3
\$\begingroup\$

This phonebook application has a menu, which sorts by name or category. I do it like this in MainActivity:

Bundle bundle = new Bundle();
Fragment lvFragSortBy = new ListViewFragment();

switch (item.getItemId()) {

    case R.id.menuSortOrderName:
        item.setChecked(true);

        bundle.putString("sortBy", "name");

        lvFragSortBy.setArguments(bundle);
        getFragmentManager()
        .beginTransaction()
        .replace(R.id.fragContainer, lvFragSortBy)
        .commit();
        return true;

    case R.id.menuSortOrderCategory:
        item.setChecked(true);

        bundle.putString("sortBy", "category");
        lvFragSortBy.setArguments(bundle);
        getFragmentManager()
        .beginTransaction()
        .replace(R.id.fragContainer, lvFragSortBy)
        .commit();

        return true;

In ListViewFragment:

Bundle sortBundle = getArguments();
if (sortBundle != null) {
    orderBy = sortBundle.getString("sortBy");
}

And then:

protected ArrayList<Contact> doInBackground(Void... params) {
    try {
        db = new DbWorker(ctx);
        contacts = (ArrayList<Contact>) db.selectAllContacts(orderBy);
    } catch (Exception e) {
        Log.e(TAG, "Exception in AsyncTask.doInBackground");
    }
    return contacts;
}

This code works fine but I would like to know if it can be done better.

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

You should use newInstance() pattern, described in this link

The profit of using this pattern is, developer cannot instantiate Fragment without providing arguments. There will be no chances that you will get default value when you try to get that value. One more thing is, whatever key is required for a fragment should be defined in public static final String so that no chances for string mismatch in key.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.