Sorry for my delay in response.

>
> What libcurl + pycurl versions are you using?

>>> print pycurl.version_info()
(2, '7.15.0', 462592, 'i486-pc-linux-gnu', 1581, 'GnuTLS/1.2.9', 0,
'1.2.3', ('tftp', 'ftp', 'gopher', 'telnet', 'dict', 'ldap', 'http',
'file', 'https', 'ftps'), None, 0, '0.5.18')

I think I have libcurl 7.9.6 but not exactly sure.

> I can't find any code in pycurl nor libcurl that contain that string.

The actual string is:
"Enter password for in <title> at <hostname>: "
Where <title> is the page / authentication title string, and
<hostname> is the host. There is two spaces between "password" and
"for" suggesting that something was supposed to go there, too.

I've tried setting a dummy username/password in pycurl via

curl.setopt(pycurl.USERPWD, "anonymous:anonymous")

but that doesn't affect it.

On Feb 14, 2006, at 4:54 PM, Daniel Stenberg wrote:

> On Tue, 14 Feb 2006, Brian Whitman wrote:
>
>> I am maintaining an app that crawls a set of web pages using
>> pycurl. I have
>> been recently running into a set of pages that require
>> authentication, and I
>> would like to treat those as 'timed out' or unreadable pages.
>> Instead, the
>> thread that reads those pages is being blocked with a "Enter
>> password for
>> hostname: " request that waits for keyboard input. After enough
>> time all of
>> my threads end up blocking, waiting for a password.
> I can't find any code in pycurl nor libcurl that contain that string.
>
> --
> Commercial curl and libcurl Technical Support: http://haxx.se/
> curl.html
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--
Brian Whitman. http://variogr.am/
The Echo Nest Corporation
brian_at_echonest.com
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