Return-Path: Delivered-To: apmail-jakarta-commons-user-archive@www.apache.org Received: (qmail 26932 invoked from network); 3 Feb 2006 07:59:19 -0000 Received: from hermes.apache.org (HELO mail.apache.org) (209.237.227.199) by minotaur.apache.org with SMTP; 3 Feb 2006 07:59:19 -0000 Received: (qmail 80631 invoked by uid 500); 3 Feb 2006 07:59:14 -0000 Delivered-To: apmail-jakarta-commons-user-archive@jakarta.apache.org Received: (qmail 80612 invoked by uid 500); 3 Feb 2006 07:59:14 -0000 Mailing-List: contact commons-user-help@jakarta.apache.org; run by ezmlm Precedence: bulk List-Unsubscribe: List-Help: List-Post: List-Id: "Jakarta Commons Users List" Reply-To: "Jakarta Commons Users List" Delivered-To: mailing list commons-user@jakarta.apache.org Received: (qmail 80599 invoked by uid 99); 3 Feb 2006 07:59:14 -0000 Received: from asf.osuosl.org (HELO asf.osuosl.org) (140.211.166.49) by apache.org (qpsmtpd/0.29) with ESMTP; Thu, 02 Feb 2006 23:59:14 -0800 X-ASF-Spam-Status: No, hits=-0.0 required=10.0 tests=SPF_PASS X-Spam-Check-By: apache.org Received-SPF: pass (asf.osuosl.org: domain of e.dominguez@ticino.com designates 195.190.166.60 as permitted sender) Received: from [195.190.166.60] (HELO mail1.ticino.com) (195.190.166.60) by apache.org (qpsmtpd/0.29) with ESMTP; Thu, 02 Feb 2006 23:59:13 -0800 X-Spam-Score: 1 Received: from localhost.localdomain (unverified [164.128.109.120]) by ticino.com (Rockliffe SMTPRA 6.1.20) with ESMTP id for ; Fri, 3 Feb 2006 08:58:51 +0100 Date: Fri, 3 Feb 2006 08:58:44 +0100 From: "e.dominguez" To: commons-user@jakarta.apache.org Subject: [NET] ftp requires 2 passwords (a proxy password) Message-Id: <20060203085844.31269f58.e.dominguez@ticino.com> X-Mailer: Sylpheed version 2.2.0beta7 (GTK+ 2.8.6; i686-pc-linux-gnu) Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable X-Virus-Checked: Checked by ClamAV on apache.org X-Spam-Rating: minotaur.apache.org 1.6.2 0/1000/N Hi. I trying to connect to an ftp server through a proxy by a customer. the enterprise has a proxy server, so for the connection I have received th= is 'connection example' (it's in german but i think is understandable): Sobald Sie die UserID und das Passwort f=FCr den Technical Account erhaltet= haben, m=FCssen Sie folgende Syntax anwenden: (Beispiel basiert auf W32 FTP-Client). =20 c:\>ftp proxy.acme.ch Verbindung mit proxy.acme.ch wurde hergestellt. 220 Blue Coat FTP Service Benutzer (proxy.acme.ch:(none)): far-username@ftp.far-server.ch proxy_tech_= userename 331 Enter password. Kennwort: far_password 332 Enter proxy password. Kennwort: proxy_tech_password Hier nochmal die technische Syntax, die angewendet werden soll: Usage: USER username@hostname proxyusername PASS password ACCT proxypassword How can i pass the second password (332/ACCT) as connection property for my= pgm? Thanx. Best regards. e.dominguez --------------------------------------------------------------------- To unsubscribe, e-mail: commons-user-unsubscribe@jakarta.apache.org For additional commands, e-mail: commons-user-help@jakarta.apache.org