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我想獲得特定名稱,姓氏等在android上的特定聯繫人,但似乎無法做到。幾個小時以來,我一直在抨擊我的頭,基本上if (nameCur.moveNext())
總是錯誤的!此代碼最初由@perborin(How to get the first name and last name from Android contacts?)提供。請幫忙!獲取具體的聯繫信息
P.S.我在AndroidManifest中添加了<uses-permission android:name="android.permission.READ_CONTACTS"/>
,所以這不是問題。
// A contact ID is fetched from ContactList
Uri resultUri = data.getData();
Cursor cont = getContentResolver().query(resultUri, null, null, null, null);
if (!cont.moveToNext()) {
Toast.makeText(this, "Cursor contains no data", Toast.LENGTH_LONG).show();
return;
}
int columnIndexForId = cont.getColumnIndex(ContactsContract.Contacts._ID);
String contactId = cont.getString(columnIndexForId);
// Fetch contact name with a specific ID
String whereName = ContactsContract.Data.MIMETYPE + " = ? AND " + ContactsContract.CommonDataKinds.StructuredName.CONTACT_ID + " = " + contactId;
String[] whereNameParams = new String[] { ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE };
Cursor nameCur = getContentResolver().query(ContactsContract.Data.CONTENT_URI, null, whereName, whereNameParams, ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME);
while (nameCur.moveToNext()) {
String given = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME));
String family = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME));
String display = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME));
Toast.makeText(this, "Name: " + given + " Family: " + family + " Displayname: " + display, Toast.LENGTH_LONG).show();
}
nameCur.close();
cont.close();