【发布时间】:2016-01-21 17:39:26
【问题描述】:
我正在尝试在 android 上获取特定联系人的名字、姓氏等,但似乎无法获得。几个小时以来我一直在努力解决这个问题,基本上if (nameCur.moveNext()) 总是错误的!此代码最初由 @perborin (How to get the first name and last name from Android contacts?) 提供。请帮忙!
附:我在 AndroidManifest 中添加了<uses-permission android:name="android.permission.READ_CONTACTS"/>,所以这不是问题。
// 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();
【问题讨论】: