Android联系人在单个数据库查询中显示名称和电话号码？

Question

我正在尝试从本机数据库中获取其显示名称和电话号码(任意或所有)的联系人列表.有许多方法可以通过对手机数据库的多个查询来获取此信息,但这会带来相当大的开销.

这是我一直在研究的查询,但结果却是

Uri uri                = ContactsContract.Contacts.CONTENT_URI;
String[] projection    = new String[] { ContactsContract.Contacts._ID,
                                        ContactsContract.Contacts.DISPLAY_NAME,
                                        ContactsContract.CommonDataKinds.Phone.NUMBER};
String selection       = ContactsContract.Contacts.HAS_PHONE_NUMBER + " = '1'";
String[] selectionArgs = null;
String sortOrder       = ContactsContract.Contacts.DISPLAY_NAME + " COLLATE LOCALIZED ASC";

Cursor people          = getContentResolver().query(uri, projection, selection, selectionArgs, sortOrder);

int index_id    = people.getColumnIndex(ContactsContract.Contacts._ID);
int indexName   = people.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME);
int indexNumber = people.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER);

people.moveToFirst();
do {
    String _id    = people.getString(index_id);
    String name   = people.getString(indexName);
    String number = people.getString(indexNumber);
    // Do work...
} while (people.moveToNext());

这是由此产生的错误.

E/AndroidRuntime(21549): Caused by: java.lang.IllegalArgumentException: Invalid column data1
E/AndroidRuntime(21549):    at android.database.DatabaseUtils.readExceptionFromParcel(DatabaseUtils.java:144)
E/AndroidRuntime(21549):    at android.database.DatabaseUtils.readExceptionFromParcel(DatabaseUtils.java:114)
E/AndroidRuntime(21549):    at android.content.ContentProviderProxy.bulkQueryInternal(ContentProviderNative.java:372)
E/AndroidRuntime(21549):    at android.content.ContentProviderProxy.query(ContentProviderNative.java:408)
E/AndroidRuntime(21549):    at android.content.ContentResolver.query(ContentResolver.java:264)

思考？我相信可能需要连接才能在单个查询中获取所有列.

Answer 1

试试这段代码:

Uri uri = ContactsContract.CommonDataKinds.Phone.CONTENT_URI;
String[] projection    = new String[] {ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME,
                ContactsContract.CommonDataKinds.Phone.NUMBER};

Cursor people = getContentResolver().query(uri, projection, null, null, null);

int indexName = people.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME);
int indexNumber = people.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER);

if(people.moveToFirst()) {
    do {
        String name   = people.getString(indexName);
        String number = people.getString(indexNumber);
        // Do work...
    } while (people.moveToNext());
}

Answer 2

联系人API非常棘手,并且有几个隐式连接.

如果您能负担得起时间,请阅读ContactContract和ContactsProvider2.

你想要什么？这些表链接如下:

• 联系1 - *原始联系人
• 原始联系人1 - *电话号码(数据表)

API的工作方式如下:您选择最底部的元素(电话号码)和隐式连接到最顶层的元素(联系人).

您想使用PHONE URI案例(ContactsProvider2/line 4377).这应该选择所有电话号码并加入联系人.

将PHONE uri与一些UI魔法(用于分组)相结合,请求DISPLAY_NAME和PHONE号码(DATA1？),您应该能够解决问题.