apache / python：文件上传未按预期工作

Question

I am working on apache/python2.7 and trying a simple file upload.

html

<form enctype="multipart/form-data"  method="POST">
   <p>Upload File: <input type="file" name="file" id="InputFile"></p>
   <p><input type="submit"  name="ws_butt" value="Create Bulk" class="bulk_submit"></p>
</form>

python

request_body_size = int(environ.get('CONTENT_LENGTH', 0))
request_body = environ['wsgi.input'].read(request_body_size)
d = parse_qs(request_body)

If I print out the value of "d", I get as :

{' name': ['"file"', '"ws_butt"\r\n\r\nCreate Bulk\r\n------WebKitFormBoundaryCXKZ4XpBfD6P2BIu--\r\n'], ' filename': ['"s3essentials-jason.csv"\r\nContent-Type: text/csv\r\n\r\n"Time","Note"\r\n"0:11:47","allows multipart upload concurrently']}

I can not access "file" key and save the file somewhere and read the contents. How to do that?

Even if I try using FieldStorage, it is empty

form = cgi.FieldStorage(environ=environ, fp=environ['wsgi.input'])

How to deal with this? Am I missing some part?

Answer 1

Take a look at this tutorial .

So basically if you're using Flask its already taking care of the uploaded files. All you need to do is to save that files somewhere on your server:

if request.method == 'POST':
        file = request.files['file']
        if file:
                file.save('/where/to/save', file.filename)

I've stripped the tutorial example, to make it ease to see what's going on, but, please, refer to tutorial to see how to secure filename.