pymongo错误:筛选器必须是dict、bson.son.son或从collections.Mapping继承的其他类型的实例

2024-09-24 00:34:05 发布

您现在位置:Python中文网/ 问答频道 /正文

我认为这个问题是正确的,但仍然是个错误。

    findQ = {"fromid": wordid}, {"toid":1}        
    res= self.db.wordhidden.find(findQ)

然而,找到一个(findQ)的作品。所以我找不到错的地方。 我使用python 3.6和pymongo。 这是我的代码:

 def getallhiddenids(self,wordids,urlids):
  l1={}
  for wordid in wordids:
    findQ = {"fromid": wordid}, {"toid":1}
    res= self.db.wordhidden.find(findQ)
    for row in res: l1[row[0]]=1
  for urlid in urlids:
    findQ = {"toid": urlid}, {"fromid":1}
    res= self.db.hiddenurl.find(findQ)

这是一个错误:

    Traceback (most recent call last):
    File "C:\Users\green\Desktop\example.py", line 9, in <module>
    neuralnet.trainquery([online], possible, notspam)
  File "C:\Users\green\Desktop\nn.py", line 177, in trainquery
    self.setupnetwork(wordids,urlids)
  File "C:\Users\green\Desktop\nn.py", line 105, in setupnetwork
    self.hiddenids=self.getallhiddenids(wordids,urlids)
  File "C:\Users\green\Desktop\nn.py", line 93, in getallhiddenids
    res= self.db.wordhidden.find(findQ)
  File "C:\Users\green\AppData\Local\Programs\Python\Python36-32\lib\site-
packages\pymongo\collection.py", line 1279, in find
    return Cursor(self, *args, **kwargs)
  File "C:\Users\green\AppData\Local\Programs\Python\Python36-32\lib\site-
packages\pymongo\cursor.py", line 128, in __init__
    validate_is_mapping("filter", spec)
  File "C:\Users\green\AppData\Local\Programs\Python\Python36-32\lib\site-
packages\pymongo\common.py", line 400, in validate_is_mapping
    "collections.Mapping" % (option,))
TypeError: filter must be an instance of dict, bson.son.SON, or other type 
that inherits from collections.Mapping

Tags: inpyselfdblineresgreenfind
1条回答
网友
1楼 · 发布于 2024-09-24 00:34:05

find_one(findQ) works

错误是因为PyMongo find()需要字典或bson.son对象。您传入的是一个Python元组对象,其形式是({"fromid": wordid}, {"toid":1})。您可以通过如下调用find()方法来更正此问题:

db.wordhidden.find({"fromid": wordid}, {"toid": 1})

从技术上讲,调用find_one()也不起作用。只是参数过滤器被find_one()更改了。见find_one() L1006-1008。基本上将元组过滤器格式化为:

{'_id': ({"fromid": wordid}, {"toid":1}) } 

上面的代码不会(不应该)返回集合中的任何匹配项。

除了要执行的操作外,还可以将filter参数存储到两个变量中,例如:

filterQ = {"fromid": wordid}
projectionQ = {"toid": 1}
cursor = db.wordhidden.find(filterQ, projectionQ)

相关问题 更多 >