一尘不染

猫鼬填充嵌套数组

node.js

假设以下3个模型:

var CarSchema = new Schema({
  name: {type: String},
  partIds: [{type: Schema.Types.ObjectId, ref: 'Part'}],
});

var PartSchema = new Schema({
  name: {type: String},
  otherIds: [{type: Schema.Types.ObjectId, ref: 'Other'}],
});

var OtherSchema = new Schema({
  name: {type: String}
});

当我查询汽车时,我可以填充零件:

Car.find().populate('partIds').exec(function(err, cars) {
  // list of cars with partIds populated
});

猫鼬是否有办法在所有汽车的嵌套零件对象中填充otherIds。

Car.find().populate('partIds').exec(function(err, cars) {
  // list of cars with partIds populated
  // Try an populate nested
  Part.populate(cars, {path: 'partIds.otherIds'}, function(err, cars) {
    // This does not populate all the otherIds within each part for each car
  });
});

我可能可以遍历每辆车并尝试填充:

Car.find().populate('partIds').exec(function(err, cars) {
  // list of cars with partIds populated

  // Iterate all cars
  cars.forEach(function(car) {
     Part.populate(car, {path: 'partIds.otherIds'}, function(err, cars) {
       // This does not populate all the otherIds within each part for each car
     });
  });
});

问题是我必须使用一个像async这样的库来对每个对象进行填充调用,然后等到所有操作完成后再返回。

可以在不循环所有汽车的情况下做?


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2020-07-07

共1个答案

一尘不染

更新: 请参阅以获取在Mongoose
4中添加的更紧凑的版本。摘要如下:

Car
  .find()
  .populate({
    path: 'partIds',
    model: 'Part',
    populate: {
      path: 'otherIds',
      model: 'Other'
    }
  })

猫鼬3及以下:

Car
  .find()
  .populate('partIds')
  .exec(function(err, docs) {
    if(err) return callback(err);
    Car.populate(docs, {
      path: 'partIds.otherIds',
      model: 'Other'
    },
    function(err, cars) {
      if(err) return callback(err);
      console.log(cars); // This object should now be populated accordingly.
    });
  });

对于这样的嵌套种群,您必须告诉猫鼬要从中填充的架构。

2020-07-07