如何在Python中将十六进制字符串转换为ObjectId

Question

我是一个ObjectId的列表,我正在迭代到dict中的查找值,其中键是ObjectId的.

email_count = 0
# user_id_list is a list of ObjectId's
for user_id in user_id_list:
    # UuserIdemailCountD is a dict where they keys are objectIds
    email_count +=  UuserIdemailCountD[user_id]

我一直收到以下错误:

email_count +=  UuserIdemailCountD[user_id]
KeyError: ObjectId('54a9c84ebf2e4e5b258b5412')

当我迭代user_id_list并只打印id时,我得到一个像这样的普通字符串54a9c84ebf2e4e5b258b5412.

将字符串转换为ObjectId的答案是什么？如果是这样,怎么样？

Answer 1

导入ObjectId:

from bson import ObjectId

从ObjectId到string:

oid = ObjectId()
oid_str = str(oid)
# oid_str is now '555fc7956cda204928c9dbab'

从string到ObjectId:

oid_str = '555fc7956cda204928c9dbab'
oid2 = ObjectId(oid_str)
print(repr(oid2))
# ObjectId('555fc7956cda204928c9dbab')