CODEWITHSUNDEEP
linuxshell
cattail
cattail

Reputation: 367

shell string and program execute

The following code is used to create a file in exactly location.However result of running test.sh abc is

touch: missing file operand

#! /bin/sh
#test.sh
location='~/configuration/'$1
echo $location
touch `$location`

Is there anything wrong?Thanks!

Upvotes: 1

Views: 7008

Answers (4)

V H
V H

Reputation: 8587

Is location a folder or file ? if file -f if folder -d in below script --- bash should understand ~

#! /bin/bash
#test.sh
var1=$1;
location=~/configuration/$var1
echo $location

if [ ! -f $location ]; then
 echo "not found touching $location"
 touch `$location`
fi

Upvotes: 2

glenn jackman
glenn jackman

Reputation: 246847

When you put ~ in quotes, it loses its special meaning. You need:

#! /bin/sh
#test.sh
location=~/configuration/"$1"
echo "$location"
touch "$location"

You should get into the habit of always putting variables in double quotes (until you learn the specific times when you don't want quotes).

Upvotes: 2

Julian
Julian

Reputation: 852

There are several ways to do this, but the first fix is to change to

touch `echo $location`

However I think it might be simpler to do

touch ~/configuration/$1

Upvotes: 0

Brian Agnew
Brian Agnew

Reputation: 272297

By putting $location in backticks, you're trying to execute the value of $location and use that as the argument to touch (see here, section 3.4.5) Just do:

touch $location

If you run #!/bin/sh with the -x value, you'll see more clearly what bash is doing in your shell script. It's a very useful means to debug scripts.

Upvotes: 3

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