variable assignment in shell script

Question

Our directory structure looks like

/long/dir/name/bin/exec.sh
/long/dir/name/logs
/long/dir/name/input
/long/dir/name/output.

in exec.sh I want to retrieve the root directory (/long/dir/name in this case), store this in a variable so I can use it to refer to $DIR/output, $DIR/input etc.

I got as far as [exec.sh]:

#!/bin/sh
export DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )" | sed -e 's,/*[^/]\+/*$,,'
echo "$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )" | sed -e 's,/*[^/]\+/*$,,'
echo "My Dir: '$DIR'"

This outputs:

/long/dir/name   <-- Which is what I want
My Dir: ''

What is going wrong when assigning it to the DIR variable?

...mind you, using a pipeline (involving sed?!) for this is silly (inefficient, hard-to-read, error-prone); fixing John's advice to use $BASH_SOURCE rather than $0 makes more sense. — Charles Duffy
– Charles Duffy, Commented Nov 7, 2014 at 14:13

Community · Accepted Answer · 2017-05-23 10:25:21Z

3

To answer the "what went wrong" question.

You stuck the pipe to sed outside the subshell for the assignment (and technically export var=... isn't an assignment it is a call to export (see ksh get exit status in assignment as an example of this).

Anyway, what happens on that first line therefore is the shell sees

export DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )" | sed -e 's,/*[^/]\+/*$,,'

(note the whitespace around the pipe there). And pipelines execute in sub-shells so your export is happening in a su-shell and then being lost.

Stick the sed in the sub-shell and you fix the problem.

export DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd | sed -e 's,/*[^/]\+/*$,,')"

edited May 23, 2017 at 10:25

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answered Nov 7, 2014 at 14:06

Etan Reisner

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1 Comment

John Zwinck Over a year ago

You're right about the direct problem with the original code, but it's still way more complicated than it needs to be (see my answer).

John Zwinck · Accepted Answer · 2014-11-07 14:42:38Z

2

I think you want this:

DIR=${BASH_SOURCE%/*/*}

That is, strip the last two path parts from the script's own path, and use that. Run it through realpath or readlink -f if you need to get a canonical path.

edited Nov 7, 2014 at 14:42

answered Nov 7, 2014 at 13:54

John Zwinck

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4 Comments

Etan Reisner Over a year ago

If you are going to suggest readlink it might be better to just suggest readlink $path/../.. and let readlink do the work.

dr jerry Over a year ago

@John DIR=${0%/*/*} gives me 'exec.sh'?

Charles Duffy Over a year ago

${BASH_SOURCE%/*/*}, rather. $0 isn't a good fit for this use -- as @drjerry's issue above demonstrates.

dr jerry Over a year ago

echo ${BASH_SOURCE%/*/*} in exec.sh still gives ./exec.sh?

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