Question

我已经为结果启动了一个新的意图活动代码：

    Intent contactPickerIntent = new Intent(Intent.ACTION_PICK, Phone.CONTENT_URI);  

    startActivityForResult(contactPickerIntent, 1);

现在我想拿到电话号码：

protected void onActivityResult(int requestCode, int resultCode, Intent data) {

    if (requestCode == 1) {

        if(resultCode == RESULT_OK){

            Uri contactUri = data.getData();

            String[] pN = {Phone.NUMBER};

            String[] pNa = {Phone.CONTACT_ID};//idk

            Cursor cP = getContentResolver().query(contactUri, pN, null, null, null);
            cP.moveToFirst();

            Cursor cPa = getContentResolver().query(contactUri, pNa, null, null, null);
            cPa.moveToFirst();

            int numc = cP.getColumnIndex(Phone.NUMBER);
            String num = cP.getString(numc);

            int namec = cPa.getColumnIndex(Phone.CONTACT_ID);//idk
            String name = cPa.getString(namec);//idk

            Log.i("", name);

        }
        if (resultCode == RESULT_CANCELED) {
            //DO OTHER STUFF
        }
    }
}

电话号码很好但是我无法找回联系人的GIVEN NAME！

Answer 1

尝试使用此代码，它可能有助于进行一些细微更改...

  static final int PICK_CONTACT_=1;

  Intent intent_1 = new Intent(Intent.ACTION_PICK, ContactsContract.Contacts.CONTENT_URI);
  startActivityForResult(intent_1, PICK_CONTACT_);


  @Override
 public void onActivityResult(int reqCode, int resultCode, Intent data) {
 super.onActivityResult(reqCode, resultCode, data);

 switch (reqCode) {
 case (PICK_CONTACT_) :
   if (resultCode == Activity.RESULT_OK) {

     Uri contactData = data.getData();
     Cursor cursor =  managedQuery(contactData, null, null, null, null);
     if (cursor.moveToFirst()) {


         String id =cursor.getString(cursor.getColumnIndexOrThrow(ContactsContract.Contacts._ID));

         String hasPhone =cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER));

           if (hasPhone.equalsIgnoreCase("1")) {
          Cursor phones = getContentResolver().query( 
                       ContactsContract.CommonDataKinds.Phone.CONTENT_URI,null, 
                       ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = "+ id, 
                       null, null);
             phones.moveToFirst();
              cNumber = phones.getString(phones.getColumnIndex("data_1"));
             System.out.println("number is:"+cNumber);
           }
         String name = cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));


     }
   }
   break;
 }
 }

Answer 2

 Uri uri = data.getData();

        Cursor cursor = getContentResolver().query(uri, null, null, null, null);

        cursor.moveToFirst();
        name = cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));

        cursor.moveToFirst();
        number = cursor.getString(cursor.getColumnIndex(Phone.NUMBER));

在startActivityForResult（）之后检索联系人姓名和电话

2 个答案: