如何从安卓系统中选择独特的联系人

Question

我想选择 单一联系人 从安卓系统中，只有那些有电话号码的联系人。

ContentResolver cr = getContentResolver();
        Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI, null,
                null, null, ContactsContract.Contacts.DISPLAY_NAME);
        // Find the ListView resource.
        mainListView = (ListView) findViewById(R.id.mainListView);

        // When item is tapped, toggle checked properties of CheckBox and
        // Planet.
        mainListView
                .setOnItemClickListener(new AdapterView.OnItemClickListener()
                {
                    public void onItemClick(AdapterView<?> parent, View item,
                            int position, long id)
                    {
                        ContactsList planet = listAdapter.getItem(position);
                        planet.toggleChecked();
                        PlanetViewHolder viewHolder = (PlanetViewHolder) item
                                .getTag();
                        viewHolder.getCheckBox().setChecked(planet.isChecked());
                    }
                });

        // Create and populate planets.
        planets = (ContactsList[]) getLastNonConfigurationInstance();
        // planets = new Planet[10];
        // planets.Add("asdf");
        ArrayList<ContactsList> planetList = new ArrayList<ContactsList>();
        String phoneNumber = null;
        String phoneType = null;

        count = cur.getCount();
        contacts = new ContactsList[count];

        if (planets == null)
        {
            if (cur.getCount() > 0)
            {
                planets = new ContactsList[cur.getCount()];
                int i = 0;
                //
                while (cur.moveToNext())
                {
                    String id = cur.getString(cur
                            .getColumnIndex(ContactsContract.Contacts._ID));
                    String name = cur
                            .getString(cur
                                    .getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
                    if (Integer
                            .parseInt(cur.getString(cur
                                    .getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0)
                    {
                        // Query phone here. Covered next
                        Cursor pCur = cr
                                .query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                                        null,
                                        ContactsContract.CommonDataKinds.Phone.CONTACT_ID
                                                + " = ?", new String[]
                                        { id }, null);

                        // WHILE WE HAVE CURSOR GET THE PHONE NUMERS
                        while (pCur.moveToNext())
                        {
                            // Do something with phones
                            phoneNumber = pCur
                                    .getString(pCur
                                            .getColumnIndex(ContactsContract.CommonDataKinds.Phone.DATA));

                            phoneType = pCur
                                    .getString(pCur
                                            .getColumnIndex(ContactsContract.CommonDataKinds.Phone.TYPE));

                            Log.i("Pratik", name + "'s PHONE :" + phoneNumber);
                            Log.i("Pratik", "PHONE TYPE :" + phoneType);
                        }
                        pCur.close();
                    }

                    planets = new ContactsList[]
                    { new ContactsList(name, phoneNumber) };

                    contacts[i] = planets[0];
                    planetList.addAll(Arrays.asList(planets));

                    i++;
                }
            }

这段代码检索所有的联系人，并将其放入一个列表中，但我想选择唯一的联系人，并且只选择有电话号码的联系人。

Answer 1

我想你的意思是你得到了一些联系人的重复记录。所以你必须为你的查询添加条件。最重要的部分是联系人必须是 显性群体 和 有电话.

String selection = ContactsContract.Contacts.IN_VISIBLE_GROUP + " = '"
                + ("1") + "'";
        String sortOrder = ContactsContract.Contacts.DISPLAY_NAME
                + " COLLATE LOCALIZED ASC";
cur = context.getContentResolver().query(
                ContactsContract.Contacts.CONTENT_URI, projection, selection
                        + " AND " + ContactsContract.Contacts.HAS_PHONE_NUMBER
                        + "=1", null, sortOrder);// this query only return contacts which had phone number and not duplicated

---

2005-2020年更新

  suspend fun fetchContacts(): ArrayList<FriendItem> {
        val list = ArrayList<FriendItem>()
        val uri: Uri = ContactsContract.CommonDataKinds.Phone.CONTENT_URI
        val selection = ContactsContract.Contacts.HAS_PHONE_NUMBER
        val cursor: Cursor? = context.contentResolver.query(
            uri,
            arrayOf(
                ContactsContract.CommonDataKinds.Phone.NUMBER,
                ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME,
                ContactsContract.CommonDataKinds.Phone._ID,
                ContactsContract.Contacts._ID
            ),
            selection,
            null,
            ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME + " ASC"
        )

        cursor?.let {
            val nameIndex = cursor.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME)
            val phoneIndex = cursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER)
            while (cursor.moveToNext()) {
                val info = FriendItem(
                    friendName = cursor.getString(nameIndex),
                    friendPhoneNumber = cursor.getString(phoneIndex)
                )
                list.add(info)
            }
            cursor.close()
        }
        return list
    }

Answer 2

这对我来说是工作，以获得联系电话号码。在这里，我们查询数据表，并使用CONTACT_ID。联系供应商文件

    @Override
    public Loader<Cursor> onCreateLoader(int id, Bundle args) {

final String ORDER_BY = ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME_PRIMARY + " ASC";

    final String[] PROJECTION = {
            ContactsContract.CommonDataKinds.Phone.CONTACT_ID,
            ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME_PRIMARY,
            ContactsContract.CommonDataKinds.Phone.NUMBER
    };

return new CursorLoader(
                context,
                ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                PROJECTION,
                null,
                null,
                ORDER_BY
        );
}

Answer 3

获取电话号码和联系人姓名的简单方法

// set as global
Set<string> phonenumbersList = new HashSet<string>();

            Cursor phones = getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null,null,null, null);
            while (phones.moveToNext())
            {
            String name=phones.getString(phones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME));
            String phoneNumber = phones.getString(phones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));

            //contact has name number and phonenumber does not exists in list
            if ( phoneNumber != null && name != null && !phonenumbersList.contains(phoneNumber)){ 
                planets = new ContactsList[]{ new ContactsList(name, phoneNumber) };

                phonenumbersList.add(phoneNumber);
                planetList.addAll(Arrays.asList(planets));
                planetList.Add(phoneNumber, name);
            }
            }
            phones.close();