Django ForeignKey显示PostgreSQL的数据 [英] Django ForeignKey show data from PostgreSQL

问题描述

我在PostgreSQL中有一个数据库，它连接2个不同的表，让它们调用它们 TBL1 （TBL1是所有类型的表）和 TBL2 （TBL2是一个表例如汽车）， TBL1 和 TBL2 具有名称为类型的列，因此我已经在它们之间创建了外键。因为我不知道这是否与Django有任何关系...

所以，现在在模型中，我创建了（基于表单）变量，名为 type = models.ForeignKey（'TBL1'，db_column ='type'，related_name ='type'），它返回给我Django对象模型ForeignKey，我猜可以。我想以选择的形式显示，我不知道如何从 TBL1 获取数据。有没有办法这样做，因为我已经搜索了2-3天，并尝试其他选项，但是他们没有做到这一点？我已经发现，例如

  STATUS_CHOICES =（
（'new'，_（'New'）），
 $（$ purchase $）
 
 status = models.CharField（max_length = 200，default ='new'，choices = STATUS_CHOICES）
  

在表单中显示选择：

PS

Django版本是1.5.4和Python 2.7，我从来没有在Django或PostgreSQL中工作，所以任何帮助将不胜感激。感谢

解决方案

所以在做了一些更多的研究后，我偶然发现了链接，其中很好地解释了如何在admin.py中编辑内联选项，尽管我正在等待时间尝试添加来自model.py的数据

  class FooForm（forms.ModelForm）：
 class Meta：
 model = Foo 
 
 def __init __（self，* args，** kwargs）：
 super（FooForm，self）.__ init __（* args，** kwargs）
 current_state = self.instance.state 
 ...根据当前状态构造available_choices ... 
 self.fields ['state']。choices = available_choices 
  

它的工作就像我希望工作：）

所以感谢很多人尝试帮助，我不是python / django dev，所以这一切都不符合我的语法：）

最好的所有

i have a db in PostgreSQL which is connecting 2 different tables lets call them TBL1 ( TBL1 is a table for all types ) and TBL2( TBL2 is a table for e.g. cars ), TBL1 and TBL2 have a same column named type, so i have created the foreign key between them. Because i'm not sure if that has anything to do with Django...

So, now in model, i have created ( based on the form ) variable called type = models.ForeignKey('TBL1', db_column='type', related_name='type') and it's returning to me Django object model ForeignKey, which i guess is ok. I would like to show that in a form like select and i don't know how to get the data from TBL1. Is there any way to do this, because i have searched for 2-3 days and trying other options, but non of them did the trick ? I have find out that e.g.

STATUS_CHOICES = (
        ('new', _('New')),
        ('purchased', _('Purchased'))
      )

status = models.CharField(max_length=200, default='new', choices=STATUS_CHOICES)

Is showing the select in form :

P.S.

Django version is 1.5.4 and Python 2.7 , i have never worked in Django or PostgreSQL, so any help would be appreciated. Regards

解决方案

So after doing some more research , i have stumbled upon link , where it's nicely explained how you can edit the choices "inline" inside admin.py , although i was waisting time trying to add the data from model.py

class FooForm(forms.ModelForm):
class Meta:
    model = Foo

def __init__(self, *args, **kwargs):
    super(FooForm, self).__init__(*args, **kwargs)
    current_state = self.instance.state
    ...construct available_choices based on current state...
    self.fields['state'].choices = available_choices

And it worked just as i hoped to work :)

So thanks a lot guys for trying to help, i'm not python/django dev, so this all are not similar syntax to me at all :)

Best all

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