在特定模式后将所有内容匹配到第一个'/' [英] match everything up to first '/' after a specific pattern

问题描述

尝试在macOS上使用sed regex在bash脚本中执行此操作.我有一个带有外部驱动器目录列表的文件.

Trying to do this within a bash script using sed regex on macos. I have a file with directory listings of an external drive.

/Volumes/WD/A/Some Learning/Learning Is Great/
/Volumes/WD/A/Some Deeper/Learning Is Great/Another Learning Opportunity/
/Volumes/WD/B/More Things Here/Great Learning/
/Volumes/WD/B/More Things/

我想搜索并返回与各种模式匹配的最顶层目录.如果您搜索学习"，则输出应为:

I want to search and return the top-most directory matching various patterns. If you search for 'Learning' the output should be:

/Volumes/WD/A/Some Learning/
/Volumes/WD/A/Some Deeper/Learning Is Great/
/Volumes/WD/B/More Things Here/Great Learning/

推荐答案

这可能对您有用(GNU sed):

This might work for you (GNU sed):

sed -n 's/\(Learning[^/]*\/\).*/\1/p' file

使用-n选项关闭隐式打印.然后使用替换命令及其p标志在成功将字符串Learning替换为下一个出现的/时进行打印，并按\(...\)分组并引用\1(在替换中)为文件的末尾仅由反向引用\1组成.其他所有行将不会打印出来.

Turn off implicit printing by using the -n option. Then using the substitution command and its p flag to print on a successful replacement of the string Learning to the next occurrence of a /, grouped by \(...\) and referred to \1 (in the replacement) to the end of the file by the back reference \1 only. All other lines will not be printed out.

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