a start string in /res/values/strings.xml , of course).

Answer 2

To simplify you could do

Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(" http://www.yahoo.com ")); startActivity(intent);