如何使用 django 根据外键链接页面

Question

我试图根据外键“ckey”将国家与大陆联系起来。 我尝试使用 filter and.raw 但它没有用。 我尝试直接在 HTML 上使用它，但它说它不能赞美它。 我需要知道是否有另一种方法可以使用 Django 链接诸如“大陆-> 国家-> 城市->....”之类的页面。

楷模

from django.db import models

# Create your models here.
class Continents(models.Model):
    ckey = models.CharField(max_length=10, primary_key=True)
    continentName = models.CharField(max_length=50)

class country(models.Model):
    countryNum = models.IntegerField(primary_key=True)
    countryName = models.CharField(max_length=15)
    countryBrief= models.TextField(max_length=500)
    currency = models.CharField(max_length=15)
    cost = models.FloatField(max_length=10)
    cultures = models.TextField(max_length=300)
    rulesBrief = models.TextField(max_length=200)
    location = models.TextField(max_length=500)
    ckey = models.ForeignKey('Continents', on_delete=models.PROTECT)

视图.py

from django.shortcuts import render
from django.http import HttpResponse
from .models import Continents, country

# Create your views here.
def home(request):
    return render(request,"guide/home.html")

def continents(request):
    continentdata = Continents.objects.all()
    return render(request,"guide/Continents.html",{'Continents':continentdata})

def countrylist(request):
    countries = country.objects.all()
    first_person = country.objects.filter(ckey='as45914')
    context = {
        "first_person":first_person,
        "countries":countries,
    }
    return render(request,"guide/countrylist.html",context=context)

html代码

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>Document</title>
</head>
<body>
    {{first_person}}
</body>
</html>

运行代码时收到的奇怪消息

我如何链接这样的页面？ 例如“欧洲 > 英国 > 英国伦敦下的所有城市...”使用 Django。 我一般都知道怎么做，所有国家>所有城市，但不是那样做。

Answer 1

# Get list by Obj
continentObj = Continents.objects.get(ckey='as45914')
countryList = Country.objects.filter(ckey=cObj)

# Get List by Related Attribute
countryList = Country.objects.filter(ckey__ckey='as45914')

---

编辑

@OP 几乎你会使用相关的属性。

假设你有三个模型

class Continents():
    name = charfield

class Country():
    continents = Key(Continents)

class City():
    country = Key(Country)

现在如何获得欧洲所有城市

list_Of_Cities = City.objects.filter(country__continents__name='Europe')

Answer 2

观点

def countrylist(request):
    list_Of_Cities = country.objects.filter(ckey__ckey='as45914').values('countryName')

    context = {
         "list_Of_Cities":list_Of_Cities
  }
    return render(request,"guide/countrylist.html",context=context)

并在 HTML

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>Document</title>
</head>
<body>

    {% for i in list_Of_Cities %}
    {{i.countryName}}
    {% endfor %}

</body>
</html>