Question

try {
            Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse("com.xxx.applicationname"));
            startActivity(intent);
        } catch(Exception e) {
            startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("http://www.facebook.com")));
        }

它总是进入捕获范围。我不知道我做错了什么

Answer 1

试试这个：

protected void launchApp(String packageName) {
    Intent mIntent = getPackageManager().getLaunchIntentForPackage(packageName);

    if (mIntent != null) {
        try {
            startActivity(mIntent);
        } catch (ActivityNotFoundException err) {
            Toast t = Toast.makeText(getApplicationContext(),
                    "Not FOund", Toast.LENGTH_SHORT);
            t.show();
        }
    }
}

Answer 2

这应该有所帮助：

ProcessStartInfo psi = new ProcessStartInfo();            
psi.FileName = "netsh";            
psi.UseShellExecute = false;
psi.RedirectStandardError = true;
psi.RedirectStandardOutput = true;
psi.Arguments = "SOME_ARGUMENTS";

Process proc = Process.Start(psi);                
proc.WaitForExit();
string errorOutput = proc.StandardError.ReadToEnd();
string standardOutput = proc.StandardOutput.ReadToEnd();
if (proc.ExitCode != 0)
    throw new Exception("netsh exit code: " + proc.ExitCode.ToString() + " " + (!string.IsNullOrEmpty(errorOutput) ? " " + errorOutput : "") + " " + (!string.IsNullOrEmpty(standardOutput) ? " " + standardOutput : ""));

Answer 3

您可以尝试：

Intent intent = new Intent();
intent.setPackage("package**name");
intent.setAction(Intent.ACTION_MAIN);
intent.addCategory(Intent.CATEGORY_LAUNCHER);
startActivity(intent);

并且，您应该在调用的应用程序清单中添加到此行：

<intent-filter>
...
<category android:name="android.intent.category.CATEGORY_DEFAULT" />
</intent-filter>

如何使用另一个有意图的应用程序打开一个Android应用程序？

3 个答案: