Django URL 来自数据库文件值的路径

Question

我正在尝试创建 'project' 页面，这些页面的路径是使用 {{ project.title }} 值，而不是我当前使用整数的方法。我不太明白如何做到这一点，但感觉我很接近？

Models.py

from django.db import models

# Create your models here.
class Project(models.Model):
    title = models.CharField(max_length=100)
    description = models.TextField()
    technology = models.CharField(max_length=20)
    image = models.FilePathField(path='projects/static/img/')
    live = models.URLField()
    source = models.URLField()

    def __str__(self):
        return self.title

Urls.py

from django.urls import path
from . import views

urlpatterns = [
    path("", views.project_index, name="projects"),
    path("<int:pk>/", views.project_details, name="project_details"),  # PK for Primary Key
]

Views.py

from django.shortcuts import render
from .models import Project

# Create your views here.
def project_index(request):
    projects = Project.objects.all()
    context = {'projects': projects}
    return render(request, 'projects/project_index.html', context)

def project_details(request, pk):
    project = Project.objects.get(pk=pk)
    context = {'project': project}
    return render(request, 'projects/project_details.html', context)

我认为 path("<int:pk>/", 需要成为一个 slug，但我就是不知道如何绑定数据库数据。可能 context = {'project': project}?

目前 url 是 http://127.0.0.1:8000/projects/1/ - I am looking for http://127.0.0.1:8000/projects/EXAMPLE/

谢谢

Answer 1

您必须将 SlugField 添加到您的 models.py 文件：

Models.py

from django.db import models
from django.utils.text import slugify

# Create your models here.
class Project(models.Model):
    title = models.CharField(max_length=100)
    description = models.TextField()
    technology = models.CharField(max_length=20)
    image = models.FilePathField(path='projects/static/img/')
    live = models.URLField()
    source = models.URLField()
    slug = models.SlugField(default="", blank=True, null=False, db_index=True)

    def __str__(self):
        return self.title

    def save(self, *args, **kwargs):
        self.slug = slugify(self.title)
        super().save(*args, **kwargs)

Views.py

from django.shortcuts import render
from .models import Project

# Create your views here.
def project_index(request):
    projects = Project.objects.all()
    context = {'projects': projects}
    return render(request, 'projects/project_index.html', context)

def project_details(request, slug):
    project = Project.objects.get(slug=slug)
    context = {'project': project}
    return render(request, 'projects/project_details.html', context)

Urls.py


from django.urls import path
from . import views

urlpatterns = [
    path("", views.project_index, name="projects"),
    path("<slug:slug>/", views.project_details, name="project_details"),
]

确保运行 makemigrations 然后 migrate.

Answer 2

urls.py

    path("<title>/", views.project_details, name="project_details"),

views.py


def project_details(request, title: str):
    project = Project.objects.filter(title=title).first()
    if project is None:
        titles = list(Project.objects.all().values_list('title', flat=True))
        msg = f'Project(title=title) not found. Exist titles are: {titles}'
        raise Exception(msg)
    ...

Django URL 来自数据库文件值的路径

Django URL Path from DB file value

python

django

django-templates

django-views