Question

我的架构如下：

foodtruck.js

var mongoose = require('mongoose');
var Schema = mongoose.Schema;
var Items = require('./items.js');


var FoodTruckSchema = new Schema({
    foodtruck_name:String,
    foodtruck_location:String,
    foodtruck_rating:{type:Number,default:5},
    foodtruck_total_votes:{type:Number,default:0},
    foodtruck_tag:String,
    foodtruck_timing:String,
    foodtruck_cusine:String,
    foodtruck_img:String,
    foodtruck_logo:String,
    item_list: [ {type : mongoose.Schema.ObjectId, ref : 'items'}]
},{ versionKey: false });



module.exports = mongoose.model('foodtruck',FoodTruckSchema);

items.js

var mongoose = require('mongoose');
var Schema = mongoose.Schema;

var ItemSchema = new Schema({
    no_of_times_ordered:Number,
    item_name:String,
    item_tag:String,
    item_category:String,
    item_illustrations:[String],
    item_stock:Number,   //0 available 1 last 5 items 2 not available
    item_quantity_ordered:{type:Number,default:0},
    item_price:Number,
    item_img:String,
    no_of_likes:{type:Number,default:0}
},{ versionKey: false });

module.exports = mongoose.model('items',ItemSchema);

现在，我希望foodtruck有这样的项目列表，其中的所有项目都应该 no_of_times_ordered 大于4000.为此，我提出问题如下：

 foodtr.find({
        'item_list.no_of_times_ordered': {
            $gt: 4000
        }
    }, function(err, items) {
        if (err) res.json({
            status: '404',
            message: 'there is no data available'
        });
        res.json({
            status: '200',
            message: 'popular items list',
            data: items
        });
    });
};

但是在这里，我得到空列表，那么如何根据上述要求提供项目清单？

Answer 1

您查询错误，您无法使用item_list.no_of_times_ordered作为参考文档。您需要使用populate，并且仅针对选择性项目，使用match query populate内的条件。

试试这个：

foodtr.find({}).populate({
    path : 'item_list',
    match : {no_of_times_ordered : { $gt :4000}}
}).exec(function(err, items) {
    ...
});

如何在条件集合中引用文档

1 个答案: