Question

我的目标是创建超链接，该超链接将把一个关键字扔进一个views函数中，然后将查询从我的数据库拉到页面上。

目标：按超链接，这将向我查询特定专业。

我试图使用转换器，所以目标是，1是第一步，3是最后一步。这可能吗？

1) Click the hyperlink -> Major = Accounting

2)URL.py 
path(<str:Accounting/, views.Major, name=Major)

3)Views.py

    def Major(request, Accounting):
       major_choice = professor.objects.filter(Major = Accounting)
       return render(request, 'locate/major.html', {'major_choice': major_choice})

注意：我将变量替换为希望包含“ Accounting”的变量，您会在底部views.py内部注意到它称为“ Major”。

Index.html

<a href="{% url 'locate:Major' 'Accounting' %}">Accounting</a>

major.html

<ul>
{% for major in major_choice %}
    <li>{{major.ProfessorName}}</li>
{%endfor%}
</ul>

urls.py

from django.urls import path

from . import views

    urlpatterns = [
        path('', views.index, name='index'),
        path('<str:Major/', views.Major, name='Major')
    ]

models.py

from django.db import models

class professor(models.Model):
    ProfessorIDS = models.IntegerField()
    ProfessorName = models.CharField(max_length=100)
    ProfessorRating = models.DecimalField(decimal_places=2,max_digits=4)
    NumberofRatings = models.CharField(max_length=50)
    Major = models.CharField(max_length=50)

    def __str__(self):
        return self.ProfessorName

views.py

from django.http import HttpResponse
from django.shortcuts import render
from .models import professor


def index(request):
    professors = professor.objects.all()
    return render(request, 'locate/index.html', {'professors': professors})

def Major(request, major):
    major_choice = professor.objects.filter(Major = major)
    return render(request, 'locate/major.html', {'major_choice': major_choice})

Answer 1

请将您的网址路径更新为此：

path('<str:Major>/', views.Major, name='Major')

在您的html中：

 <a href="{% url 'Major' 'Accounting' %}">Accounting</a>

观看次数：

def Major(request, Major):
   ....

Django 2.0 NoReverseMatch：不是注册的名称空间

1 个答案: