Shell script on passing arguments

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

No need to use shout at us (don't use caps lock). People are less inclined to help you that way. — Femaref
– Femaref, Commented Aug 25, 2012 at 13:41

victorsavu3 · Accepted Answer · 2012-08-25 13:48:13Z

3

Try ${dir}/${wavPath}/${featPath}

answered Aug 25, 2012 at 13:48

victorsavu3

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Comments

Colin Pickard · Accepted Answer · 2012-08-25 13:48:17Z

0

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.

answered Aug 25, 2012 at 13:48

Colin Pickard

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1 Comment

victorsavu3 Over a year ago

${dir/$wavPath/$featPath} is the problem itself, a variable is in the form ${name}

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