Android - How to find out the name of the application? (Not package name)

Question

In my manifest, I have:

<application android:name=".MyApp" android:icon="@drawable/ic_launcher_icon" android:label="@string/app_name" android:debuggable="true"> 

How to get a label item?

Note. My code runs inside someone else's code, so I do not have access to @ string / app_name

Answer 1

It is easier than other answers that do not require explicit resource specification or concern about exceptions with package names. It also works if you used the string directly and not a resource.

Just do:

 public static String getApplicationName(Context context) { ApplicationInfo applicationInfo = context.getApplicationInfo(); int stringId = applicationInfo.labelRes; return stringId == 0 ? applicationInfo.nonLocalizedLabel.toString() : context.getString(stringId); } 

Hope this helps.

Edit

In light of the comment from Snicolas, I modified the above so that it does not try to resolve the identifier if it is 0. Instead, it uses nonLocalizedLabel as a deferment. There is no need for a wrapper in try / catch.

Answer 2

Unless specified in the strings.xml / hardcoded file in AndroidManifest.xml for any reason, such as android: label = "MyApp"

 public String getAppLable(Context context) { PackageManager packageManager = context.getPackageManager(); ApplicationInfo applicationInfo = null; try { applicationInfo = packageManager.getApplicationInfo(context.getApplicationInfo().packageName, 0); } catch (final NameNotFoundException e) { } return (String) (applicationInfo != null ? packageManager.getApplicationLabel(applicationInfo) : "Unknown"); } 

Or, if you know the String resource identifier, you can directly get it through

 getString(R.string.appNameID); 

Answer 3

 public static String getApplicationName(Context context) { return context.getApplicationInfo().loadLabel(context.getPackageManager()); } 

Answer 4

From any use of context:

 getApplicationInfo().loadLabel(getPackageManager()).toString(); 

Answer 5

If you know the package name, use the following snippet

 ApplicationInfo ai; try { ai = pm.getApplicationInfo(packageName, 0); } catch (final NameNotFoundException e) { ai = null; } final String applicationName = (String) (ai != null ? pm.getApplicationLabel(ai) : "(unknown)"); 

Answer 6

If you only need the name of the application and not the name of the package, just write this code.

  String app_name = packageInfo.applicationInfo.loadLabel(getPackageManager()).toString(); 

Answer 7

Get application name using RunningAppProcessInfo as:

 ActivityManager am = (ActivityManager)this.getSystemService(ACTIVITY_SERVICE); List l = am.getRunningAppProcesses(); Iterator i = l.iterator(); PackageManager pm = this.getPackageManager(); while(i.hasNext()) { ActivityManager.RunningAppProcessInfo info = (ActivityManager.RunningAppProcessInfo)(i.next()); try { CharSequence c = pm.getApplicationLabel(pm.getApplicationInfo(info.processName, PackageManager.GET_META_DATA)); Log.w("LABEL", c.toString()); }catch(Exception e) { //Name Not FOund Exception } } 

Answer 8

Ok guys, another smooth option is

Application.Context.ApplicationInfo.NonLocalizedLabel

Checked for Android hard coded label in application element.

<application android:label="Big App"></application>

Link: http://developer.android.com/reference/android/content/pm/PackageItemInfo.html#nonLocalizedLabel

Answer 9

In Kotlin, use the following codes to get the Application Name:

  // Get App Name var appName: String = "" val applicationInfo = this.getApplicationInfo() val stringId = applicationInfo.labelRes if (stringId == 0) { appName = applicationInfo.nonLocalizedLabel.toString() } else { appName = this.getString(stringId) } 

Answer 10

Have you tried using the PackageManager#getActivityInfo() method? There will be a field that should contain the name of the application.

See the answer to a very similar question here .

Answer 11

The original comment added to NonLocalizedLabel directs us to:

 return context.getPackageManager().getApplicationLabelFormatted(context.getApplicationInfo());