How do we do a file upload using Python CGI Programming?

Python Server Side Programming Programming CGI

To upload a file, the HTML form must have the enctype attribute set to multipart/form-data. The input tag with the file type creates a "Browse" button.

Example

<html>
<body>
   <form enctype = "multipart/form-data"
                     action = "save_file.py" method = "post">
   <p>File: <input type = "file" name = "filename" /></p>
   <p><input type = "submit" value = "Upload" /></p>
   </form>
</body>
</html>

Output

The result of this code is the following form −

File: Choose file
Upload

Here is the script save_file.py to handle file upload −

#!/usr/bin/python
import cgi, os
import cgitb; cgitb.enable()
form = cgi.FieldStorage()
# Get filename here.
fileitem = form['filename']
# Test if the file was uploaded
if fileitem.filename:
   # strip leading path from file name to avoid
   # directory traversal attacks
   fn = os.path.basename(fileitem.filename)
   open('/tmp/' + fn, 'wb').write(fileitem.file.read())
   message = 'The file "' + fn + '" was uploaded successfully'
 
else:
   message = 'No file was uploaded'
 
print """\
Content-Type: text/html\n
<html>
<body>
   <p>%s</p>
</body>
</html>
""" % (message,)

If you run the above script on Unix/Linux, then you need to take care of replacing file separator as follows, otherwise on your windows machine above open() statement should work fine.

fn = os.path.basename(fileitem.filename.replace("", "/" ))

Rajendra Dharmkar

Updated on: 09-Sep-2023

2K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started