How can I open a URL in Android's web browser

2018-12-31 02:46发布

How to open an URL from code in the built-in web browser rather than within my application?

I tried this:

try {
    Intent myIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(download_link));
    startActivity(myIntent);
} catch (ActivityNotFoundException e) {
    Toast.makeText(this, "No application can handle this request."
        + " Please install a webbrowser",  Toast.LENGTH_LONG).show();
    e.printStackTrace();
}

but I got an Exception:

No activity found to handle Intent{action=android.intent.action.VIEW data =www.google.com

29条回答
宁负流年不负卿
2楼-- · 2018-12-31 03:07

Basic Introduction:

https:// is using that one into the "code" so that no one in between can read them. This keeps your information safe from hackers.

http:// is using just sharing purpose, it's not secured.

About Your Problem:
XML designing:

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:tools="http://schemas.android.com/tools"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:orientation="vertical"
    tools:context="com.example.sridhar.sharedpreferencesstackoverflow.MainActivity">
   <LinearLayout
       android:orientation="horizontal"
       android:background="#228b22"
       android:layout_weight="1"
       android:layout_width="match_parent"
       android:layout_height="0dp">
      <Button
          android:id="@+id/normal_search"
          android:text="secure Search"
          android:onClick="secure"
          android:layout_weight="1"
          android:layout_width="0dp"
          android:layout_height="wrap_content" />
      <Button
          android:id="@+id/secure_search"
          android:text="Normal Search"
          android:onClick="normal"
          android:layout_weight="1"
          android:layout_width="0dp"
          android:layout_height="wrap_content" />
   </LinearLayout>

   <LinearLayout
       android:layout_weight="9"
       android:id="@+id/button_container"
       android:layout_width="match_parent"
       android:layout_height="0dp"
       android:orientation="horizontal">

      <WebView
          android:id="@+id/webView1"
          android:layout_width="match_parent"
          android:layout_height="match_parent" />

   </LinearLayout>
</LinearLayout>

Activity Designing:

public class MainActivity extends Activity {
    //securely open the browser
    public String Url_secure="https://www.stackoverflow.com";
    //normal purpouse
    public String Url_normal="https://www.stackoverflow.com";

    WebView webView;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        webView=(WebView)findViewById(R.id.webView1);

    }
    public void secure(View view){
        webView.setWebViewClient(new SecureSearch());
        webView.getSettings().setLoadsImagesAutomatically(true);
        webView.getSettings().setJavaScriptEnabled(true);
        webView.setScrollBarStyle(View.SCROLLBARS_INSIDE_OVERLAY);
        webView.loadUrl(Url_secure);
    }
    public void normal(View view){
        webView.setWebViewClient(new NormalSearch());
        webView.getSettings().setLoadsImagesAutomatically(true);
        webView.getSettings().setJavaScriptEnabled(true);
        webView.setScrollBarStyle(View.SCROLLBARS_INSIDE_OVERLAY);
        webView.loadUrl(Url_normal);

    }
    public class SecureSearch extends WebViewClient{
        @Override
        public boolean shouldOverrideUrlLoading(WebView view, String Url_secure) {
            view.loadUrl(Url_secure);
            return true;
        }
    }
    public class NormalSearch extends WebViewClient{
        @Override
        public boolean shouldOverrideUrlLoading(WebView view, String Url_normal) {
            view.loadUrl(Url_normal);
            return true;
        }
    }
}

Android Manifest.Xml permissions:

<uses-permission android:name="android.permission.INTERNET"/>

You face Problems when implementing this:

  1. getting The Manifest permissions
  2. excess space's between url
  3. Check your url's correct or not
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梦该遗忘
3楼-- · 2018-12-31 03:08

Try this:

Uri uri = Uri.parse("https://www.google.com");
startActivity(new Intent(Intent.ACTION_VIEW, uri));

or if you want then web browser open in your activity then do like this:

WebView webView = (WebView) findViewById(R.id.webView1);
WebSettings settings = webview.getSettings();
settings.setJavaScriptEnabled(true);
webView.loadUrl(URL);

and if you want to use zoom control in your browser then you can use:

settings.setSupportZoom(true);
settings.setBuiltInZoomControls(true);
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明月照影归
4楼-- · 2018-12-31 03:09

I think this is the best

openBrowser(context, "http://www.google.com")

Put below code into global class

    public static void openBrowser(Context context, String url) {

        if (!url.startsWith("http://") && !url.startsWith("https://"))
            url = "http://" + url;

        Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
        context.startActivity(browserIntent);
    }
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爱死公子算了
5楼-- · 2018-12-31 03:10

Try this:

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

That works fine for me.

As for the missing "http://" I'd just do something like this:

if (!url.startsWith("http://") && !url.startsWith("https://"))
   url = "http://" + url;

I would also probably pre-populate your EditText that the user is typing a URL in with "http://".

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倾城一夜雪
6楼-- · 2018-12-31 03:13

a common way to achieve this is with the next code:

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

that could be changed to a short code version ...

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

or :

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

the shortest! :

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

happy coding!

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姐姐魅力值爆表
7楼-- · 2018-12-31 03:13

A short code version...

 if (!strUrl.startsWith("http://") && !strUrl.startsWith("https://")){
     strUrl= "http://" + strUrl;
 }


 startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse(strUrl)));
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