使用python通过sftp上传文件
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/33751854/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Upload file via sftp with python
提问by user781486
I wrote a simple code to upload a file to a sftp server in python. I am using python 2.7
我写了一个简单的代码将文件上传到python中的sftp服务器。我正在使用 python 2.7
import pysftp
srv = pysftp.Connection(host="www.destination.com", username="root",
password="password",log="./temp/pysftp.log")
srv.cd('public') #chdir to public
srv.put('C:\Users\XXX\Dropbox\test.txt') #upload file to nodejs/
# Closes the connection
srv.close()
The file did not appear on the server. However, no error message appeared. What is wrong with the code?
该文件未出现在服务器上。但是，没有出现错误消息。代码有什么问题？
EDIT: I have enabled logging. I discovered that the file is uploaded to the root folder and not under public folder. Seems like srv.cd('public')did not work.
编辑：我已启用日志记录。我发现文件上传到根文件夹而不是公共文件夹下。好像没用srv.cd('public')。
采纳答案by user781486
I found the answer to my own question.
我找到了我自己问题的答案。
import pysftp
srv = pysftp.Connection(host="www.destination.com", username="root",
password="password",log="./temp/pysftp.log")
with srv.cd('public'): #chdir to public
srv.put('C:\Users\XXX\Dropbox\test.txt') #upload file to nodejs/
# Closes the connection
srv.close()
Put the srv.putinside with srv.cd
把srv.put里面放srv.cd
回答by Amit Ghosh
import pysftp
with pysftp.Connection(host="www.destination.com", username="root",
password="password",log="./temp/pysftp.log") as sftp:
srv.cwd('/root/public'): #Write the whole path
srv.put('C:\Users\XXX\Dropbox\test.txt') #upload file to nodejs/
No srv.close()as connection closed automatically at the end of the with-block
否，srv.close()因为在 with-block 结束时连接自动关闭
I did a minor change with cdto cwd
我做了一个小的改动cd，以cwd
Syntax -
句法 -
# sftp.put('/my/local/filename') # upload file to public/ on remote
# sftp.get('remote_file') # get a remote file
