[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}]
```