根据数组元素查找对象，仅返回匹配的数组元素? [英] Find object based on array element, return only matching array element?

问题描述

我在猫鼬中有一个Person对象，并且该person对象具有多个对象(每个对象都有唯一的ID).

I have a Person object in mongoose, and that person object has multiple things (each thing has a unique ID).

person1 = {
  things[{id: 1, name: 'one'},{id:2, name: 'two'}]
}
person2 = {
  things[{id: 3, name: 'three'},{id:4, name: 'four'}]
}

然后查询:

Person.findOne({'things.id': 2},{'things.$': 1}, function(err, person) { ...

这很好用，但是我正在搜索所有Person对象(可能有很多).在这种情况下，我知道我需要的人的ID和事物"的一些唯一ID.通过id获取Person可能要快得多:

This works great but I am searching through all Person objects (which there could be a lot of). In this case I know the id of the Person I need and some unique id of a 'thing'. Its probably a lot faster to get the Person by id:

Person.findById(personId, function(err, person) { ...

然后遍历所有事物以找到正确的事物:

Then loop over all the things to find the right one:

var thing
person.things.forEach(function(t) {
  if (t.id == thingId) {
    thing = t;
  }
});

我想知道的是，是否有更好的方法. IE.我可以通过id查询Person集合以仅得到一个Person，然后仅过滤出我要查找的内容(没有丑陋的循环)吗?

What I am wondering is if there is a better way. I.E. can I query the Person collection by id to get just one Person then filter out just the thing I am looking for (without the ugly loop)?

推荐答案

您可以在一个查询中同时包含两个id项，并且单个元素投影仍将起作用:

You can include both id terms in a single query and the single element projection will still work:

Person.findOne({_id: personId, 'things.id': 2}, {'things.$': 1}, 
    function(err, person) { ...

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