如何用Mongoose填充MongoDB嵌套数据，让评论显示用户名？

定义的Schema结构 #

BlogSchema #

const blogSchema = new mongoose.Schema({
  title: { type: String, required: true, minLength: 5 },
  author: String,
  user: {
    type: mongoose.Schema.Types.ObjectId,
    ref: "User",
  },
  comments: [
    {
      type: mongoose.Schema.Types.ObjectId,
      ref: "Comment",
    },
  ],
});

UserSchema #

const userSchema = new mongoose.Schema({
  username: { type: String, required: true, unique: true, minLength: 4 },
  name: String,
  passwordHash: String,
  blogs: [
    {
      type: mongoose.Schema.Types.ObjectId,
      ref: "Blog",
    },
  ],
});

CommentSchema #

const commentSchema = new mongoose.Schema({
  comment: { type: String, minLength: 5 },
  user: {
    type: mongoose.Schema.Types.ObjectId,
    ref: "User",
  },
  blog: {
    type: mongoose.Schema.Types.ObjectId,
    ref: "Blog",
    required: true,
  },
});

期望的查询结果格式 #

blog = {
  "title" : ".....",
  "author": "......",
  "user" : "user-Id",
  "comments": [
    {
      "comment":"....",
      "user": "username"
    }
  ]
}

当前尝试的代码 #

const blogs = await Blog.find({})
  .populate("user", { id: 1 })
  .populate("comments", { comment: 1, user: 1 });

问题 #

执行上述代码后，评论中的user字段显示的是用户ID，而非预期的用户名。

解决方案 #

需要使用嵌套Populate，在填充comments的同时，进一步填充其关联的user字段，并指定只返回用户名：

const blogs = await Blog.find({})
  .populate("user", { _id: 1 }) // 若你使用virtual id可替换为{ id: 1 }
  .populate({
    path: "comments",
    select: "comment user", // 仅保留comment和user字段
    populate: {
      path: "user",
      select: "username -_id" // 仅返回username并排除_id，使user直接为用户名字符串
    }
  });

说明 #

• 嵌套Populate通过对象形式配置populate参数，先指定comments的填充规则，再在其中配置关联user的填充规则。
• select: "username -_id"的作用是只选取username字段，同时排除_id，让返回的user字段直接是用户名的字符串值，而非包含_id的用户对象。

内容的提问来源于stack exchange，提问作者Tajpuriya