我想从Android中选择唯一的联系人,只有那些有电话号码的联系人。我正在使用此代码
ContentResolver cr = getContentResolver();
Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI, null,
null, null, ContactsContract.Contacts.DISPLAY_NAME);
// Find the ListView resource.
mainListView = (ListView) findViewById(R.id.mainListView);
// When item is tapped, toggle checked properties of CheckBox and
// Planet.
mainListView
.setOnItemClickListener(new AdapterView.OnItemClickListener()
{
public void onItemClick(AdapterView<?> parent, View item,
int position, long id)
{
ContactsList planet = listAdapter.getItem(position);
planet.toggleChecked();
PlanetViewHolder viewHolder = (PlanetViewHolder) item
.getTag();
viewHolder.getCheckBox().setChecked(planet.isChecked());
}
});
// Create and populate planets.
planets = (ContactsList[]) getLastNonConfigurationInstance();
// planets = new Planet[10];
// planets.Add("asdf");
ArrayList<ContactsList> planetList = new ArrayList<ContactsList>();
String phoneNumber = null;
String phoneType = null;
count = cur.getCount();
contacts = new ContactsList[count];
if (planets == null)
{
if (cur.getCount() > 0)
{
planets = new ContactsList[cur.getCount()];
int i = 0;
//
while (cur.moveToNext())
{
String id = cur.getString(cur
.getColumnIndex(ContactsContract.Contacts._ID));
String name = cur
.getString(cur
.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
if (Integer
.parseInt(cur.getString(cur
.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0)
{
// Query phone here. Covered next
Cursor pCur = cr
.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
null,
ContactsContract.CommonDataKinds.Phone.CONTACT_ID
+ " = ?", new String[]
{ id }, null);
// WHILE WE HAVE CURSOR GET THE PHONE NUMERS
while (pCur.moveToNext())
{
// Do something with phones
phoneNumber = pCur
.getString(pCur
.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DATA));
phoneType = pCur
.getString(pCur
.getColumnIndex(ContactsContract.CommonDataKinds.Phone.TYPE));
Log.i("Pratik", name + "'s PHONE :" + phoneNumber);
Log.i("Pratik", "PHONE TYPE :" + phoneType);
}
pCur.close();
}
planets = new ContactsList[]
{ new ContactsList(name, phoneNumber) };
contacts[i] = planets[0];
planetList.addAll(Arrays.asList(planets));
i++;
}
}
此代码检索所有联系人并将其放入列表中。但我想要独特的联系,只有那些没有电话的人。我怎样才能做到这一点??是否有任何方法可以在查询中传递一些参数来仅选择唯一的联系人???
答案 0 :(得分:10)
我认为你的意思是你有一些联系人的重复记录。因此,您必须为查询添加条件。关键部分是联系人必须在可见组和有电话号码。
String selection = ContactsContract.Contacts.IN_VISIBLE_GROUP + " = '"
+ ("1") + "'";
String sortOrder = ContactsContract.Contacts.DISPLAY_NAME
+ " COLLATE LOCALIZED ASC";
cur = context.getContentResolver().query(
ContactsContract.Contacts.CONTENT_URI, projection, selection
+ " AND " + ContactsContract.Contacts.HAS_PHONE_NUMBER
+ "=1", null, sortOrder);// this query only return contacts which had phone number and not duplicated
答案 1 :(得分:1)
这对我来说可以联系电话号码。在这里,我们查询数据表,并使用CONTACT_ID contact provider documentation
@Override
public Loader<Cursor> onCreateLoader(int id, Bundle args) {
final String ORDER_BY = ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME_PRIMARY + " ASC";
final String[] PROJECTION = {
ContactsContract.CommonDataKinds.Phone.CONTACT_ID,
ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME_PRIMARY,
ContactsContract.CommonDataKinds.Phone.NUMBER
};
return new CursorLoader(
context,
ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
PROJECTION,
null,
null,
ORDER_BY
);
}
答案 2 :(得分:0)
获取电话号码和联系人姓名的简便方法
// set as global
Set<string> phonenumbersList = new HashSet<string>();
Cursor phones = getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null,null,null, null);
while (phones.moveToNext())
{
String name=phones.getString(phones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME));
String phoneNumber = phones.getString(phones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
//contact has name number and phonenumber does not exists in list
if ( phoneNumber != null && name != null && !phonenumbersList.contains(phoneNumber)){
planets = new ContactsList[]{ new ContactsList(name, phoneNumber) };
phonenumbersList.add(phoneNumber);
planetList.addAll(Arrays.asList(planets));
planetList.Add(phoneNumber, name);
}
}
phones.close();