python - 为什么我收到无反向匹配错误

Question

您好，正在使用 Django 在一个新站点上工作。似乎无法弄清楚这个错误。首先，预期站点的结构如下:
主页 - 显示不同的语言，并带有指向您可以学习这些语言的国家/地区的链接。
单击主页上的国家/地区 -> 国家/地区页面
国家/地区页面将列出在该国家/地区教授该语言的所有学校。
单击国家页面上的学校 -> 学校页面。
我发现 Country 页面中的这行代码导致了这个问题，但我不知道如何解决它

<a href="{% url 'schools' schools.id %}">
    <div class="left"><h4>Test Link</h4></div>
</a>

这是我在浏览器中得到的错误

NoReverseMatch at /5/ Reverse for 'schools' with arguments '('',)' not found. 1 pattern(s) tried: ['(?P<country_id>[0-9]+)/(?P<schools_id>[0-9]+)/$']

项目中的代码:
模型.py

from django.db import models

class Languages(models.Model):
    language = models.CharField(max_length=100, blank=True, null=True)
    country = models.ManyToManyField('Country', blank=True)
    image = models.CharField(max_length=100, blank=True)

    def __str__(self):
        return self.language

class Country(models.Model):
    country_name = models.CharField(max_length=50)
    country_image = models.CharField(max_length=100, blank=True)
    country_city = models.ManyToManyField('City', blank=True)

    def __str__(self):
        return self.country_name

class City(models.Model):
    city_name = models.CharField(max_length=50)
    city_image = models.CharField(max_length=1000, blank=True)
    city_schools_in = models.ManyToManyField('Schools', blank=True)

    def __str__(self):
        return self.city_name

class Schools(models.Model):
    school_name = models.CharField(max_length=100)
    school_image = models.CharField(max_length=1000, blank=True)
    school_country = models.ManyToManyField('Country', blank=True)
    school_city = models.ManyToManyField('City', blank=True)
    school_description = models.TextField()
    school_accreditations = models.ManyToManyField('Accreditations', blank=True)
    school_lpw = models.IntegerField()
    school_cs = models.IntegerField()
    school_course = models.ManyToManyField('CoursePrices', blank=True)

    def __str__(self):
        return self.school_name

class Accreditations(models.Model):
    accreditation_name = models.CharField(max_length=50)

    def __str__(self):
        return self.accreditation_name

class CoursePrices(models.Model):
    course_title = models.CharField(max_length=50, blank=True)
    lessons_per_week = models.IntegerField()
    class_size = models.IntegerField()
    weekly_price =  models.IntegerField()
    language_level = models.CharField(max_length=50, blank=True)

    def __str__(self):
        return self.course_title

View .py

from django.shortcuts import render
from django.http import Http404

from .models import Schools, CoursePrices, Languages, Country, City


def home(request):
    languages = Languages.objects.all()
    return render(request, 'home.html', {
        'languages': languages,
    })

def country(request, country_id):
    try:
        country = Country.objects.get(id=country_id)
    except Languages.DoesNotExist:
        raise Http404('Language not found')
    return render(request, 'country.html', {
        'country': country,
    })

def schools(request, country_id, schools_id):
    try:
        schools = Schools.objects.get(id=schools_id)
    except Schools.DoesNotExist:
        raise Http404('school not found')
    return render(request, 'schools.html', {
        'schools': schools,
    })

网址.py

from django.contrib import admin
from django.urls import path

from schools import views

urlpatterns = [
    path('admin/', admin.site.urls),
    path('', views.home, name='home'),
    path('<int:country_id>/', views.country, name='country'),
    path('<int:country_id>/<int:schools_id>/', views.schools, name='schools'),
]

项目结构截图

任何帮助将不胜感激。还请原谅发布或编码中的任何错误，对于编码和堆栈溢出/一般来说仍然很陌生:)

Answer 1

您的路径名为 schools需要两个参数，您只传递了一个，因此无法匹配任何路由。
您对 Rafael 的回答(即使提供了两个参数)仍然不匹配路径的评论是因为 urls.py 中路径的顺序文件。
Django 会按顺序遍历路径列表，因为路径只有一个参数 country_id首先找到它尝试但失败，因为您有第二个参数。
要解决这个问题，请更改路径的顺序

urlpatterns = [
    path('admin/', admin.site.urls),
    path('', views.home, name='home'),
    # next two lines have been swapped. 
    path('<int:country_id>/<int:schools_id>/', views.schools, name='schools'),
    path('<int:country_id>/', views.country, name='country'),
]