我想从
Android只选择具有电话号码的联系人的唯一联系人.我正在使用此代码
ContentResolver cr = getContentResolver();
Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI,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,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？？有什么方法可以在查询中传递一些参数来只选择唯一的联系人???
解决方法
我想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中并且有电话号码.
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",sortOrder);// this query only return contacts which had phone number and not duplicated
