Upload Files to Web Server

Demonstrates how to upload files from an application to a web server. Note: There must be something provided on the server-side to consume the upload. It could be an ASP or ASP.NET script, Perl, C++ CGI, PHP, Ruby, etc,. However, uploading via HTTP is not the same as uploading via FTP where the FTP server handles receiving the file and saving to disk.

The C# code behind http://www.chilkatsoft.com/receiveUpload.aspx may be found at this URL: C# to Receive Upload in ASP.NET.

Downloads for Windows/Linux and Install Instructions

require 'rubygems'
require 'chilkat'


http = Chilkat::CkHttp.new()

#  Any string unlocks the component for the 1st 30-days.
success = http.UnlockComponent("Anything for 30-day trial")
if (success != true)
    print http.lastErrorText() + "\n"
    exit
end

#  Build a HTTP request with the files to be uploaded:
req = Chilkat::CkHttpRequest.new()
req.UseUpload()

#  The URL we'll be posting to is:
#  http://www.chilkatsoft.com/receiveUpload.aspx
#  Therefore, the path part of the URL is:
req.put_Path("/receiveUpload.aspx")

#  Note: You may test uploads to this URL, but anything over 80K will be aborted on the receiving side.

#  Add some files to the request:
#  The 1st argument is an arbitrary name.  It's the POST form field name.
#  The 2nd argument is the filename currently existing on
#  the local filesystem.  It may include an absolute or relative
#  path, or no path at all if it's in the current working directory.
success = req.AddFileForUpload("file1","hamlet.xml")
if (success != true)
    print req.lastErrorText() + "\n"
    exit
end

success = req.AddFileForUpload("file2","dude.gif")
if (success != true)
    print req.lastErrorText() + "\n"
    exit
end

#  Send the HTTP POST and get the response.  Note: This is a blocking call.
#  The method does not return until the full HTTP response is received.

domain = "www.chilkatsoft.com"
port = 80
ssl = false

resp = http.SynchronousRequest(domain,port,ssl,req)
if (resp == nil )
    print http.lastErrorText() + "\n";
else
    #  Display the HTML source of the page returned.
    print resp.bodyStr() + "\n";
end