传递参数的Shell脚本

Question

My code was like this I'm passing 4 arguments to a script

ex.sh "wavpath" "featpath"

"ex.sh" code is

#!/bin/bash

wavPath=$1
featPath=$2
rm -f $scpFile

echo $wavPath
echo $featPath


for dir in `ls -R $wavPath|grep ":"|cut -d':' -f1`
do
    mkdir -p ${dir/$wavPath/$featPath}
done

The error message:

bad substitution

and it is at ${dir/$wavPath/$featPath} and its showing both the paths

can anyone help

Answer 1

尝试$ {dir} / $ {wavPath} / $ {featPath}

Answer 2

maybe you meant $dir/$wavPath/$featPath

try changing

mkdir -p ${dir/$wavPath/$featPath}

to

echo $dir/$wavPath/$featPath

and see if the output is what you expected for the input of mkdir .

Also, you're not setting a value for the variable $scpFile before you use it.