how to simply handle the request with python

Question

I'm trying to do some testing in html5 uploading files,here is the simple html codes:

var fileInput=document.getElementById("the-file")

fileInput.addEventListener('change',function(e){

var file=e.target.files[0]

var xhr=new XMLHttpRequest()

xhr.open('post','upload/handlecode',true)

xhr.send(file)
}

I use python -m SimpleHTTPServer to test this section the problem is that I DON'T know how to write the handlecode file to handle the http request,

the handlecode file only save the file in the upload floder,no more complicated tasks

Although I don'k know whether testing in this way is right or not

Can someone give some hints or offer me the right way to test ,Thank you all very much!

Zach Kelling · Accepted Answer · 2011-06-17 07:20:26Z

5

Use bottle:

from bottle import request, route, run

@route('/upload', method='POST')
def do_upload():
    data = request.files.get('data')
    raw = data.file.read()
    filename = data.filename
    return "You uploaded %s (%d bytes)." % (filename, len(raw))

run(host='localhost', port=8080)

answered Jun 17, 2011 at 7:20

Zach Kelling

54.1k15 gold badges112 silver badges108 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

how to simply handle the request with python

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related