Question

我有一个这样的架构：

var resortSchema = new mongoose.Schema ({
name: String,
price: String,
reviews: [
    {
        type: mongoose.Schema.Types.ObjectId,
        ref: "Review"
    }
]});


var reviewSchema = new mongoose.Schema({
    rating: {
        // Setting the field type
        type: Number,
        // Defining min and max values
        min: 1,
        max: 5,
        // Adding validation to see if the entry is an integer
        validate: {
            // validator accepts a function definition which it uses for validation
            validator: Number.isInteger,
            message: "{VALUE} is not an integer value."
        }
    },
    // author id and username fields
    author: {
        id: {
            type: mongoose.Schema.Types.ObjectId,
            ref: "User"
        },
        username: String
    },
    // resort associated with the review
    resort: {
        type: mongoose.Schema.Types.ObjectId,
        ref: "Resort"
    }
});

在代码中，我在resorts文档中填充了评论数组，我只想从resort.reviews数组中删除一些特定的评论（基于作者ID）。像这样：

resort.reviews.pull({"author.id": someSpecificUserID});
resort.save();

但这不起作用。我该怎么办？

编辑：我可以通过特定的用户ID查找所有评论，但是我不知道该如何按度假胜地进行过滤。我尝试过：

console.log("req.user._id: " + typeof req.user._id); // object
console.log("req.params.id: " + typeof req.params.id); // string
console.log("ObjectId: " + typeof mongoose.Types.ObjectId(req.params.id));

var resortObjectID = mongoose.Types.ObjectId(req.params.id);

// Get ids of reviews matching a particular author
const reviews = await Review.find({ "author.id": req.user._id, "resort": resortObjectID });
console.log("reviews: " + reviews);     // reviews is []. if I remove "resort": resortObjectID I do get all the reviews of that user in all resorts. But I need to filter to only one resort.

//Pass review ids to $pull     // I am not sure if this works yet
await Resort.updateOne(
    { _id: req.params.id },
    { $pull: { reviews: { $in: reviews.map(x=>x._id) } } },
    { multi: true })

Answer 1

由于两个集合（srcStageMask都不同，因此基于其他集合的值从一个集合中提取值根本行不通。因此，您首先必须从resort and review集合中获取要删除的评论的ID，然后将其传递给review

$pull

猫鼬基于一些过滤器从对象数组中删除对象

1 个答案: