使用聯繫人卡片選擇電話號碼

Question

我想從聯繫人中選擇多個電話號碼的電話號碼。 這可以通過Android自己的聯繫人列表完成嗎？ 我設法列出我的所有聯繫人：

Intent intent = new Intent(Intent.ACTION_PICK, Contacts.CONTENT_URI);  
startActivityForResult(intent, PICK_CONTACT); 

我選擇，讓我們說，「李四」的，我知道，有3個電話號碼。 有沒有反正讓Android本身讓我看看那3個電話號碼來選擇使用intent？

Answer 1

首先，let's使得被叫爲選擇器

Intent intent = new Intent(Intent.ACTION_PICK, ContactsContract.Contacts.CONTENT_URI); 
startActivityForResult(intent, PICK_CONTACT); 

然後在onActivityResult（）讓我們檢查多少數量已選定的聯繫人，如果人數> 1，則可以顯示一個對話框，選擇一個：

Uri result = data.getData(); 
Log.v(TAG, "Got a result: " + result.toString()); 

// get the contact id from the Uri 
String id = result.getLastPathSegment(); 

// query for phone numbers for the selected contact id 
c = getContentResolver().query(
    Phone.CONTENT_URI, null, 
    Phone.CONTACT_ID + "=?", 
    new String[]{id}, null); 

int phoneIdx = c.getColumnIndex(Phone.NUMBER); 
int phoneType = c.getColumnIndex(Phone.TYPE); 

if(c.getCount() > 1) { // contact has multiple phone numbers 
    final CharSequence[] numbers = new CharSequence[c.getCount()]; 
    int i=0; 
    if(c.moveToFirst()) { 
     while(!c.isAfterLast()) { // for each phone number, add it to the numbers array 
      String type = (String) Phone.getTypeLabel(this.getResources(), c.getInt(phoneType), ""); // insert a type string in front of the number 
      String number = type + ": " + c.getString(phoneIdx); 
      numbers[i++] = number; 
      c.moveToNext(); 
     } 
     // build and show a simple dialog that allows the user to select a number 
     AlertDialog.Builder builder = new AlertDialog.Builder(this); 
     builder.setTitle(R.string.select_contact_phone_number_and_type); 
     builder.setItems(numbers, new DialogInterface.OnClickListener() { 

      @Override 
      public void onClick(DialogInterface dialog, int item) { 
       String number = (String) numbers[item]; 
       int index = number.indexOf(":"); 
       number = number.substring(index + 2); 
       loadContactInfo(number); // do something with the selected number 
      } 
     }); 
     AlertDialog alert = builder.create(); 
     alert.setOwnerActivity(this); 
     alert.show(); 

    } else Log.w(TAG, "No results"); 
} else if(c.getCount() == 1) { 
    // contact has a single phone number, so there's no need to display a second dialog 
}