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试图过滤掉任何对象？在朋友数组中？不匹配字段“friends.username”与“one@one.com”的值

      final userDoc = await store.find({
    '_id': ObjectId.fromHexString(authDetails.subject),
    'friends.username': 'one@one.com',
    '\$filter': {'friends.username': 'one@one.com'}
  }).forEach((element) {
    print(element);
  });

我知道这是可能的，因为我可以使用

      Object newReceived = {
    'username': myUsername,
    'timestamp': addedTime,
  };

  //add your name to their friend request received list
  await store.updateOne({
    '_id': theirId
  }, {
    '\$push': {'friendRequestReceived': newReceived}
  });

mongodb数据库结构如下：

{
"_id" : ObjectId("60b2faa2de1f7691268bf907"),
"email" : "nine5@one.com",
"emailverified" : false,
"username" : "nine5@one.com",
"displayName" : "complete n00b",
"dob" : "",
"points" : 0,
"attended" : 0,
"hosted" : 0,

"friends" : [
    {
        "username" : "one@one.com",
        "timestamp" : "2021-05-29 19:38:58.758325"
    },
    {
        "username" : "nine4@one.com",
        "timestamp" : "2021-05-29 19:39:54.915636"
    }
]

}

我只想获取与“_id”关联的“朋友”数组下具有用户名“one@one.com”的任何文档：ObjectId（“60b2faa2de1f7691268bf907”）。并过滤掉任何不这样做的用户名，我希望最终使用此结构仅返回来自某个用户的消息，该用户在名为“_id”的消息数组下找到：ObjectId（“60b2faa2de1f7691268bf907”）。谢谢

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