Open Application in Android by label name, and unknowing package name. - java

First of all, I searching for my question many times in Google but I didn't find what I want.
My Question is: Is there any way to open application in Android by "label name" where on the other hand the package name is "knowing" ?
My approach that I used to do this is by creating two ArrayList and store the package name and label name in each one, then, I matching the label name with package name after I converted them to lowercase in order to be the same letters to see if there are matching, if matching , then I launch the application by getting its package name.
the problem that I faced it, the package name sometimes didn't have or matching the same name in label name, based on the application developer what he had write the package name.
for example: if I want to open gmail , based on my method I cannot because the package name is com.google.android.gm and label name is gmail , I apologized to the prolongation ,thanks in advanced.
My code:
private void getAllApps() {
final Intent mainIntent = new Intent(Intent.ACTION_MAIN, null);
mainIntent.addCategory(Intent.CATEGORY_LAUNCHER);
List<ResolveInfo> activities = getPackageManager()
.queryIntentActivities(mainIntent, 0);
for (ResolveInfo resolveInfo : activities) {
appsName.add(resolveInfo.loadLabel(getPackageManager()).toString()
.toLowerCase());
pkgsName.add(resolveInfo.activityInfo.packageName.toString());
}
}
private void openApplication(String appName) {
String packageName = null, appNameLowerCase = null, pkgNameLowerCase = null;
// matching the package name with label name
if (appsName.contains(appName)) {
appNameLowerCase = appName.trim().replace(" ", "");
for (int i = 0; i < pkgsName.size(); i++) {
pkgNameLowerCase = pkgsName.get(i).trim().toLowerCase();
if (pkgNameLowerCase
.matches("(.*)" + appNameLowerCase + "(.*)")) {
packageName = pkgsName.get(i);
break;
}
}
}
// to launch the application
Intent i;
PackageManager manager = getPackageManager();
try {
i = manager.getLaunchIntentForPackage(packageName);
if (i == null)
throw new PackageManager.NameNotFoundException();
i.addCategory(Intent.CATEGORY_LAUNCHER);
startActivity(i);
} catch (PackageManager.NameNotFoundException e) {
}
}

thanks for all replies, i solve it by #CommonsWare solution and it works correctly :
class Applications {
private String packageName;
private String labelName;
}
private void getAllApps() {
final Intent mainIntent = new Intent(Intent.ACTION_MAIN, null);
mainIntent.addCategory(Intent.CATEGORY_LAUNCHER);
List<ResolveInfo> activities = getPackageManager()
.queryIntentActivities(mainIntent, 0);
for (ResolveInfo resolveInfo : activities) {
Applications applications = new Applications();
applications.labelName = resolveInfo.loadLabel(getPackageManager())
.toString().toLowerCase();
applications.packageName = resolveInfo.activityInfo.packageName
.toString();
applicationsArrayList.add(applications);
}
}
private void openApplication(String appName) {
String packageName = null;
// matching the package name with label name
for (int i = 0; i < applicationsArrayList.size(); i++) {
if (applicationsArrayList.get(i).labelName.trim().equals(
appName.trim())) {
packageName = applicationsArrayList.get(i).packageName;
break;
}
}
// to launch the application
Intent i;
PackageManager manager = getPackageManager();
try {
i = manager.getLaunchIntentForPackage(packageName);
if (i == null)
throw new PackageManager.NameNotFoundException();
i.addCategory(Intent.CATEGORY_LAUNCHER);
startActivity(i);
} catch (PackageManager.NameNotFoundException e) {
}
}

Try this
public List<ApplicationInfo> getApplicationList(Context con){
PackageManager p = con.getPackageManager();
List<ApplicationInfo> info = p.getInstalledApplications(0);
String example = info.get(0).packageName.toString();
return info;
}

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