DjangoDjango 如何支持 分组查询统计?
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代码:
from django.db.models import Sum alarm_sum_group_items = models.FILE_PROTECT_ALARM.objects.filter(customer_id=customer_id).values(\'device_hash\').annotate(alarm_sum=Sum(\'cmn_merge_count\')) data = [] for item in alarm_sum_group_items: tmp = {} tmp[\'customer_id\'] = customer_id tmp[\'device_hash\'] = item[\'device_hash\'] tmp[\'alarm_sum\'] = item[\'alarm_sum\'] data.append(tmp) return APIResponse(status=status_code.success, data=data)
参考资料:
https://www.douban.com/group/topic/18930354/?type=like
http://www.tuicool.com/articles/aiUBVrn
https://www.douban.com/group/topic/18930354/?type=like
http://stackoverflow.com/questions/19101665/django-how-to-do-select-count-group-by-and-order-by
https://my.oschina.net/duoduo3369/blog/199190
http://stackoverflow.com/questions/327807/django-equivalent-for-count-and-group-by
https://segmentfault.com/q/1010000005828240
http://www.cnblogs.com/BeginMan/archive/2013/05/27/3101322.html
http://blog.csdn.net/iloveyin/article/details/46380619
http://www.jb51.net/article/69774.htm
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