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Borrowing graciously from this StackOverflow answer, I want to find the directory a script is running in so I can load relative paths for it on login:

The script is pretty small right now:

SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ] ; do SOURCE="$(readlink "$SOURCE")"; done
DIR=$( cd -P "$( dirname "$SOURCE" )" && pwd )

But when I log in or call source <this_script.sh>, I get:

-bash: script.sh: line 4: syntax error: unexpected end of file

Couple questions:

  1. What is it hanging up on?
  2. What's the fix?

This is running:

-bash-3.2$ bash --version
GNU bash, version 3.2.25(1)-release (x86_64-redhat-linux-gnu)
Copyright (C) 2005 Free Software Foundation, Inc.

Edit 1:

trying Gile's solution:

parent="$(dirname "$SOURCE")"
echo "PARENT IS $parent"
DIR="$(cd -P "$parent" && pwd)"
echo "DIR IS $DIR"

I get the following when trying to source it or log in:

Last login: Mon Mar 12 08:07:13 2012 from ....
PARENT IS .
: No such file or directory
DIR IS
: command not found
: command not found
: command not found
: command not found
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The answer has been updated and is by now: DIR="$( cd -P "$( dirname "$SOURCE" )" && pwd )". Any reason why you don't use that? –  user unknown Mar 8 '12 at 16:22
 
When doing variable assignment to the output of command substitution or to the output of a variable reference, there is no reason whatsoever to use quotes. Proof: run x="my name is ryran" and then a=$x or a=${x/ryran/nick klauer} or a=$(echo $x) –  ryran Mar 8 '12 at 18:48
 
Why not just re-use $PWD ? (not worth an 'answer') –  laebshade Mar 10 '12 at 0:23
 
@laebshade: The issue is that this script needs to know what directory the file is in. calling the script from a symlink $PWD will print where the symlink is, not where the source file is. I want to load in sources relative to that directory, and all of the above is just to try to find the symlink's real directory to load other files. –  Nick Klauer Mar 12 '12 at 13:33
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2 Answers

up vote 1 down vote

You're probably missing a newline at the end of your very last line.

share|improve this answer
 
I'm not sure what you mean, but if I just add another new line, i get -bash: script.sh: line 5: syntax error: unexpected end of file instead of line 4, so did you mean for me to add some \r character somewhere? –  Nick Klauer Mar 8 '12 at 19:58
 
@NickKlauer I don't get an error in bash 4, perhaps there's a bash4 feature that it isn't parsing right –  Kevin Mar 8 '12 at 20:05
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up vote 0 down vote

I cannot reproduce this behavior with bash 3.2.25(2) compiled from source.

There was a change somewhere in the 3.2 series in the way $(…) was parsed.

uuu. Bash now parses command substitutions according to Posix rules: parsing the command contained in $() to find the closing delimiter.

I don't know why I couldn't reproduce it if that's the case, but you might be hitting the old, buggy parsing behavior. Try simplifying the command.

parent=$(dirname "$SOURCE")
DIR=$(cd -P "$parent" && pwd)

I think adding double quotes around the outer command substitution might help. Although double quotes should not make a difference when you use them around the value part of an assignment, I remember a shell version that had a buggy parser that broke on foo=$(something) but coped with foo="$(something)"; this may well have been it.

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