我正在学习猫鼬,看着文档http://mongoosejs.com/docs/populate.html
然后我做了:
# initialize some TodoLists and their todos
l1 = new TodoList
name: "List 1"
l2 = new TodoList
name: "List 2"
t1 = new Todo
title: "Todo 1"
desc: "Hello 1"
list: l1._id
t2 = new Todo
title: "Todo 2"
desc: "Hello 1"
list: l1._id
t3 = new Todo
title: "Todo 3"
dueOn: new Date 2012,6,1
completedOn: new Date 2012,6,1
list: l1._id
t4 = new Todo
title: "Todo 4"
desc: "Hello 4"
list: l2._id
# save all
async.parallel [
(done) -> l1.save(done)
(done) -> l2.save(done)
(done) -> t1.save(done)
(done) -> t2.save(done)
(done) -> t3.save(done)
(done) -> t4.save(done)
], (err) ->
if !err
# debugging
Todo.find {}, (err, todos) ->
console.log todos
# the populate
TodoList.findOne({ name: "List 2" })
.populate("todos")
.exec (err, list) ->
console.log list
done()
else
done err
我得到的是:
[ { desc: 'Hello 1',
list: 5026321c684d0a6b08000007,
_id: 5026321c684d0a6b08000009,
__v: 0 },
{ desc: 'Hello 1',
list: 5026321c684d0a6b08000007,
_id: 5026321c684d0a6b0800000a,
__v: 0 },
{ dueOn: Sun Jul 01 2012 00:00:00 GMT+0800 (SGT),
completedOn: Sun Jul 01 2012 00:00:00 GMT+0800 (SGT),
list: 5026321c684d0a6b08000007,
_id: 5026321c684d0a6b0800000b,
__v: 0 },
{ desc: 'Hello 4',
list: 5026321c684d0a6b08000008,
_id: 5026321c684d0a6b0800000c,
__v: 0 } ]
{ name: 'List 2',
_id: 5026321c684d0a6b08000008,
__v: 0,
todos: [] }
Noice最后一行,todos没有填充,我希望它有Todo 4?
答案 0 :(得分:2)
参考不是双向的。您已将引用保存到Todo
'父级List
,但您不会将Todo
的引用存储到List.todos
数组中。
假设您的列表架构类似于:
var ListSchema = new Schema({
...
todos : [{ type: Schema.Types.ObjectId, ref: 'Todo' }]
});
然后你需要做一个list.todos.push(todo._id)
。
答案 1 :(得分:0)
通过使用以下代码,您无需在ListSchema中添加todos密钥。在定义ListSchema之后,只需输入以下代码:
ListSchema.virtual('todos', {
ref: 'Post',
localField: '_id',
foreignField: 'list'
});
现在填写列表集合中的待办事项。
ListModel.find({}).populate('todos').exec((err, lists) => {
if (err) return res.status(500).json(err);
return res.status(200).json(lists);
});