使用pymongo返回ObjectID的.str

2 回答

• 2

  python中获取对象字符串表示的标准方法是使用 str 内置函数：

  id = bson.objectid.ObjectId()
  str(id)
  => '5190666674d3cc747cc12e61'
  

  回复于 2024-04-27T19:55:35+08:00
• 0

  试试这个：

  for post in db.votes.find({'user_id':userQuery['_id']}):
              posts += str(post['_id'])
  

  顺便说一句，您可以使用MongoKit来处理特殊的bson数据结构 .

  from bson.objectid import ObjectId
  
  
  class CustomObjectId(CustomType):
  mongo_type = ObjectId  # optional, just for more validation
  python_type = str
  init_type = None  # optional, fill the first empty value
  
  def to_bson(self, value):
      """convert type to a mongodb type"""
      return ObjectId(value)
  
  def to_python(self, value):
      """convert type to a python type"""
      return str(value)
  
  def validate(self, value, path):
      """OPTIONAL : useful to add a validation layer"""
      if value is not None:
          pass  # ... do something here
  

  这个自定义ObjectId可以将bson ObjectId 变为python str .

  有关更多信息，请访问http://mongokit.readthedocs.org/mapper.html#the-structure .

  回复于 2024-04-27T19:55:35+08:00