java 打开浏览器到网页 Android App
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/6166522/
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
Open browser to web page Android App
提问by Denoteone
Trying to grasp Java and Android would like help with a simple task of opening a users browser after they click a button.
试图掌握 Java 和 Android 需要帮助完成一个简单的任务，即在用户单击按钮后打开浏览器。
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.
过去两天我一直在做教程，但如果我只是尝试一下并获得反馈可能会有所帮助。提前感谢您的帮助。
main.xml:
主文件：
<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:
获取URL.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);
}
}
});
}
}
回答by Sven Viking
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 WinMergeto see exactly what changed.
它很接近，但有些东西在错误的地方或丢失了。下面的代码有效——我试图进行最少的必要改动。您可以将两个版本都加载到WinMerge 之类的内容中，以准确查看发生了什么变化。
main.xml:
主文件：
<?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:
获取URL.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.pngfile in /res/drawable/and a startstring in /res/values/strings.xml, of course).
（当然，您还需要一个bgimage2.png文件 in/res/drawable/和一个start字符串 in /res/values/strings.xml）。
回答by Dave S
To simplify you could do
为了简化你可以做
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse("http://www.yahoo.com")); startActivity(intent);
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(" http://www.yahoo.com")); 开始活动（意图）；
