使用带有条件的ls列出文件，仅使用空格的process/grep文件 [英] List file using ls with a condition and process/grep files that only whitespaces

问题描述

我在一个文件夹中有一个文件列表，其中一些文件的文件名中有空格. 我需要用_替换空格，但首先，我需要列出条件为ls *_[1-4]*[A-c]*的文件.过滤文件后，某些文件具有空白且没有固定位置(前，中间，结束位置). ls命令后如何替换空白?

I have a list of files in a folder which some of the files have spaces in the filename. I need to replace the whitespace with _ but first, i need to list the file with condition ls *_[1-4]*[A-c]* . After filter the files, some of the files have whitespace with no fixed position(front, middle, end position). How can i replace the whitespace after ls command?

推荐答案

您不想处理来自ls .只需循环匹配的文件即可.

You don't want to process the output from ls. Simply loop over the matching files.

for file in *_[1-4]*[A-c]*; do
    # Skip files which do not contain any whitespace
    case $file in *\ *) ;; *) continue;; esac
    echo mv -n "$file" "${file// /_}"
done

echo可以作为保障措施；如果输出看起来正确，则将其取出.

The echo is there as a safeguard; take it out if the output looks correct.

case和替换查找空格(ASCII 32)；如果您还想匹配选项卡，换页等，请进行相应的调整. bash允许使用$[\t ]之类的内容来匹配制表符或空格，但这不能移植到其他Bourne Shell实现中

The case and the substitution looks for a space (ASCII 32); if you also want to match tabs, form feeds, etc, adapt accordingly. bash allows for something like $[\t ] to match a tab or space, but this is not portable to other Bourne shell implementations

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