重复的联系人不显示在listiview

Question

我在写一个应用程序，以显示在ListView联系人号码，但它给重复的联系人条目我想只显示一个联系人单独

private void getContactList() { 
    ContentResolver cr = getContentResolver(); 
    Cursor cur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null, null, null, ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME + " ASC"); 

    if ((cur != null ? cur.getCount() : 0) > 0) { 
     while (cur != null && cur.moveToNext()) { 
      String id = cur.getString(
        cur.getColumnIndex(ContactsContract.Contacts._ID)); 
      String name = cur.getString(cur.getColumnIndex(
        ContactsContract.Contacts.DISPLAY_NAME)); 

      if (cur.getInt(cur.getColumnIndex(
        ContactsContract.Contacts.HAS_PHONE_NUMBER)) > 0) { 
       Cursor pCur = cr.query(
         ContactsContract.CommonDataKinds.Phone.CONTENT_URI, 
         null, 
         ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?", 
         new String[]{id}, null); 
       while (pCur.moveToNext()) { 
        String phoneNo = pCur.getString(pCur.getColumnIndex(
          ContactsContract.CommonDataKinds.Phone.NUMBER)); 
        ContactsPlacer obj = new ContactsPlacer(); 
        obj.setContactname(name); 
        obj.setContactnumber(phoneNo); 
        names.add(obj); 

       } 
       pCur.close(); 
      } 
     } 
    } 
    if(cur!=null){ 
     cur.close(); 
    } 




} 

帮我一下的是，重复的联系人列表视图中不显示

Answer 1

boolean checkexsistance(ContactsPlacer obj) 
{ 
boolean test= true; 
for(int i = 0 ; i<names.size();i++) 
{ 
if(names.get(i).getContactname().equals(obj.getContactname())) 
{ 
    test = false; 
    break; 

} 

} 


return test; 
} 



private void getContactList() { 
    ContentResolver cr = getContentResolver(); 
    Cursor cur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null, null, null, ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME + " ASC"); 

    if ((cur != null ? cur.getCount() : 0) > 0) { 
     while (cur != null && cur.moveToNext()) { 
      String id = cur.getString(
        cur.getColumnIndex(ContactsContract.Contacts._ID)); 
      String name = cur.getString(cur.getColumnIndex(
        ContactsContract.Contacts.DISPLAY_NAME)); 

      if (cur.getInt(cur.getColumnIndex(
        ContactsContract.Contacts.HAS_PHONE_NUMBER)) > 0) { 
       Cursor pCur = cr.query(
         ContactsContract.CommonDataKinds.Phone.CONTENT_URI, 
         null, 
         ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?", 
         new String[]{id}, null); 
       while (pCur.moveToNext()) { 
        String phoneNo = pCur.getString(pCur.getColumnIndex(
          ContactsContract.CommonDataKinds.Phone.NUMBER)); 


        ContactsPlacer obj = new ContactsPlacer(); 
        obj.setContactname(name); 
        obj.setContactnumber(phoneNo); 
        if(checkexsistance(obj)) 
         names.add(obj); 

       } 
       pCur.close(); 
      } 
     } 
    } 
    if(cur!=null){ 
     cur.close(); 
    } 




} 

Answer 2

试试这个

if(email!=null){ 
    if(!names.contains(name)){ 
     names.add(name); 
    } 
} 

Answer 3

在这里使用ArrayList中的方法包含这样

public ArrayList<String> getNameEmailDetails(){ 
    ArrayList<String> names = new ArrayList<String>(); 
    ContentResolver cr = getContentResolver(); 
    Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI,null, null, null, null); 
    if (cur.getCount() > 0) { 
     while (cur.moveToNext()) { 
      String id = cur.getString(cur.getColumnIndex(ContactsContract.Contacts._ID)); 
      Cursor cur1 = cr.query(
        ContactsContract.CommonDataKinds.Email.CONTENT_URI, null, 
        ContactsContract.CommonDataKinds.Email.CONTACT_ID + " = ?", 
        new String[]{id}, null); 
      while (cur1.moveToNext()) { 
       //to get the contact names 
       String name=cur1.getString(cur1.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME)); 
       Log.e(TAG+"Name :", name); 
       String email = cur1.getString(cur1.getColumnIndex(ContactsContract.CommonDataKinds.Email.DATA)); 
       Log.e(TAG+"Email", email); 
       if(email!=null){ 
        if(!names.contains(name)) 
        names.add(name); 
       } 
      } 
      cur1.close(); 
     } 
    } 
    return names; 
} 

Answer 4

我有同样的问题，前几天，并使用此代码解决一样， 请给它一个尝试，让我知道，如果再需要帮助。

不要忘记索要许可，并在清单 “android.permission.READ_CONTACTS”提

ContentResolver cr = getActivity().getContentResolver(); 
     String[] PROJECTION = new String[]{ContactsContract.RawContacts._ID, 
       ContactsContract.Contacts.DISPLAY_NAME, 
       ContactsContract.Contacts.PHOTO_ID, 
       ContactsContract.CommonDataKinds.Email.DATA, 
       ContactsContract.CommonDataKinds.Phone.NUMBER, 
       ContactsContract.RawContacts.VERSION}; 
     String order = "CASE WHEN " 
       + ContactsContract.Contacts.DISPLAY_NAME 
       + " NOT LIKE '%@%' THEN 1 ELSE 2 END, " 
       + ContactsContract.Contacts.DISPLAY_NAME 
       + ", " 
       + ContactsContract.CommonDataKinds.Phone.DATA 
       + " COLLATE NOCASE"; 
     String filter = ""+ ContactsContract.Contacts.HAS_PHONE_NUMBER + " > 0 and " + ContactsContract.CommonDataKinds.Phone.TYPE +"=" + ContactsContract.CommonDataKinds.Phone.TYPE_MOBILE; 
     Cursor cur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, PROJECTION, filter, null, order); 

     for (int i = 0; i < cur.getColumnCount(); i++) { 
      Timber.e("column " + i + "=" + cur.getColumnName(i)); 
     } 

     if (cur.moveToFirst()) { 
      do { 
       String name = cur.getString(1); 
       String number = cur.getString(4); 

       number = Utils.removeExtraCharFromString(number); 

       if (number.length() > 8) { 
        mContactNumbers = mContactNumbers + number + ","; 
        PhoneContacts phoneContacts = new PhoneContacts(); 
        try { 
         phoneContacts.setName(URLEncoder.encode(name, "UTF-8")); 
         phoneContacts.setNumber(number); 
         phoneContacts.setNameInitial(String.valueOf(phoneContacts.getName().charAt(0))); 

         listOfContactsRaw.add(phoneContacts); 
        } catch (UnsupportedEncodingException e) { 
         e.printStackTrace(); 
        } 
       } 
      } while (cur.moveToNext()); 
     } 
     cur.close(); 

     new Handler().postDelayed(new Runnable() { 
      @Override 
      public void run() { 
       Set set = new TreeSet(new Comparator() { 
        @Override 
        public int compare(Object o1, Object o2) { 

         PhoneContacts phoneContacts1 = (PhoneContacts) o1; 
         PhoneContacts phoneContacts2 = (PhoneContacts) o2; 
         if (phoneContacts1.getNumber().equalsIgnoreCase(phoneContacts2.getNumber())) { 
          return 0; 
         } 
         return 1; 
        } 
       }); 
       set.addAll(listOfContactsRaw); 

       System.out.println("\n***** After removing duplicates *******\n"); 

       listOfContacts = new ArrayList(set); 

       // Set listOfContacts with your recycler view 
      } 
     }, 200);