[英]Run s3cmd sync via python script & cron
我几天来一直在努力解决这个问题,希望能有所帮助-
基本上,我编写了以下Python脚本
import os, sys
# =__=__=__=__=__=__=__ START MAIN =__=__=__=__=__=__=__
if __name__ == '__main__':
# initialize variables
all_files = []
# directory to download data siphon files to
dDir = '/path/to/download/directory/'
# my S3 bucket
s3bucket = "com.mybucket/"
foldername = "test"
# get a list of available feeds
feeds = <huge JSON object with URLs to feeds>
for item in range(feeds['count']):
# ...check if the directory exists, and if not, create the directory...
if not os.path.exists(folderName):
os.makedirs(folderName)
... ... ...
# Loop through all the splits
for s in dsSplits:
... ... ...
location = requestFeedLocation(name, timestamp)
... ... ...
downloadFeed(location[0], folderName, nameNotGZ)
# THIS IS WHERE I AM HAVING PROBLEMS!!!!!!!!!!!
cmd = 's3cmd sync 'dDir+folderName+'/ s3://'+s3bucket+'/'
os.system(cmd)
代码中的所有内容都可以正常运行...当我从命令行直接运行时,所有内容均按预期运行...但是,当我通过cron执行它时,以下代码不执行(其他所有操作都可以)
# THIS IS WHERE I AM HAVING PROBLEMS!!!!!!!!!!!
cmd = 's3cmd sync 'dDir+folderName+'/ s3://'+s3bucket+'/'
os.system(cmd)
为了回答一些问题,我以root用户身份运行cron,为root用户配置了s3cmd,操作系统是Ubuntu 12.04,python版本是2.7,所有必需的目录都具有读/写权限...
我想念什么?
因此,我希望命令如下所示。
cmd ='s3cmd sync'+ dDir +文件夹名称+'/ s3://'+ s3bucket +'/'
使用os.system(CMD)
在这里,我发现了更多s3cmd同步帮助: http : //tecadmin.net/s3cmd-file-sync-with-s3bucket/
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