优化查询
Room
id 1 2 .. 1000 User: id 1 .. 10000 Booking: user_id room_id time_id date 1 1 8:00 2017-11-11 1 2 8:00 2017-11-11 1 3 8:00 2017-11-11 1 4 8:00 2017-11-11 1 5 8:00 2017-11-11
二、 需求:获取2017-11-11所有预定信息:
打印:用户名称,会议室名称, 预定时间段
# 解决方案一:执行11次sql语句 bk = models.Booking.objects.filter(date=2017-11-11) for item in bk: print(item.time_id, item.room.caption, item.user.user) # 解决方案二:执行1次 #select * from ... left join user ... join room bk = models.Booking.objects.filter(date=2017-11-11).select_related('user','room') for item in bk: print(item.time_id, item.room.caption, item.user.user) # 解决方案三:执行3次 #select * from booking where date=2017-11-11 #select * from user where id in [1,] #select * from room where id in [1,2,3,4,5] bk = models.Booking.objects.filter(date=2017-11-11).prefetch_related('user','room') for item in bk: print(item.time_id, item.room.caption, item.user.user)
总结:以后对于SQL语句的优化要加上selsect_releated或者prefetch_releated,这只是对于跨表做的优化,如果是单表的话就没有必要进行优化查询了
那么什么时候用selsect_releated,什么时候用prefetch_releated呢?这个按情况而定,
selsect_releated是主动连表,执行一次SQL
prefetch_releated不连表执行3次SQL
二、Q查询的第二种方式
remove_booking = Q() for room_id, time_id_list in booking_info['del'].items(): for time_id in time_id_list: temp = Q() #实例化一个Q对象 temp.connector = 'AND' #以and的方式连接 # user_id是一个字段,后面的是一个字段对应的值 temp.children.append(('user_id', request.session['user_info']['id'],)) temp.children.append(('booking_date', booking_date,)) temp.children.append(('room_id', room_id,)) temp.children.append(('booking_time', time_id,)) remove_booking.add(temp, 'OR') #以or的方式添加到temp
转载请注来源:https://github.com/fugui0310