Django 中的 Slugfield URL 实现

Question

所以，我在尝试对我的模型中的标题字段进行 slugify 并且仍然拥有 return 正确信息时遇到了一些困难。

目前，如果用户帐户中的列表存在于此正则表达式下，则用户可以关注 url：

url(r'^user/(?P<username>\w+)/list/(?P<listname>\w+)/$', mylistpage, name='lists'),

我面临的问题是用户可以有一个包含空格的列表，但正则表达式将他们的 url 基于他们的列表名称。我想实现一个 slug url，但仍然让它检索正确的 model/object 信息。

我试图有一个 slug 字段，然后 pre-populate 它基于列表名称，但我不知道这个实现应该如何工作。任何有见识的人都提前表示感谢。

型号

class newlist(models.Model):


    user = models.ForeignKey(User)
    list_name = models.CharField(max_length = 100,)
    picture = models.ImageField(upload_to='profiles/', default = "/media/profiles/default.jpg")
    slugurl = models.SlugField(default = slugurl(self))

    def __str__(self):
        return self.list_name

    def slugurl(self):
        return slugify(self.list_name)

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def mylistpage(request, username, listname):

    context = RequestContext(request)

    #make sure that the user is authenticated
    if username == request.user.username:
        #If the user is authenticated, then perform the following functions to the page
        if request.user.is_authenticated():
            #Store the current user request object into a variable
            user = User.objects.get(username=username)

            #Store the list name to the item that starts with the url input
            listname = request.user.newlist_set.filter(list_name__iexact=listname)

            listitems = request.user.newlist_set.all()
            if not listname:
                return redirect('/notfound')
    else:
        return redirect('/notfound')

    return render_to_response('listview.html', {'lista': listname}, context)

Answer 1

我用过django-autoslug to great success. You can find a live example here.

SlugField 只是一个带有 little syntactic sugar.

的字符字段

您需要将您的 slug 命名为 slug 以便 django 可以在 URL 分辨率中自动找到它并将正确的参数传递给视图。

您修改后的代码如下所示：

from autoslug import AutoSlugField
from django.db import models

class Newlist(models.Model): # Classes start with uppercase names by default
    user = models.ForeignKey(User)
    list_name = models.CharField(max_length = 100,)
    picture = models.ImageField(upload_to='profiles/', default = "/media/profiles/default.jpg")
    slug = AutoSlugField(populate_from='list_name')
    def __str__(self):
        return self.list_name

您的观点：

def mylistpage(request,username, slug):

    context = RequestContext(request)

    #make sure that the user is authenticated
    if username == request.user.username:
        #If the user is authenticated, then perform the following functions to the page
        if request.user.is_authenticated():
            #Store the current user request object into a variable
            user = User.objects.get(username=username)

            #Store the list name to the item that starts with the url input
            listname = request.user.newlist_set.filter(slug=slug)

            listitems = request.user.newlist_set.all()
            if not listname:
                return redirect('/notfound')
    else:
        return redirect('/notfound')

    return render_to_response('listview.html', {'lista': listname}, context)

urls.py

url(r'^user/(?P<username>\w+)/list/(?P<slug>[\w-]+)/$', mylistpage, name='lists'),

Django 中的 Slugfield URL 实现

Slugfield URL implementation in Django

django

slug

型号

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