如何从游标中删除唯一值

Question

我正在开发一个应仅向用户显示重复电话联系人，以便他们可以删除重复联系人的应用程序。Duplicates

到目前为止，我无法使用以下代码显示所有联系人：

主要活动：

private void showContacts() {
Cursor cursor = ContactHelper.getContactCursor(getContentResolver(),"");
String[] fields = new String[]{ContactsContract.Data.DISPLAY_NAME};

adapter =new SimpleCursorAdapter(this,android.R.layout.simple_list_item_multiple_choice,cursor,fields,new  int[]{android.R.id.text1});
listView.setAdapter(adapter);
adapter.notifyDataSetChanged(); }

ContactHelper.GetContactsCursor函数：

public static Cursor getContactCursor(ContentResolver contactHelper,
                                      String startsWith) {

    String[] projection = { ContactsContract.CommonDataKinds.Phone._ID,
            ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME,
            ContactsContract.CommonDataKinds.Phone.NUMBER };
    Cursor cur = null;

    try {
        if (startsWith != null && !startsWith.equals("")) {
            cur = contactHelper.query(
                    ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                    projection,
                    ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME
                            + " like \"" + startsWith + "%\"", null,
                    ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME
                            + " ASC");
        } else {
            cur = contactHelper.query(
                    ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                    projection, null, null,
                    ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME
                            + " ASC");
        }
        cur.moveToFirst();
    } catch (Exception e) {
        e.printStackTrace();
    }

    return cur;
}

private static long getContactID(ContentResolver contactHelper,
                                 String number) {
    Uri contactUri = Uri.withAppendedPath(ContactsContract.PhoneLookup.CONTENT_FILTER_URI,
            Uri.encode(number));

    String[] projection = { ContactsContract.PhoneLookup._ID };
    Cursor cursor = null;

    try {
        cursor = contactHelper.query(contactUri, projection, null, null,
                null);

        if (cursor.moveToFirst()) {
            int personID = cursor.getColumnIndex(ContactsContract.PhoneLookup._ID);
            return cursor.getLong(personID);
        }

        return -1;
    } catch (Exception e) {
        e.printStackTrace();
    } finally {
        if (cursor != null) {
            cursor.close();
            cursor = null;
        }
    }

    return -1; }

如何从游标中过滤唯一值？并且仅在可能的情况下仅根据电话号码保留重复的联系人？

Answer 1

也许这会起作用

    try {
        if (startsWith != null && !startsWith.equals("")) {
            cur = contactHelper.query(
                    ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                    projection,
                    ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME
                            + " like \"" + startsWith + "%\""
                            + " AND "
                            + ContactsContract.CommonDataKinds.Phone.NUMBER
                            + " IN (SELECT " + ContactsContract.CommonDataKinds.Phone.NUMBER
                            + " FROM view_data GROUP BY "
                            + ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME + ")",
                    null,
                    ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME
                            + " ASC");
        } else {
            cur = contactHelper.query(
                    ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                    projection,
                    ContactsContract.CommonDataKinds.Phone.NUMBER
                            + " IN (SELECT " + ContactsContract.CommonDataKinds.Phone.NUMBER
                            + " FROM view_data GROUP BY "
                            + ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME + ")",
                    null,
                    ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME
                            + " ASC");
        }
        cur.moveToFirst();
    } catch (Exception e) {
        e.printStackTrace();
    }