Question

我正在尝试远程获取FTP用户的目录结构。我是使用[spur] [1]模块从python脚本执行此操作。我使用cron作业调用此脚本。但我得到以下

Traceback (most recent call last):
  File "/mnt/voylla-staging/releases/20140717193920/voylla_scripts/snapdeal/GetOrders/getOrders.py", line 54, in <module>
    feeds = getFeeds()
  File "/mnt/voylla-staging/releases/20140717193920/voylla_scripts/snapdeal/GetOrders/getOrders.py", line 26, in getFeeds
    result = shell.run(["ls", FEED_LOCATION])
  File "/usr/local/lib/python2.7/dist-packages/spur/ssh.py", line 73, in run
    return self.spawn(*args, **kwargs).wait_for_result()
  File "/usr/local/lib/python2.7/dist-packages/spur/ssh.py", line 83, in spawn
    channel = self._get_ssh_transport().open_session()
  File "/usr/local/lib/python2.7/dist-packages/spur/ssh.py", line 190, in _get_ssh_transport
    raise self._connection_error(error)
spur.ssh.ConnectionError: Error creating SSH connection
Original error: Authentication failed.

如果我在不使用cron的情况下手动运行scipt，它的运行完美！请有人帮忙。

由于

相关守则：

FTP_SERVER = "abc.example.com"
FTP_USER = "root"
FEED_LOCATION = "/home/xyz/abc"
PROCESSED_FEED_LOCATION = "/home/xyz/def"
PREFIX = "alpha"

def getFeeds():                                                                 ####returns the list of feeds in FEED_LOCATION
    shell = spur.SshShell(hostname=FTP_SERVER, username=FTP_USER)
    with shell:
        result = shell.run(["ls", FEED_LOCATION])
        feeds = result.output.decode().split("\n")
    return feeds

feeds = getFeeds()

Answer 1

您必须使用private_key_file选项指定私钥文件的路径。以下是documentation的示例。

# Use a private key
spur.SshShell(
    hostname="localhost",
    username="bob",
    private_key_file="path/to/private.key"
)

无法通过cron作业运行访问EC2服务器

1 个答案: