问题
I am trying to upload many files at once to my CherryPy server.
I am following this tutorial that shows PHP code on the server side.
The JavaScript part is simple. Here is a summary of what it does:
function FileSelectHandler(e) {
var files = e.target.files || e.dataTransfer.files;
for (var i = 0, f; f = files[i]; i++) {
var xhr = new XMLHttpRequest();
xhr.open("POST", "upload", true);
xhr.setRequestHeader("X_FILENAME", file.name);
xhr.send(file);
}
I translated the upload.php
described in the tutorial into something like this:
def upload(self):
[...]
When the server receives the request I can see that cherrypy.request.headers['Content-Length'] == 5676
which is the length of the file I'm trying to upload, so I assume the whole file has been sent to the server.
How do I get the content of the file?
回答1:
At its minimum it looks like the following. Tested in Firefox and Chromium. If you need to support legacy browsers I'd look at some JavaScript library for polyfills and fallback.
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import os
import shutil
import cherrypy
config = {
'global' : {
'server.socket_host' : '127.0.0.1',
'server.socket_port' : 8080,
'server.thread_pool' : 8,
}
}
class App:
@cherrypy.expose
def index(self):
return '''<!DOCTYPE html>
<html>
<head>
<title>CherryPy Async Upload</title>
</head>
<body>
<form id='upload' action=''>
<label for='fileselect'>Files to upload:</label>
<input type='file' id='fileselect' multiple='multiple' />
</form>
<script type='text/javascript'>
function upload(file)
{
var xhr = new XMLHttpRequest();
xhr.upload.addEventListener('progress', function(event)
{
console.log('progess', file.name, event.loaded, event.total);
});
xhr.addEventListener('readystatechange', function(event)
{
console.log(
'ready state',
file.name,
xhr.readyState,
xhr.readyState == 4 && xhr.status
);
});
xhr.open('POST', '/upload', true);
xhr.setRequestHeader('X-Filename', file.name);
console.log('sending', file.name, file);
xhr.send(file);
}
var select = document.getElementById('fileselect');
var form = document.getElementById('upload')
select.addEventListener('change', function(event)
{
for(var i = 0; i < event.target.files.length; i += 1)
{
upload(event.target.files[i]);
}
form.reset();
});
</script>
</body>
</html>
'''
@cherrypy.expose
def upload(self):
'''Handle non-multipart upload'''
filename = os.path.basename(cherrypy.request.headers['x-filename'])
destination = os.path.join('/home/user', filename)
with open(destination, 'wb') as f:
shutil.copyfileobj(cherrypy.request.body, f)
if __name__ == '__main__':
cherrypy.quickstart(App(), '/', config)
来源:https://stackoverflow.com/questions/26876695/uploading-a-file-in-ajax-to-cherrypy