如何修复 Django：没有返回 HttpResponse 对象。它返回 None 而不是？ [复制]

Posted 2023-02-26

【中文标题】如何修复 Django：没有返回 HttpResponse 对象。它返回 None 而不是？ [复制]

【英文标题】：How to fix Django : didn't return an HttpResponse object. It returned None instead? [duplicate]

【发布时间】：2019-02-15 12:19:56

【问题描述】：

我使用过 python3.7 和 django 2.1，对于 django 应用程序模型和视图目录有单独的文件夹和文件。

项目名称：融合 应用名称：admin_lte

出现以下错误

/csv/ 处的值错误 视图 django_adminlte.views.file_csv_view.file_csv_show.file_csv_show 没有返回 HttpResponse 对象。它返回 None 。

Traceback :

Traceback (most recent call last):
  File "C:\Users\parth\Desktop\admin_lte\lib\site-packages\django\core\handlers\exception.py", line 34, in inner
    response = get_response(request)
  File "C:\Users\parth\Desktop\admin_lte\lib\site-packages\django\core\handlers\base.py", line 137, in _get_response
    "returned None instead." % (callback.__module__, view_name)
ValueError: The view django_adminlte.views.file_csv_view.file_csv_show.file_csv_show didn't return an HttpResponse object. It returned None instead.
[15/Feb/2019 17:23:57] "POST /csv/ HTTP/1.1" 500 67126

我已经为单独的文件夹制作了所有 init.py 文件。

View file : file_csv_show.py


   from django.shortcuts import render,redirect
   from django_adminlte.forms import File_csvForm
   from django_adminlte.models import file_csv

def file_csv_show(request):
    if request.method == 'POST':
        form = File_csvForm(request.POST,request.FILES) #request.POST = geting title/auth request.FILES = for book upload
        if form.is_valid():
            try:
                form.save()
                return redirect("file_csv_show")
            except:
                pass
    else:
        form = File_csvForm()
        return render(request,'adminlte/file_csv_index.html','form' : form)

models file : file_csv.py

from django.db import models

# Create your models here.
class File_csv(models.Model):
    title = models.CharField(max_length=50)
    csv_files = models.FileField(upload_to='files/csv')

    class Meta:
        db_table = "csv"

    def __str__(self):
        return self.title

html file

% extends 'adminlte/base.html' %
% load crispy_forms_tags %
% block content %

% if user.is_authenticated %
    <h2>Upload</h2>
    <form method="post" enctype="multipart/form-data" action="% url "file_csv" %">
        % csrf_token %
        <input type="file" name="document">
        <button type="submit">Upload File</button>
    </form>
    % if url %
        <p>Uploaded File : <a href="url">url</a></p>
    % endif %
% endif %
% endblock %

urls.py file

from django_adminlte import views
from django_adminlte.views.file_csv_view import file_csv_show

urlpatterns = [
      path('csv/',views.file_csv_view.file_csv_show,name='file_csv'),
]



I want to when file has been uploaded and after page redirect to index page

【问题讨论】：

发布要重定向到的索引函数及其网址

@ans2human file_csv_show.py 是视图文件，包含索引函数

请不要做except: pass。事实上，无论如何都没有理由尝试/除了那里。

【参考方案1】：

你的观点有问题。当表单无效时，它不会返回任何内容。

def file_csv_show(request):
    if request.method == 'POST':
        form = File_csvForm(request.POST,request.FILES) #request.POST = geting title/auth request.FILES = for book upload
        if form.is_valid():
            try:
                form.save()
                return redirect("file_csv_show")
            except:
                # If the form is invalid, 
                pass
        # This part is missing:
        # Here you need an else for when the form is invalid.
        else:
            return render(request,'adminlte/file_csv_index.html','form' : form)
    else:
        form = File_csvForm()
        return render(request,'adminlte/file_csv_index.html','form' : form)