排除文件的grep数组参数 [英] grep array parameter of excluded files

问题描述

我想从我的 grep 命令中排除一些文件，为此我使用了参数：

   -  exclude = excluded_file.ext 
  

阅读我想用排除文件的bash数组：

  EXCLUDED_FILES =（excluded_file.ext）
  

然后将$ {EXCLUDED_FILES}数组传递给 grep ，如：

  grep -Rnimy text--exclude = $ {EXCLUDED_FILES} 
  

如何将数组作为参数传递给 grep 命令？

   
 
 grep -Rni$ {excluded_files [@] /＃/  -  exclude =}my text
  

参数扩展$ {excluded_files [@] /＃/ - exclude =}将扩展到数组的扩展 excluded_files ，每个字段的前缀为 - exclude = 。看：

  $ excluded_files =（'* .txt''* .stuff'）
 $ printf'％s \\\
'$ {excluded_files [@] /＃/  -  exclude =}
 --exclude = *。txt 
 --exclude = *。stuff 
  

I want to exclude some files from my grep command, for this I'm using parameter:

--exclude=excluded_file.ext

To make more easy to read I want to use a bash array with excluded files:

EXCLUDED_FILES=("excluded_file.ext")

Then pass ${EXCLUDED_FILES} array to grep, like:

grep -Rni "my text" --exclude=${EXCLUDED_FILES}

How I can pass an array as parameter to grep command?

解决方案

I guess you're looking for this:

grep -Rni "${excluded_files[@]/#/--exclude=}" "my text"

The parameter expansion "${excluded_files[@]/#/--exclude=}" will expand to the expansion of the array excluded_files with each field prefixed with --exclude=. Look:

$ excluded_files=( '*.txt' '*.stuff' )
$ printf '%s\n' "${excluded_files[@]/#/--exclude=}"
--exclude=*.txt
--exclude=*.stuff

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