如何在 mongoose 中排除多个特定文档？

Question

在这里，我想排除正在请求响应的用户的文档以及用户已经发送了好友请求的所有用户以及已经与用户成为好友的用户，我该如何实现呢？

从下面的代码中，我只能排除用户数据而不是其他数据。

try {
      
      const {id} = req.params
      const user = await userModel.findById(id)
   
      const newresponse = await userModel.find({_id:user.friendRequests}).select("_id")
      
      const response = await userModel.find({_id:{"$ne":id}}).sort({creation_date:-1}).limit(5)
      return res.json({succcess:true, response, newresponse})
    } catch (error) {
        console.log(error); 
    }

model：

const mongoose = require("mongoose");

const userSchema = new mongoose.Schema({
  firstname: String,
  lastname: String,
  password: String,
  email: {
    type: String,
    unique: true,
  },
  birthday: Date,
  creation_date: {
    type: Date,
    default: Date.now,
  },
  gender: String,
  profile_picture: {
    type: String,
    default: "",
  },
  friends:[{
    type:mongoose.ObjectId,
    ref:"userModel"
  }],
  friendRequests:[{
    type:mongoose.ObjectId,
    ref:"userModel"
  }]

});

module.exports = mongoose.model("userModel", userSchema);

Answer 1

您可以连接friends和朋友friendRequests并将其传递给过滤器：

const { id } = req.params;

const user = await userModel.findById(id, 'friends friendRequests');

const excludeUsersArray = user.friends.concatenate(user.friendRequests);
   
const response = await userModel.find({ _id: { "$nin": excludeUsersArray }}).sort({ creation_date:-1 });
      
return res.json({succcess:true, response})