1
這裏是PymongoPymongo失敗,但不會放棄例外
import mong #just my library for initializing
collection_1 = mong.init(collect="col_1")
collection_2 = mong.init(collect="col_2")
for name in collection_2.find({"field1":{"$exists":0}}):
try:
to_query = name['something']
actual_id = collection_1.find_one({"something":to_query})['_id']
crap_id = name['_id']
collection_2.update({"_id":id},{"$set":{"new_name":actual_id}},upset=True)
except:
open('couldn_find_id.txt','a').write(name)
所有這一切都做的是從一個集合場,發現現場的ID和更新另一個集合的ID查詢。它適用於大約1000-5000次迭代,但定期失敗,然後我必須重新啓動腳本。
> Traceback (most recent call last):
File "my_query.py", line 6, in <module>
for name in collection_2.find({"field1":{"$exists":0}}):
File "/home/user/python_mods/pymongo/pymongo/cursor.py", line 814, in next
if len(self.__data) or self._refresh():
File "/home/user/python_mods/pymongo/pymongo/cursor.py", line 776, in _refresh
limit, self.__id))
File "/home/user/python_mods/pymongo/pymongo/cursor.py", line 720, in __send_message
self.__uuid_subtype)
File "/home/user/python_mods/pymongo/pymongo/helpers.py", line 98, in _unpack_response
cursor_id)
pymongo.errors.OperationFailure: cursor id '7578200897189065658' not valid at server
^C
bye
有沒有人有任何想法,這是什麼故障,我怎麼可以把它變成一個例外,繼續我的劇本即使在這個失敗嗎?
感謝