如何从Bash中的文件列表中隐藏多个文件? [英] How to cat multiple files from a list of files in Bash?

问题描述

我有一个保存文件列表的文本文件.我想一起cat他们的内容.做这个的最好方式是什么?我正在做这样的事情，但看起来太复杂了:

I have a text file that holds a list of files. I want to cat their contents together. What is the best way to do this? I was doing something like this but it seems overly complex:

let count=0
while read -r LINE
do
    [[ "$line" =~ ^#.*$ ]] && continue
    if [ $count -ne 0 ] ; then
        file="$LINE"
        while read PLINE
        do
            echo $PLINE | cat - myfilejs > /tmp/out && mv /tmp/out myfile.js
        done < $file
    fi
    let count++
done < tmp

我当时跳过了注释行并遇到了问题.必须有一个更好的方法来做到这一点，而无需两个循环.谢谢！

I was skipping commented lines and running into issues. There has to be a better way to do this, without two loops. Thanks!

推荐答案

或通过简单的命令

cat $(grep -v '^#' files) > output

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