将命令作为参数传递给bash脚本 [英] pass a command as an argument to bash script

问题描述

如何将命令作为参数传递给bash脚本?在以下脚本中，我尝试执行此操作，但是它不起作用！

How do I pass a command as an argument to a bash script? In the following script, I attempted to do that, but it's not working!

#! /bin/sh

if [ $# -ne 2 ]
then
    echo "Usage: $0 <dir> <command to execute>"
    exit 1;
fi;

while read line
do
    $($2) $line
done < $(ls $1);

echo "All Done"

此脚本的示例用法为

./myscript thisDir echo

执行上面的调用应该回显 thisDir 目录中所有文件的名称.

Executing the call above ought to echo the name of all files in the thisDir directory.

推荐答案

您的命令"echo"命令已从$ line中的子段隐藏"在子外壳中.

your command "echo" command is "hidden" inside a sub-shell from its argments in $line.

我想我理解您尝试使用 $($ 2)进行的操作，但是它可能会过大，除非不是全部，所以

I think I understand what your attempting in with $($2), but its probably overkill, unless this isn't the whole story, so

 while read line ; do
    $2 $line
 done < $(ls $1)

应与 thisDir echo 一起用于您的示例.如果您确实需要cmd替换和subshel​​l，请给您输入参数，以便他们可以互相看到:

should work for your example with thisDir echo. If you really need the cmd-substitution and the subshell, then put you arguments so they can see each other:

   $($2 $line)

正如D.S.所提到的，您可能需要 eval 之前的任何一个.

And as D.S. mentions, you might need eval before either of these.

IHTH

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