[django]django models最佳实战

models

class People(models.Model):

    name = models.CharField(max_length=30)

    age = models.CharField(max_length=30)

    def __str__(self):

        return self.name

insert into app03_people (name,age) values('m1','1');

查出最小的age

使用annotate(Min('age')出错

In [1]: from django.db.models import Max,Min,Sum,Avg,Count

In [2]: from app03.models import People

In [3]: People.objects.all()

Out[3]: <QuerySet [<People: m1>, <People: m2>, <People: m3>, <People: m4>]>

In [6]: People.objects.annotate(Min('age'))

Out[6]: <QuerySet [<People: m1>, <People: m2>, <People: m3>, <People: m4>]>

In [10]: print(People.objects.annotate(Min('age')).query)

SELECT `app03_people`.`id`, `app03_people`.`name`, `app03_people`.`age`, MIN(`app03_people`.`age`) AS `age__min` FROM `app03_people` GROUP BY `app03_people`.`id` ORDER BY NULL

小结: People.objects.annotate(Min('age'))默认group by id;

使用aggregate完美解决

In [13]: print(People.objects.aggregate(Min('age'))) # 主这里无法query, 因为他返回的是字典类型.

{'age__min': '1'}

等同的sql,(聚合函数可以单独使用)

mysql> SELECT  MIN(`app03_people`.`age`) AS `age__min` FROM `app03_people`;

+----------+

| age__min |

+----------+

| 1        |

+----------+

1 row in set (0.00 sec)

查询age最大最小

aggregate解决

In [15]: People.objects.aggregate(Min('age'),Max('age'))

Out[15]: {'age__min': '1', 'age__max': '4'}

等价的sql

mysql> SELECT  MIN(`app03_people`.`age`) AS `age__min`,MAX(`app03_people`.`age`) AS `age__max` FROM `app03_people`;

+----------+----------+

| age__min | age__max |

+----------+----------+

| 1        | 4        |

+----------+----------+

1 row in set (0.00 sec)

annotate分组查询

In [6]: People.objects.values('part').annotate(Min('age'))

Out[6]: <QuerySet [{'part': 'UI', 'age__min': '10'}, {'part': 'python', 'age__min': '30'}, {'part': 'java', 'age__min': '40'}]>

In [7]: print(People.objects.values('part').annotate(Min('age')).query)

SELECT `app03_people`.`part`, MIN(`app03_people`.`age`) AS `age__min` FROM `app03_people` GROUP BY `app03_people`.`part` ORDER BY NULL

默认值显示两个字段(正常)

mysql> SELECT `app03_people`.`part`, MIN(`app03_people`.`age`) AS `age__min` FROM `app03_people` GROUP BY `app03_people`.`part` ORDER BY NULL

    -> ;

+--------+----------+

| part   | age__min |

+--------+----------+

| UI     | 10       |

| python | 30       |

| java   | 40       |

+--------+----------+

3 rows in set (0.00 sec)

In [17]: print(People.objects.values('part').annotate(Min('age'),Max('age')).query);

SELECT `app03_people`.`part`, MIN(`app03_people`.`age`) AS `age__min`, MAX(`app03_people`.`age`) AS `age__max` FROM `app03_people` GROUP BY `app03_people`.`part` ORDER BY NULL

mysql> SELECT `app03_people`.`part`, MIN(`app03_people`.`age`) AS `age__min`, MAX(`app03_people`.`age`) AS `age__max` FROM `app03_people` GROUP BY `app03_people`.`part` ORDER BY NULL;

+--------+----------+----------+

| part   | age__min | age__max |

+--------+----------+----------+

| UI     | 10       | 20       |

| python | 30       | 50       |

| java   | 40       | 40       |

+--------+----------+----------+

3 rows in set (0.00 sec)

In [19]: print(People.objects.values('part').annotate(Min('age')).annotate(Max('age')).query);

SELECT `app03_people`.`part`, MIN(`app03_people`.`age`) AS `age__min`, MAX(`app03_people`.`age`) AS `age__max` FROM `app03_people` GROUP BY `app03_people`.`part` ORDER BY NULL

mysql> SELECT `app03_people`.`part`, MIN(`app03_people`.`age`) AS `age__min`, MAX(`app03_people`.`age`) AS `age__max` FROM `app03_people` GROUP BY `app03_people`.`part` ORDER BY NULL;

+--------+----------+----------+

| part   | age__min | age__max |

+--------+----------+----------+

| UI     | 10       | 20       |

| python | 30       | 50       |

| java   | 40       | 40       |

+--------+----------+----------+

3 rows in set (0.00 sec)

以2个字段为准 分类

model

In [157]: print ChainLog.objects.values('src_svc_id', 'dst_svc_id').annotate(

     ...:         is_success_total_count=Count('is_success')).query

SELECT `home_application_chainlog`.`src_svc_id`, `home_application_chainlog`.`dst_svc_id`, COUNT(`home_application_chainlog`.`is_success`) AS `is_success_total_count` FROM `home_application_chainlog` GROUP BY `home_application_chainlog`.`src_svc_id`, `home_application_chainlog`.`dst_svc_id` ORDER BY NULL

mysql> SELECT `home_application_chainlog`.`src_svc_id`, `home_application_chainlog`.`dst_svc_id`, COUNT(`home_application_chainlog`.`is_success`) AS `is_success_total_count` FROM `home_application_chainlog` GROUP BY `home_application_chainlog`.`src_svc_id`, `home_application_chainlog`.`dst_svc_id` ORDER BY NULL;

+------------+------------+------------------------+

| src_svc_id | dst_svc_id | is_success_total_count |

+------------+------------+------------------------+

| 1          | 1          |                      2 |

| 1          | 2          |                      1 |

| 2          | 3          |                      1 |

| 2          | 4          |                      1 |

| 3          | 5          |                      1 |

+------------+------------+------------------------+

5 rows in set (0.00 sec)

如果是py字典,以2个字段为维度,不好统计

total = [

    {'from':1,'to':2,'is_success':1},

    {'from':1,'to':3,'is_success':0},

    {'from':1,'to':4,'is_success':1}

]

faild = [

    {'from':1,'to':2,'is_success':1},

    {'from':1,'to':3,'is_success':0},

    {'from':1,'to':4,'is_success':1}

]