2
因此,我從網上轉移我的實例,下載了一個安裝它的密鑰對,一切都運行良好大約一天。然後我重新啓動了我的電腦,並且無法再通過SSH連接。我嘗試redownloading一個密鑰對並重新安裝它,但我不斷獲得權限被拒絕(publickey)錯誤。無法SSH進入EC2實例
debug1: Reading configuration data /etc/ssh_config
debug1: Applying options for *
debug1: Connecting to ec2-184-73-218-40.compute-1.amazonaws.com [184.73.218.40] port 22.
debug1: Connection established.
debug1: identity file /Users/tigger/.ec2/domainpolish type 1
debug1: identity file /Users/tigger/.ec2/domainpolish-cert type -1
debug1: Remote protocol version 2.0, remote software version OpenSSH_5.3p1 Debian-3ubuntu7
debug1: match: OpenSSH_5.3p1 Debian-3ubuntu7 pat OpenSSH*
debug1: Enabling compatibility mode for protocol 2.0
debug1: Local version string SSH-2.0-OpenSSH_5.6
debug1: SSH2_MSG_KEXINIT sent
debug1: SSH2_MSG_KEXINIT received
debug1: kex: server->client aes128-ctr hmac-md5 none
debug1: kex: client->server aes128-ctr hmac-md5 none
debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent
debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP
debug1: SSH2_MSG_KEX_DH_GEX_INIT sent
debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY
debug1: Host 'ec2-184-73-218-40.compute-1.amazonaws.com' is known and matches the RSA host key.
debug1: Found key in /Users/tigger/.ssh/known_hosts:11
debug1: ssh_rsa_verify: signature correct
debug1: SSH2_MSG_NEWKEYS sent
debug1: expecting SSH2_MSG_NEWKEYS
debug1: SSH2_MSG_NEWKEYS received
debug1: Roaming not allowed by server
debug1: SSH2_MSG_SERVICE_REQUEST sent
debug1: SSH2_MSG_SERVICE_ACCEPT received
debug1: Authentications that can continue: publickey
debug1: Next authentication method: publickey
debug1: Offering RSA public key: /Users/tigger/.ec2/domainpolish
debug1: Authentications that can continue: publickey
debug1: Offering RSA public key: domainpolish
debug1: Authentications that can continue: publickey
debug1: No more authentication methods to try.
Permission denied (publickey).
我也試圖重新啓動實例:
這是詳細輸出(我使用-i /路徑/到/密鑰對的選項,以及沒有運氣嘗試)。任何人都可以嘗試其他任何東西?非常感謝!
我會在亞馬遜論壇上發佈相同的內容:https://forums.aws.amazon.com/forum.jspa?forumID=30 AWS員工有一個絕招,即絕大多數SO用戶不會:他們可以看到發生了什麼事。 :) – Iterator
&然後如果你得到答案,請回答在答案部分它可能是很好的幫助。 –