node.js 根据数组元素查找对象，只返回匹配的数组元素？
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Find object based on array element, return only matching array element?
提问by lostintranslation
I have a Person object in mongoose, and that person object has multiple things (each thing has a unique ID).
我在 mongoose 中有一个 Person 对象，并且那个 person 对象有多个东西（每个东西都有一个唯一的 ID）。
person1 = {
things[{id: 1, name: 'one'},{id:2, name: 'two'}]
}
person2 = {
things[{id: 3, name: 'three'},{id:4, name: 'four'}]
}
then query:
然后查询：
Person.findOne({'things.id': 2},{'things.$': 1}, function(err, person) { ...
This works great but I am searching through all Person objects (which there could be a lot of). In this case I know the id of the Person I need and some unique id of a 'thing'. Its probably a lot faster to get the Person by id:
这很好用，但我正在搜索所有 Person 对象（可能有很多）。在这种情况下，我知道我需要的人的 id 和“事物”的一些唯一 id。通过 id 获取 Person 可能要快得多：
Person.findById(personId, function(err, person) { ...
Then loop over all the things to find the right one:
然后遍历所有内容以找到正确的内容：
var thing
person.things.forEach(function(t) {
if (t.id == thingId) {
thing = t;
}
});
What I am wondering is if there is a better way. I.E. can I query the Person collection by id to get just one Person then filter out just the thing I am looking for (without the ugly loop)?
我想知道是否有更好的方法。IE 是否可以通过 id 查询 Person 集合以获得一个 Person 然后过滤掉我正在寻找的东西（没有丑陋的循环）？
回答by JohnnyHK
You can include both id terms in a single query and the single element projection will still work:
您可以在单个查询中包含两个 id 术语，并且单元素投影仍然有效：
Person.findOne({_id: personId, 'things.id': 2}, {'things.$': 1},
function(err, person) { ...
