从查找中排除文件列表 [英] Exclude list of files from find

问题描述

如果在运行 find 时要排除的文本文件中有文件名列表，该怎么办?例如，我想做类似的事情:

If I have a list of filenames in a text file that I want to exclude when I run find, how can I do that? For example, I want to do something like:

find /dir -name "*.gz" -exclude_from skip_files

并获得/dir中的所有.gz文件，但skip_files中列出的文件除外.但是find没有 -exclude_from 标志.如何跳过 skip_files 中的所有文件?

and get all the .gz files in /dir except for the files listed in skip_files. But find has no -exclude_from flag. How can I skip all the files in skip_files?

推荐答案

我不认为 find 有这样的选项，您可以使用 printf 来构建命令和您的排除列表:

I don't think find has an option like this, you could build a command using printf and your exclude list:

find /dir -name "*.gz" $(printf "! -name %s " $(cat skip_files))

与做的相同:

find /dir -name "*.gz" ! -name first_skip ! -name second_skip .... etc

或者，您也可以从 find 到 grep :

find /dir -name "*.gz" | grep -vFf skip_files

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