[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}
]