在 startActivityForResult() 之后检索联系人姓名和电话 [重复]

Posted 2023-04-15

【中文标题】在 startActivityForResult() 之后检索联系人姓名和电话 [重复]

【英文标题】：Retrieve contact name and phone after startActivityForResult() [duplicate]

【发布时间】：2014-07-05 18:36:58

【问题描述】：

我已经为结果开始了一个新的意图活动 代码：

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

    startActivityForResult(contactPickerIntent, 1);

现在我想获取电话和号码：

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
        
    
   

电话号码没问题，但我无法检索联系人的 GIVEN NAME！

【问题讨论】：

联系人的名字是DISPLAY_NAME，而不是CONTACT_ID

【参考方案1】：

 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));

【参考方案2】：

试试这个代码，它可能有助于做一些小的改动...

  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;
 
 

【讨论】：

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

这是因为，在你的情况下它是 'Number' 而不是 'HAS_PHONE_NUMBER' ，所以你应该做一些小的改变来适应你的需求。

好的等我测试

没有ContactsContract.Contacts.NUMBER这样的东西