如何从内置web浏览器而不是应用程序中的代码打开URL?

我试过了:

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();
}

但我有个例外:

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

当前回答

只需使用简短的一个,即可在浏览器中打开您的Url:

Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("YourUrlHere"));
startActivity(browserIntent);

其他回答

//OnClick侦听器

  @Override
      public void onClick(View v) {
        String webUrl = news.getNewsURL();
        if(webUrl!="")
        Utils.intentWebURL(mContext, webUrl);
      }

//你的Util方法

public static void intentWebURL(Context context, String url) {
        if (!url.startsWith("http://") && !url.startsWith("https://")) {
            url = "http://" + url;
        }
        boolean flag = isURL(url);
        if (flag) {
            Intent browserIntent = new Intent(Intent.ACTION_VIEW,
                    Uri.parse(url));
            context.startActivity(browserIntent);
        }

    }

其他选项使用Webview在同一应用程序中加载Url

webView = (WebView) findViewById(R.id.webView1);
webView.getSettings().setJavaScriptEnabled(true);
webView.loadUrl("http://www.google.com");

在Android 11中打开URL链接的新的更好方法。

  try {
        val intent = Intent(ACTION_VIEW, Uri.parse(url)).apply {
            // The URL should either launch directly in a non-browser app
            // (if it’s the default), or in the disambiguation dialog
            addCategory(CATEGORY_BROWSABLE)
            flags = FLAG_ACTIVITY_NEW_TASK or FLAG_ACTIVITY_REQUIRE_NON_BROWSER or
                    FLAG_ACTIVITY_REQUIRE_DEFAULT
        }
        startActivity(intent)
    } catch (e: ActivityNotFoundException) {
        // Only browser apps are available, or a browser is the default app for this intent
        // This code executes in one of the following cases:
        // 1. Only browser apps can handle the intent.
        // 2. The user has set a browser app as the default app.
        // 3. The user hasn't set any app as the default for handling this URL.
        openInCustomTabs(url)
    }

参考文献:

https://medium.com/androiddevelopers/package-visibility-in-android-11-cc857f221cd9和https://developer.android.com/training/package-visibility/use-cases#avoid-a-消除歧义对话框

实现这一点的常见方法是使用以下代码:

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

可以更改为短代码版本。。。

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);

最短!:

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

我认为这是最好的

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

将以下代码放入全局类

    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);
    }