这是我的views.py文件

from django.shortcuts import render

def home(request):
    return render(request, 'home.html',{'name':'irtiza'})

def add(request):

val1 = int(request.GET['num1'])
val2 = int(request.GET['num2'])
res = val1 + val2

return render(request, "result.html",{'result': res})

当我运行此代码时，我收到了这个错误，

错误

然后我尝试了这段代码：

def add(request):

    val1 = int(request.GET.get(['num1']))
    val2 = int(request.GET.get(['num2']))
    res = val1 + val2

    return render(request, "result.html",{'result': res})

出现此错误。

在此处输入图像描述

URL。py文件

from django.urls import path 
from . import views

urlpatterns = [
    path("",views.home,name="home"),
    path("add", views.add, name="add")
]

home.html档案

{% extends 'base.html' %}

{% block content %}

<h1> hello  {{name}} </h1>

<form action="add">

    {% csrf_token %}

    Enter 1st number : <input type="text" name="num1"><br>
    Enter 2nd number : <input type="text" name="num2"><br>
    <input type="submit">

</form>

{% endblock %}

Result.html档案

{% extends 'base.html' %}

{% block content %}

    Result : {{result}}

{% endblock %}

当我尝试POST功能时，它给出了下面提到的错误。

from django.shortcuts import render

# Create your views here.

def home(request):
    return render(request, 'home.html',{'name':'irtiza'})

def add(request):

    val1 = int(request.POST['num1'])
    val2 = int(request.POST['num2'])
    res = val1 + val2

    return render(request, "result.html",{'result': res})

/add“num1”请求方法处的MultiValueDictKeyError：GET请求URL：http://127.0.0.1:7000/add?csrfmiddlewaretoken=t6mVWboKI37vxW3zDbVmpnB3j3g89EzjFhikDDMrB4qVEfrNyqG5pebeXgDLAFE0

我现在应该怎么做才能解决这个错误。这只是一个简单的加法函数。请帮帮我。