<!--upload page-->
<form method="POST" enctype="multipart/form-data"
action="test.rpy">
<input type="file"
name="filename">
<input type="submit">
</form>
#rpy script
from twisted.web.resource import Resource
class
test(Resource):
def
render_POST(self,request):
f = open("/root/basic_daemon/htdocs/test/test.txt","wb") #absolute
path,the path must use chmod modify
authorization
f.write(request.args.get('filename')[0])
f.close()
return "<html><body>" + 'success' +
"</body></html>"
resource = test()
----by Leon----
----- Original Message -----
Sent: Monday, November 29, 2004 4:56
PM
Subject: FW: [Twisted-Python] How to use
.rpy do upload ?
If you got answer.
Why not answer yourself onto the
post?
singuan
I am a beginner,how can use .rpy do upload
???