ValueError：具有基数10外键Django的int（）的无效文字

I'm learning Django. I have two models as follows:

我正在学习Django。我有两个模型如下:

class IdentificationAddress(models.Model):
    id_ident_address = models.AutoField(primary_key=True)
    ident = models.ForeignKey('Ident', models.DO_NOTHING, db_column='ident')
    address = models.TextField()
    time = models.DateTimeField()

    class Meta:
        managed = False
        db_table = 'identification_address'


class IdentC(models.Model):
    id_ident = models.AutoField(primary_key=True)
    ident = models.TextField(unique=True)
    name = models.TextField()

    class Meta:
        managed = False
        db_table = 'ident'

I am using the Django debugger:

我正在使用Django调试器:

python manage.py shell

and import the model

并导入模型

from invent.models import IdentC as identity
from invent.models import IdentificationAddress as ident_Address

I can access the name of a specific identity

我可以访问特定身份的名称

indentity.objects.filter(ident='00LD').values('name')

it returns

<QuerySet [{'name': u'Joss'}]>

Then I can access information of a specific address:

然后我可以访问特定地址的信息:

ident_Address.objects.filter(address='LOC23').values().last()

it returns

{'times': u'{"2017-07-16"}', u'ident_id': u'00LD', 'address': u'LOC23', 'id_ident_address': 217}

but I get an error when I try to use any identity

但是当我尝试使用任何身份时,我收到错误

ident_Address.objects.filter(ident='00LD').values()

ident_Address.objects.filter(ident_id='00LD').values()

error:

ValueError: invalid literal for int() with base 10: '00LD'

If I try this in postgres

如果我在postgres中尝试这个

SELECT * FROM identification_address WHERE ident= '00LD'

I don't have any problem.

我没有任何问题。

I appreciate any help.

我感谢任何帮助。

Thank you in advance NOTE: For the sake of clarity, I've changed the name of the Ident model to IdentC

提前谢谢注意:为了清楚起见,我已将Ident模型的名称更改为IdentC

3 个解决方案

#1

You try to match '00LD' with the field id of Ident.
Or in your first example you wrote :
indentity.objects.filter(ident='00LD').values('name')
'00LD' match with ident field of Ident

您尝试将'00LD'与Ident的字段ID匹配。或者在你写的第一个例子中:indentity.objects.filter(ident ='00LD')。values('name')'00LD'与Ident的ident字段匹配

So you can filter like this :
ident_Address.objects.filter(ident__ident='00LD').values()
First ident represent your model, and the second one, the field of Ident.

所以你可以像这样过滤:ident_Address.objects.filter(ident__ident ='00LD')。values()第一个ident表示你的模型,第二个表示Ident的字段。

If you want to filter ident_Address by Ident name : ident_Address.objects.filter(ident__name='Joss').values()

如果你想通过Ident名称过滤ident_Address:ident_Address.objects.filter(ident__name ='Joss')。values()

#2

In Django, when you want to access some field in ForeignKey table, you should use __ (Double underscore). For your case, it will be

在Django中,当您想要访问ForeignKey表中的某个字段时,您应该使用__(双下划线)。对于你的情况,它会

ident_Address.objects.filter(ident__id_ident='00LD')

#3

Thanks guys. Sorry @Exprator I'm learning how to use the answer field, as well. So I started debugging how Django queries are translated to SQL queries. Then I found the solution:

多谢你们。对不起@Exprator我也在学习如何使用答案字段。所以我开始调试如何将Django查询转换为SQL查询。然后我找到了解决方案:

ident_Address.objects.filter(ident__id_ident__iexact='00LD').values()

Where first ident means field ident in IdentificationAddress model and id_ident field in identC model

其中first ident表示IdentificationAddress模型中的字段ident和identC模型中的id_ident字段

#1