从 shell 脚本执行命令失败

Question

I'm trying read a file which contains lines like this:

 Run COMMAND with options "OPTIONS" and arguments "ARGUMENTS"

Then I want to execute this command with given options and arguments. For example I'd like to execute these commands:

 Run pwd with options "" and arguments ""
 Run ls with options "-al" and arguments "$HOME"
 Run ls with options "-al" and arguments "Example: \"strange folder name\""

This is my code

#!/bin/bash

while read -r line
do
 COMMAND=$(echo "$line" | cut -d" " -f 2)
 OPTIONS=$(echo "$line" | cut -d" " -f 5 | tr -d '"')
 ARGUMENTS=$(echo "$line" | cut -d" " -f 8)

 $COMMAND $OPTIONS $ARGUMENTS
 done <$1

First example is working as it should, second one is giving me error ls: cannot access $HOME: No such file or directory' and third one is not storing the name of the folder to $ARGUMENTS correctly.

Answer 1

second one is giving me error ls: cannot access $HOME: No such file or directory'

This is because the folder named $HOME does not exist. I am not talking about the value of $HOME variable, but the string literal. The shell does not execute the parameter expansion in your situation.

third one is not storing the name of the folder to $ARGUMENTS correctly

This is because -f 8 only extract column 8, try -f 8- to extract the 8th column and all the others until the end of line.

You can give a try to this version below:

while read -r line; do
  COMMAND=$(printf "%s" "${line}" | cut -d" " -f 2)
  OPTIONS=$(printf "%s" "${line}" | cut -d" " -f 5 | tr -d '"')
  ARGUMENTS=$(printf "%s" "${line}" | cut -d" " -f 8-)
  $COMMAND $OPTIONS "$(eval printf \"%s\" "$ARGUMENTS")"
done < "${1}"

The eval is a shell built-in command which is used to enable parameter expansion of ARGUMENTS , if applicable.

I have to warn you that the eval is usualy say risky to use.