Question

我正在开发一种使用restify和mongoose作为odm的服务。下面是DB调用后得到的服务响应。

[
    {
        "_id": "5a548b7c025cfcffdd286e0f",
        "createdAt": "2018-01-09T09:29:32.515Z",
        "updatedAt": "2018-01-09T10:59:49.159Z",
        "subject": "Hello buddy.",
        "body": "i am good and fine.",
        "category": "email service",
        "category_id": "232",
        "author": "5a5485834c2274ce5d5e54f8",
        "__v": 0,
        "images": [
            "https://google.com"
        ],
        "email content": [
            {
                "_id": "5a549dcb80a16b2e822357ae",
                "createdAt": "2018-01-09T10:47:39.883Z",
                "updatedAt": "2018-01-09T10:47:39.883Z",
                "body": "my first comment",
                "name": "asdf"
                "password":"312312"
                "__v": 0,
                "photos": []
            }
        ]
}]

Emails.find({_id: emaiId, author_id: author_id}).
                populate('emailcontent').exec(function(err, listOfEmails) {
                if (err) {
                    console.error(err);
                    return next(
                        new errors.InvalidContentError(err.errors.name.message),
                    );
                }
                res.send(listOfEmails);
                next();

这只是我回复的一小部分。现在，我不想将所有响应发送回客户端。例如，用户名，密码，电话号码等。所以我有很多我不想发回的数据。我怎么在这里做到这一点？

Answer 1

在发回数据之前手动删除这些字段：

Proposal

Answer 2

您可以使用map()创建一个作为响应发送的新阵列。使用delete删除字段。

const newListOfEmails = listOfEmails.map((email) => {
  delete email._id;
  delete email.createdAt;
  delete email.updatedAt;

  for (let eIdx = 0; eIdx < email["email content"].length; eIdx += 1) {
    delete email["email content"][eIdx].password;
  }

  return email;
});

res.send(newListOfEmails);

Answer 3

上面的答案是关于黑名单字段，但您也可以尝试列入白名单，例如

const croppedEmails = emails.map( (email) => { body: email.body, ...(rest of needed fields) });

使用这种方法，当您添加新的保密字段时，您不必添加新的删除，另一方面，在添加新的公共字段时，您还需要将其添加到上述功能。

哦，还有投影Mongoose, find, return specific properties

Answer 4

@Mikas我尝试使用map并在res.send（）中返回响应。但响应属性不会被删除。

Posts.find({_id: catId}).populate('author')
.exec(function(err, list) {
  if(err) {
    return next(new errors.InvalidContentError(err.errors.name.message),);};
    res.send(200, massageResponse(list, createResponse));
    next();
  });
});
let createResponse = function (list) {
  const trimmedList = list.map((element) => {
    delete element.createdAt;
      return element;
  });
  return trimmedList;
}
let massageResponse = function (list, callback) {
  return callback(list);
}

从Response中过滤掉不需要的数据

4 个答案: