bash 使用 cut 从一堆文件名中删除前 n 个字符
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Remove first n character from bunch of file names with cut
提问by Ridalgo
I am using
我在用
ls | cut -c 5-
ls | cut -c 5-
This does return a list of the file names in the format i want them, but doesn't actually perform the action. Please advise.
这确实以我想要的格式返回文件名列表，但实际上并不执行操作。请指教。
回答by Aman
rename -n 's/.{5}(.*)//' *
The -nis for simulating; remove it to get the actual result.
的-n是用于模拟; 删除它以获得实际结果。
回答by armtatoo
you can use the following command when you are in the folder where you want to make the renaming:
当您在要进行重命名的文件夹中时，可以使用以下命令：
rename -n -v 's/^(.{5})//' *
-nis for no action and -vto show what will be the changes. if you are satisfied with the results you can remove both of them
-n是为了不采取行动并-v显示将发生什么变化。如果您对结果感到满意，您可以将它们都删除
rename 's/^(.{5})//' *
回答by legoscia
Something like this should work:
这样的事情应该工作：
for x in *; do
echo mv $x `echo $x | cut -c 5-`
done
Note that this could be destructive, so run it this way first, and then remove the leading echoonce you're confident that it does what you want.
请注意，这可能具有破坏性，因此请先以这种方式运行它，然后在echo您确信它可以满足您的要求时删除前导。
回答by Saumil
If you get an error message saying,
如果您收到一条错误消息，说明
rename is not recognized as the name of a cmdlet
重命名未被识别为 cmdlet 的名称
This might work for you,
这可能对你有用，
get-childitem * | rename-item -newname { [string]($_.name).substring(5) }
