i create unique index like this:
self.db_database[co_name].ensure_index([('src_md5',-1),('src_time',-1),('src_size',-1)],unique=True)
self.db_database[co_name].ensure_index(('notification'),unique=True)
self.db_database[co_name].ensure_index(('version'),unique=True)`
before insert i creat a record as follows:
self.db_database[co_name].insert({"notification":"yes","file_md5":-1,"file_size":-1,"file_time":-1,"bypass":0,"server_list":[],"ok_to_download":0,"force_to_download":-1,"idx":0},safe=True)`
then i insert some info like this:
collection.insert({"src_host":src_host,"src_path":src_path,"src_name":src_name,"src_md5":src_md5,"src_time":src_time,"src_size":src_size,"version":idx},safe=True)`
and it raise a error :
DuplicateKeyError: E11000 duplicate key error index: data_transfer.nova_mon_test.log.small_20120517202918765432.$notification_1 dup key: { : null }
WHY?
You probably already have a document in your collection which either has
notification: NULL
or a document that doesn't have the notification field set. If a field is not set, then it's regarded as null. Because a unique index only allows one value per field, you can not have two documents that don't have a field set. You can get around this by also using thesparse
option while creating an index. Something like this should work (after dropping the already existing index onnotification
:See also: sparse indexes and null values in mongo