Question

您好我有以下代码。我只需要将电话号码输入自动填充文本框......如何操作

  @Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    mPeopleList = new ArrayList<Map<String, String>>();
    PopulatePeopleList();
    mTxtPhoneNo = (AutoCompleteTextView) findViewById(R.id.mmWhoNo);

    mAdapter = new SimpleAdapter(this, mPeopleList, R.layout.custcontview,
            new String[] {"Name","Phone","Type"}, new int[] {R.id.ccontName,R.id.ccontNo,R.id.ccontType});

    mTxtPhoneNo.setAdapter(mAdapter);

}

public void PopulatePeopleList() {

    mPeopleList.clear();

    Cursor people = getContentResolver().query(
            ContactsContract.Contacts.CONTENT_URI, null, null, null, null);

    while (people.moveToNext()) {
        String contactName = people.getString(people
                .getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));

        String contactId = people.getString(people
                .getColumnIndex(ContactsContract.Contacts._ID));
        String hasPhone = people
                .getString(people
                        .getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER));

        if ((Integer.parseInt(hasPhone) > 0)) {

            // You know have the number so now query it like this
            Cursor phones = getContentResolver().query(
                    ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                    null,
                    ContactsContract.CommonDataKinds.Phone.CONTACT_ID
                            + " = " + contactId, null, null);
            while (phones.moveToNext()) {

                // store numbers and display a dialog letting the user
                // select which.
                String phoneNumber = phones
                        .getString(phones
                                .getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));

                String numberType = phones
                        .getString(phones
                                .getColumnIndex(ContactsContract.CommonDataKinds.Phone.TYPE));

                Map<String, String> NamePhoneType = new HashMap<String, String>();

                NamePhoneType.put("Name", contactName);
                NamePhoneType.put("Phone", phoneNumber);

                if (numberType.equals("0"))
                    NamePhoneType.put("Type", "Work");
                else if (numberType.equals("1"))
                    NamePhoneType.put("Type", "Home");
                else if (numberType.equals("2"))
                    NamePhoneType.put("Type", "Mobile");
                else
                    NamePhoneType.put("Type", "Other");

                // Then add this map to the list.
                mPeopleList.add(NamePhoneType);
            }
            phones.close();
        }
    }
    people.close();

    startManagingCursor(people);
}

}

以此格式返回类似Name = xyz.Type = Mobile，Number = 1234 .... 我只需要使用该号码并使用该号码进一步为我的程序添加代码以获取号码。

Answer 1

仅使用此行

String phoneNumber = phones
                        .getString(phones
                                .getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));


            NamePhoneType.put("Phone", phoneNumber);

您要输入所有数据，包括姓名，类型..只能拨打电话并查看

Answer 2

* 您可以尝试此代码*

<AutoCompleteTextView
android:id="@+id/AfromACT"
android:layout_width="150px"
android:layout_height="40px"
android:layout_marginLeft="10px"
android:layout_marginTop="15px"
android:background="#FFFFFF"
android:textColor="#000000" 
android:singleLine="true"
android:inputType="textVisiblePassword"
android:textCursorDrawable="@null"
android:paddingLeft="5px"

机器人：的inputType = “手机” android：textSize =“17px”/＆gt;

仅选择电话号码

2 个答案: