Question

{
 "teachers" : [
               {"name": "Lucy", "id": 3, course: "history"}, 
               {"name": "Mark", "id": 6, "course": "maths"}, 
               {"name": "Joan", "id": 20, course: "French"} 
               ] 
}

此文件位于“学校”系列中。我一直试图使用

访问这些嵌入的文档

db.school.find({teachers:{id:3}})

我也试过

db.school.find({teacher.id:3})

但我知道它不起作用，因为mongo无法查看嵌入式数组。因此，我想将这些嵌入的文档转换为单独的文档。也就是说，删除嵌入和“教师”键，为每位教师创建单独的文档。

最终的“学校”系列将是

{"name": "Lucy", "id": 3, "course": "history"}, 
{"name": "Mark", "id": 6, "course": "maths"}, 
{"name": "Joan", "id": 20, "course": "French"}

我想用python做它并将新文档保存到集合中。

修改的

这就是我现在想出来的：

import pymongo
import sys

connection = pymongo.Connection("mongodb://localhost", safe=True)

db = connection.hello
shows = db.school

for doc in db.school:
    for indiv in "teachers":
            try:
            db.individual.insert(indiv)
        except:
            print "Unexpected error", sys.exc_info()[0]

Answer 1

顺便说一句，Mongodb可以找到数组中的嵌入式文档：

db.school.find({ 'teachers.id' : 3 });

您可以在mongodb documentation了解有关点符号的更多信息。

如果目标是仅返回嵌入文档，则可以使用aggregate请求：

db.school.aggregate(
  {$match: { 'teachers.id' : 3 }}, 
  {$unwind : '$teachers'}, 
  {$project: { 
    _id: 0, 
    name: '$teachers.name',
    id: '$teachers.id', 
    course: '$teachers.course' 
  }}, 
  {$match: {id:3}}
);

Answer 2

school_records = db.school.find()
for i in school_records:
    for teacher in i['teachers']:
        db.individual.insert(teacher)

这个怎么样？

Answer 3

您可以使用mongo v2.2 +中的聚合命令（它位于pymongo中）：

    fagg=db.school.aggregate([{$unwind: "$teachers"},
    {$project: {name: "$teachers.name", 
    id: "$teachers.id", course: "$teachers.course"}}])
    fagg.result.forEach(function(o){
    db.teachers.insert({_id: o.id, name: o.name, course: o.course})})

Mongodb：如何将一组文档转换为单独的文档？

3 个答案: