elozin
March 24, 2016, 11:55am
1
Hi!
I receive an error:
The client lacks sufficient authorization :: Error parsing key authorization file: Invalid key authorization: 6 parts .
But when i create a file manual, all is given to normal.
example/public_html/.well-known/acme-challenge [0] # curl 'http://example.ru/.well-known/acme-challenge/testfile ’
test
example/public_html/.well-known/acme-challenge [0] # cat testfile
test
Where is the mistake ?
what command line are you running ? have you got he webroot directory correct ?
elozin
March 24, 2016, 12:47pm
3
Yep, directory is correct. And command line is correctly too. Failed requests 6 out of 1,000.
Without seeing the command line you used, or knowing the domain it becomes very difficult to help.
The most common reason is that the webroot path is wrong
The next is either a folder / file permission issue or a htaccess / redirect issue.
elozin
March 24, 2016, 12:53pm
5
Command like :
" certonly --webroot {0} --text --renew-by-default --email {1} --break-my-certs --agree-tos".format(command,LETSENCRYPT_EMAIL)
I don’t think what reason in this command. Domain 3d-advisor.ru .
http://3d-advisor.ru/.well-known/acme-challenge/testfile isn't returning a plain text file for me, you may need to change your config for that location
< html>< head>< script>function set_cookie(){var now = new Date();var time = now.getTime();time += 19360000 * 1000;now.setTime(time);document.cookie='beget=begetok'+'; expires='+now.toGMTString()+'; path=/';}set_cookie();location.reload();;< /script>< /head>< body>< /body>< /html>
elozin
March 24, 2016, 1:19pm
7
Please tell me that it is for “Invalid key authorization: 6 parts .” ? What the “6 parts”?
elozin
March 24, 2016, 1:20pm
8
And can u tell me what has in logs about it?
What’s the key expected and received ?
You need to turn on debug, and paste your full log for me to tell exactly what the 6 parts were;
You can use the -v, --verbose flag can be used multiple times to incrementally increase the verbosity of output
The key ( token) is effectively a random string for the file name which includes another random string within it ( hence why you need to return that as plain text ). In reality they aren't completely random, but in effect it looks that way.
elozin
March 24, 2016, 1:29pm
10
How i can turn on debug ?
Osiris
March 24, 2016, 1:33pm
11
Add -vv to the command line.