我在 celery 任务期间尝试保存对模型的更改时遇到了一些行为，但更改未提交。我有一个模型可以记录用户上传的内容，一旦文件上传，就会运行 celery 任务来处理 csv 并将处理结果保存到数据库中，例如处理状态、处理时间、处理的记录数等等
在models.py中，我有以下方法：
def import_records_data(self):
total_records = 0
with self.filepath.file as csvfile:
reader = csv.reader(csvfile, delimiter=',')
next(reader, None) # skip header
for row in reader:
# process record
total_records += 1
return total_records
def process(self):
self.date_start_processing = datetime.datetime.utcnow().replace(tzinfo=utc)
try:
# process upload data,
records_processed = self.import_records_data()
except Exception, e:
self._mark_failed(unicode(e))
else:
self._mark_processed(num_records=records_processed)
def _mark_processed(self, num_records, description=None):
self.status = self.PROCESSED
self.date_end_processing = datetime.datetime.utcnow().replace(tzinfo=utc)
self.num_records = num_records
self.processing_description = description
self.save()
def _mark_failed(self, description):
self.status = self.FAILED
self.processing_description = description
self.save()
def was_processing_successful(self):
return self.status == self.PROCESSED
当调用 _mark_processed 或 _mark_failed 时，调用时所做的更改不会保存到数据库self.save()中。此方法从 tasks.py 调用：
@task(name='csv-process-upload')
def process_upload(upload_id):
upload = Upload.objects.get(id=upload_id)
upload.process()
if upload.was_processing_successful():
message_user(
upload.user,
"Your upload '%s' was processed successfully, %s records processed" % (
upload.filename,
upload.num_records))
else:
message_user(
upload.user,
"Your upload '%s' could not be processed, error message: %s" % (
upload.filename,
upload.processing_description,))
什么可能阻止模型保存？_mark_processed当我在 shell 和 type 中调试时self.save()，更改会反映在数据库中。