New submission from chen wu <wuche...@huawei.com>: when I tried to request a url like "https://[fc00:0a08::2]:35357"， I got 400. The code is like: import requests requests.get("https://[fc00:0a08::2]:35357";, verify=False) And the apache logs: vhost.c(889): [client fc00:ac1c::9a5:58692] AH00550: Client sent malformed Host header: [[fc00::0a08::2]]:35357 If user no set "Host" in header, httpslib will pase it from url and set it. The paser function is urllib3.util.url.pase_url. When url is "https://[fc00:0a08::2]:35357";, we got host [fc00:0a08::2]. And then httplib sets host in putrequest, "[" and "]" will be added to [fc00:0a08::2], which is not a valid format for host. The part of codes are: 974 # Wrap the IPv6 Host Header with [] (RFC 2732) 975 if host_enc.find(':') >= 0: 976 host_enc = "[" + host_enc + "]"
maybe the judgement condition for wrap ipv6 host header with [] is not very well? ---------- components: Library (Lib) messages: 324149 nosy: visionwun priority: normal severity: normal status: open title: httplib sets unbefitting "Host" in request header when requests an ipv6 format url. type: behavior versions: Python 2.7 _______________________________________ Python tracker <rep...@bugs.python.org> <https://bugs.python.org/issue34516> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
