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django怎样获得框架自动定义的自增id字段
django自定义字段类型,实现非主键字段的自增
# -*- encoding: utf-8 -*-from django.db.models.fields import Field, IntegerFieldfrom django.core import checks, exceptionsfrom django.utils.translation import ugettext_lazy as _class AutoIncreField(Field):
description = _("Integer")
empty_strings_allowed = False
default_error_messages = { 'invalid': _("'%(value)s' value must be an integer."),
} def __init__(self, *args, **kwargs):
kwargs['blank'] = True
super(AutoIncreField, self).__init__(*args, **kwargs) def check(self, **kwargs):
errors = super(AutoIncreField, self).check(**kwargs) # 每张表只能设置一个字段为自增长字段，这个字段可以是主键，也可以不是主键，如果不是主键，则必须设置为一种“键(key)”
# (primary key)也是键(key)的一种，key还包括外键(foreign key)、唯一键(unique key)
errors.extend(self._check_key()) return errors def _check_key(self):
if not self.unique: return [
checks.Error( 'AutoIncreFields must set key(unique=True).',
obj=self,
id='fields.E100',
),
] else: return [] def deconstruct(self):
name, path, args, kwargs = super(AutoIncreField, self).deconstruct() del kwargs['blank']
kwargs['unique'] = True
return name, path, args, kwargs def get_internal_type(self):
return "AutoIncreField"
def to_python(self, value):
if value is None: return value try: return int(value) except (TypeError, ValueError): raise exceptions.ValidationError(
self.error_messages['invalid'],
code='invalid',
params={'value': value},
) def db_type(self, connection):
return 'bigint AUTO_INCREMENT'
def rel_db_type(self, connection):
return IntegerField().db_type(connection=connection) def validate(self, value, model_instance):
pass
def get_db_prep_value(self, value, connection, prepared=False):
if not prepared:
value = self.get_prep_value(value)
value = connection.ops.validate_autopk_value(value) return value def get_prep_value(self, value):
value = super(AutoIncreField, self).get_prep_value(value) if value is None: return None
return int(value) def contribute_to_class(self, cls, name, **kwargs):
assert not cls._meta.auto_field, "A model can't have more than one AutoIncreField."
super(AutoIncreField, self).contribute_to_class(cls, name, **kwargs)
cls._meta.auto_field = self def formfield(self, **kwargs):
return None
Django使用save新增一条数据之后如何获取自增id
back = models.Back.objects.get(bookname=bookname) # 查询你保存的那本书的记录
print(back.id) # 打印出你要的新增记录的id值
Django 插入数据后，获取该数据id的方法
'''
class Car(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=100)
top_speed = models.IntegerField()
'''
'''
car = Car(name=car_name, top_speed=top_speed)
try:
car.save()
response = json.dumps([{ 'Success': 'Car added successfully!', 'id': car.id}])
'''
关于django中取得当前登录用户id的问题
Django的做法，是在artivle里面定义一个指向user的外键，这样就不用获得用户id，直接把用户对象赋值给那个外键属性就行了。
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