Question

我想在列表中显示所有本机联系人，并让用户将列表中的联系人（多个联系人）添加到我的应用程序数据库。如何点击这可以让任何人给我构思或分享一些代码。提前谢谢..

Answer 1

我在Android 2.1上使用了这段代码。它会删除任何拥有电话号码（由String SELECTION变量定义）并返回Person类型的列表。人是保存用户姓名和电话号码的对象。您必须实现Person对象才能使用此代码，但它可以完美地运行：

public List<Person> getContactList(){
        ArrayList<Person> contactList = new ArrayList<Person>();

        Uri contactUri = ContactsContract.Contacts.CONTENT_URI;
        String[] PROJECTION = new String[] {
                ContactsContract.Contacts._ID,
                ContactsContract.Contacts.DISPLAY_NAME,
                ContactsContract.Contacts.HAS_PHONE_NUMBER,
        };
        String SELECTION = ContactsContract.Contacts.HAS_PHONE_NUMBER + "='1'";
        Cursor contacts = getContentResolver().query(ContactsContract.Contacts.CONTENT_URI, PROJECTION, SELECTION, null, null);


        if (contacts.getCount() > 0)
        {
            while(contacts.moveToNext()) {
                Person aContact = new Person();
                int idFieldColumnIndex = 0;
                int nameFieldColumnIndex = 0;
                int numberFieldColumnIndex = 0;

                String contactId = contacts.getString(contacts.getColumnIndex(ContactsContract.Contacts._ID));

                nameFieldColumnIndex = contacts.getColumnIndex(PhoneLookup.DISPLAY_NAME);
                if (nameFieldColumnIndex > -1)
                {
                    aContact.setName(contacts.getString(nameFieldColumnIndex));
                }

                PROJECTION = new String[] {Phone.NUMBER};
                final Cursor phone = managedQuery(Phone.CONTENT_URI, PROJECTION, Data.CONTACT_ID + "=?", new String[]{String.valueOf(contactId)}, null);
                if(phone.moveToFirst()) {
                    while(!phone.isAfterLast())
                    {
                        numberFieldColumnIndex = phone.getColumnIndex(Phone.NUMBER);
                        if (numberFieldColumnIndex > -1)
                        {
                            aContact.setPhoneNum(phone.getString(numberFieldColumnIndex));
                            phone.moveToNext();
                            TelephonyManager mTelephonyMgr;
                            mTelephonyMgr = (TelephonyManager) getSystemService(Context.TELEPHONY_SERVICE);
                            if (!mTelephonyMgr.getLine1Number().contains(aContact.getPhoneNum()))
                            {
                                contactList.add(aContact);  
                            }
                        }
                    }
                }
                phone.close();
            }

            contacts.close();
        }

        return contactList;
    }

编辑：一个基本的人类：

public class Person {
    String myName = "";
    String myNumber = "";

    public String getName() {
        return myName;
    }

    public void setName(String name) {
        myName = name;
    }

    public String getPhoneNum() {
        return myNumber;
    }

    public void setPhoneNum(String number) {
        myNumber = number;
    }
}

Answer 2

此代码在android 4.2中完美运行。它的工作速度要快得多，因为您不会为每个联系人进行额外的查询

private static final String CONTACT_ID = ContactsContract.Contacts._ID;
private static final String DISPLAY_NAME = ContactsContract.Contacts.DISPLAY_NAME;
private static final String HAS_PHONE_NUMBER = ContactsContract.Contacts.HAS_PHONE_NUMBER;

private static final String PHONE_NUMBER = ContactsContract.CommonDataKinds.Phone.NUMBER;
private static final String PHONE_CONTACT_ID = ContactsContract.CommonDataKinds.Phone.CONTACT_ID;

public static ArrayList<Contact> getAll(Context context) {
    ContentResolver cr = context.getContentResolver();

    Cursor pCur = cr.query(
            ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
            new String[]{PHONE_NUMBER, PHONE_CONTACT_ID},
            null,
            null,
            null
    );
    if(pCur != null){
        if(pCur.getCount() > 0) {
            HashMap<Integer, ArrayList<String>> phones = new HashMap<>();
            while (pCur.moveToNext()) {
                Integer contactId = pCur.getInt(pCur.getColumnIndex(PHONE_CONTACT_ID));
                ArrayList<String> curPhones = new ArrayList<>();
                if (phones.containsKey(contactId)) {
                    curPhones = phones.get(contactId);
                }
                curPhones.add(pCur.getString(pCur.getColumnIndex(PHONE_CONTACT_ID)));
                phones.put(contactId, curPhones);
            }
            Cursor cur = cr.query(
                    ContactsContract.Contacts.CONTENT_URI,
                    new String[]{CONTACT_ID, DISPLAY_NAME, HAS_PHONE_NUMBER},
                    HAS_PHONE_NUMBER + " > 0",
                    null,
                    DISPLAY_NAME + " ASC");
            if (cur != null) {
                if (cur.getCount() > 0) {
                    ArrayList<Contact> contacts = new ArrayList<>();
                    while (cur.moveToNext()) {
                        int id = cur.getInt(cur.getColumnIndex(CONTACT_ID));
                        if(phones.containsKey(id)) {
                            Contact con = new Contact();
                            con.setMyId(id);
                            con.setName(cur.getString(cur.getColumnIndex(DISPLAY_NAME)));
                            con.setPhone(TextUtils.join(",", phones.get(id).toArray()));
                            contacts.add(con);
                        }
                    }
                    return contacts;
                }
                cur.close();
            }
        }
        pCur.close();
    }
    return null;
}

班级Contact与答案中的班级Person类似。

如何从Android中的本机电话簿获取联系人

2 个答案: