我终于弄清楚发生了什么。我发现该--manual
标志以交互方式逐步完成身份验证过程。
流程中的每个阶段都会显示类似于以下内容的提示:
Make sure your web server displays the following content at
http://www.example.net/.well-known/acme-challenge/twJCKQm9SbPEapgHpyU5TdAR1ErRaiCyxEB5zhhw0w8 before continuing:
twJCKQm9SbPEapgHpyU5TdAR1ErRaiCyxEB5zhhw0w8.t7J7DDTbktMGCCu2KREoIHv1zwkvwGfJTAkJrnELb4U
If you don't have HTTP server configured, you can run the following
command on the target server (as root):
mkdir -p /tmp/letsencrypt/public_html/.well-known/acme-challenge
cd /tmp/letsencrypt/public_html
printf "%s" twJCKQm9SbPEapgHpyU5TdAR1ErRaiCyxEB5zhhw0w8.t7J7DDTbktMGCCu2KREoIHv1zwkvwGfJTAkJrnELb4U > .well-known/acme-challenge/twJCKQm9SbPEapgHpyU5TdAR1ErRaiCyxEB5zhhw0w8
# run only once per server:
$(command -v python2 || command -v python2.7 || command -v python2.6) -c \
"import BaseHTTPServer, SimpleHTTPServer; \
s = BaseHTTPServer.HTTPServer(('', 80), SimpleHTTPServer.SimpleHTTPRequestHandler); \
s.serve_forever()"
Press ENTER to continue
正如我发现的那样,尽管该进程本身以 root 身份运行,但它没有启动挑战服务器本身的权限。当然,这可能是 API 中的一个错误。
在提示符下直接运行脚本会产生如下错误:
$(command -v python2 || command -v python2.7 || command -v python2.6) -c \
> "import BaseHTTPServer, SimpleHTTPServer; \
> s = BaseHTTPServer.HTTPServer(('', 80), SimpleHTTPServer.SimpleHTTPRequestHandler); \
> s.serve_forever()"
Traceback (most recent call last):
File "<string>", line 1, in <module>
File "/usr/lib/python2.7/SocketServer.py", line 419, in __init__
self.server_bind()
File "/usr/lib/python2.7/BaseHTTPServer.py", line 108, in server_bind
SocketServer.TCPServer.server_bind(self)
File "/usr/lib/python2.7/SocketServer.py", line 430, in server_bind
self.socket.bind(self.server_address)
File "/usr/lib/python2.7/socket.py", line 224, in meth
return getattr(self._sock,name)(*args)
socket.error: [Errno 13] Permission denied
但是以 root 身份运行它(如提示本身所述)正确启动了服务器,并且可以在外部服务器查询它以完成挑战时对其进行监视:
sudo $(command -v python2 || command -v python2.7 || command -v python2.6) -c "import BaseHTTPServer, SimpleHTTPServer; \
s = BaseHTTPServer.HTTPServer(('', 80), SimpleHTTPServer.SimpleHTTPRequestHandler); \
s.serve_forever()"
66.133.109.36 - - [08/Jan/2016 21:25:10] "GET /.well-known/acme-challenge/SZ88SorxBGXBtSZCTn4FX2g7u5XjnPFOOV3f5S5DuXB HTTP/1.1" 200 -
66.133.109.36 - - [08/Jan/2016 21:25:10] "GET /.well-known/acme-challenge/twJCKQm9SbPEapgHpyU5TdAR1ErRaiCyxEB5zhhw0w8 HTTP/1.1" 200 -
这个错误需要一段时间才能诊断出来,因为很多事情本可以阻止挑战失败,并且生成的服务器在后台默默地失败。