Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 146083
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 11, 20262026-05-11T08:32:18+00:00 2026-05-11T08:32:18+00:00

I want to create a simple file upload form and I must be completely

  • 0

I want to create a simple file upload form and I must be completely incapable. I’ve read docs and tutorials,but for some reason, I’m not getting the submitted form data. I wrote the smallest amount of code I could to test and it still isn’t working. Any ideas what’s wrong?

def index():     html = '''     <html>       <body>       <form id='fileUpload' action='./result' method='post'>         <input type='file' id='file'/>         <input type='submit' value='Upload'/>       </form>       </body>     </html>     '''     return html  def result(req):     try: tmpfile = req.form['file']     except:         return 'no file!' 
  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. 2026-05-11T08:32:18+00:00Added an answer on May 11, 2026 at 8:32 am

    try putting enctype=’multipart/form-data’ in your form tag. Your mistake is not really mod_python related.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I want to create a simple form for users to upload a file, which
I want to create a simple batch file that would perform some Visual Studio
I want to create a simple file upload server. The server should be able
I want to create a simple user registration form with First / Last name,
I want to create simple file viewer. What control should i use to view
So I'm trying to create a simple file transfer method. It's completely working for
I want to create a simple C# application that uploads a text file to
Problem decription: I want to create a file upload screen using JSP. The screen
I want to create a simple site to display records from an xml file
I am trying to create a simple screen where the file selected for upload

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.