node.js Mongoose 在没有 _id 字段的情况下检索数据

Mongoose retrieving data without _id field

提问by Masiar

I would like to retrieve some data from a Mongoose setting in my Node.js application. I noticed that no matter what I write as field selection, I always get the _idfield. Is there a way not to fetch it? This is how I do right now:

我想从我的 Node.js 应用程序中的 Mongoose 设置中检索一些数据。我注意到无论我写什么作为字段选择，我总是得到_id字段。有没有办法不取？这就是我现在的做法：

Transaction.find({username : user.username}, ['uniqueId', 'timeout', 'confirmation_link', 'item_name'], function(err, txs){
        console.log("user : " + user.username + " with txs: " + txs);
        callback(txs);
});

And logs me the results which contain the _idfield.

并将包含该_id字段的结果记录给我。

回答by VisioN

Another way is to use text argument with prefix -which will exclude this or that field from the result:

另一种方法是使用带有前缀的文本参数-，它将从结果中排除这个或那个字段：

Entity.find({ ... }, '-_id field1 field2', function(err, entity) {
    console.log(entity);  // { field1: '...', field2: '...' }
});

回答by danmactough

_idmust be specifically excluded. For example,

_id必须特别排除。例如，

Transaction.find({username : user.username}, { '_id': 0, 'uniqueId' :1, 'timeout': 1, 'confirmation_link': 1, 'item_name': 1}, function(err, txs){
  console.log("user : " + user.username + " with txs: " + txs);
  callback(txs);
});

回答by Gábor Imre

Another approach:

另一种方法：

• Augment the .toJSON()of the schema that it deletes the _idand the __vfields
• Call .toJSON()on all DB objects sent to client
• Extra benefit #1: you can use item.id === 'something'because typeof id === 'string', not ObjectId.
• Extra benefit #2: When you got gan object back from the client and you want to search / update then you don't have to manually delete _idbecause there is none, just an idwhich is ignored.

• 增加 .toJSON()它删除的架构的_id和__v字段
• 调用.toJSON()发送到客户端的所有 DB 对象
• 额外好处#1：您可以使用item.id === 'something'因为typeof id === 'string'，而不是ObjectId。
• 额外好处#2：当您从客户端取回 gan 对象并且想要搜索/更新时，您不必手动删除，_id因为没有，只是id被忽略。

Augmenting JSON:

增强 JSON：

mySchema.set('toJSON', {
    virtuals: true,
    transform: (doc, ret, options) => {
        delete ret.__v;
        ret.id = ret._id.toString();
        delete ret._id;
    },
});

So you can use:

所以你可以使用：

 let item = (await MyCollection.findOne({/* search */}).exec()).toJSON();
 if (item.id === 'someString') return item;

I know it's ugly. But it's the best bad idea that I have so far.

我知道这很丑。但这是我迄今为止最好的坏主意。

回答by Durja Arai

In 5.2.13 version of Mongoose (Sept 2018)- using the query builder approach the same can be converted to

在 Mongoose 的 5.2.13 版本（2018 年 9 月）中 - 使用查询构建器方法相同可以转换为

async function getUserDetails(user) {
    try {
        if (!user || !user.name) return;
        const result = await Transaction.
        find({username : user.username}).
        select('uniqueId timeout confirmation_link item_name -_id'); 
        // Adding minus sign before the _id (like -_id) in the select string unselects the _id which is sent by default. 
        console.log(result);
    } catch(ex) {
        return ex
    }
}