Java 获取联系人姓名？

我想知道联系人的名字，但我不能。在查看之后，我尝试使用family、given和display来获取名称，但没有任何效果

  @Override
    protected void onActivityResult(int requestCode, int resultCode, Intent data) {
        if (requestCode == PICK_CONTACT && resultCode == RESULT_OK) {
            Uri contactUri = data.getData();
            Cursor cursor = getContentResolver().query(contactUri, null, null, null, null);
            cursor.moveToFirst(); //Move to first row...I actually dont know why this part is necessary, but I get an error without it... 
            int NumberColumn = cursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER); //Int column is the column of the numbers
            int NameColumn = cursor.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME);
            String contactNumber = cursor.getString(NumberColumn);
            String contactName = cursor.getString(NameColumn);
            Toast.makeText(MainActivity.this, ""+ contactNumber +"" +contactName, Toast.LENGTH_SHORT).show();

        }

/

---

请尝试以下代码以获取特定号码的联系人

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);

    if (requestCode == REQUEST_CODE_PICK_CONTACTS && resultCode == RESULT_OK) {
        Log.d(TAG, "Response: " + data.toString());
        uriContact = data.getData();

        retrieveContactName();

    }
}

 private void retrieveContactName() {

        String contactName = null;

        // querying contact data store
        Cursor cursor = getContentResolver().query(uriContact, null, null, null, null);

        if (cursor.moveToFirst()) {

            // DISPLAY_NAME = The display name for the contact.
            // HAS_PHONE_NUMBER =   An indicator of whether this contact has at least one phone number.

            contactName = cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
        }

        cursor.close();

        Log.d(TAG, "Contact Name: " + contactName);

    }

---

有关更多详细信息，请参阅下面的链接

尝试下面的代码以获取特定号码的联系人

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);

    if (requestCode == REQUEST_CODE_PICK_CONTACTS && resultCode == RESULT_OK) {
        Log.d(TAG, "Response: " + data.toString());
        uriContact = data.getData();

        retrieveContactName();

    }
}

 private void retrieveContactName() {

        String contactName = null;

        // querying contact data store
        Cursor cursor = getContentResolver().query(uriContact, null, null, null, null);

        if (cursor.moveToFirst()) {

            // DISPLAY_NAME = The display name for the contact.
            // HAS_PHONE_NUMBER =   An indicator of whether this contact has at least one phone number.

            contactName = cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
        }

        cursor.close();

        Log.d(TAG, "Contact Name: " + contactName);

    }

---

更多详细信息请参阅下面的链接

为什么

if（cursor.moveToFirst（））{

？谢谢]我想我们只有一个条目，这就是为什么我们使用cursor movetofirst。首先，它将检查cursor是否为空，如果不是，它将返回数据。您是否能够获取电话号码名称？否，因为它与我在问题中提供的代码相同。是否检查号码是否有名称？为什么

如果（cursor.movetofirst（））{

？谢谢]我想我们只有一个条目，这就是为什么我们使用游标movetofirst。首先，它将检查游标是否为空，如果不是，它将返回数据。您是否能够获取电话号码名称？否，因为它与我在问题中提供的代码相同。是否检查号码是否有名称？