Getting a partial path to a file in a bash script

Question

I have a path that is stored in a variable

$FULLPATH="/this/is/the/path/to/my/file.txt"

I also have another variable containing a partial path

$PARTIAL="/this/is/the/"

I want to remove the partial path from the full path so that I am left with:

path/to/my/file.txt

What's the best way to do this?

As is always the case, I solved it 30 seconds after posting. I am running the path through sed and removing the partial path... silly me — Tom Sweeney
– Tom Sweeney, Commented Apr 7, 2010 at 18:11

R Samuel Klatchko · Accepted Answer · 2010-04-07 18:09:25Z

3

Use bash's # pattern matching operator:

${FULLPATH#${PARTIAL}}

answered Apr 7, 2010 at 18:09

R Samuel Klatchko

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1 Comment

David Over a year ago

Is there a way to modify this so that there is an optional /? That way, if $PARTIAL doesn't have the /, it will still be removed?

Kevin Little · Accepted Answer · 2010-04-07 19:02:27Z

3

Here's a little more detail to Mr. Klatchko's excellent answer:

$ FULLPATH="/this/is/the/path/to/my/file.txt"
$ PARTIAL="/this/is/the/"
$ echo ${FULLPATH#${PARTIAL}}
path/to/my/file.txt

answered Apr 7, 2010 at 19:02

Kevin Little

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Comments

Ignacio Vazquez-Abrams · Accepted Answer · 2010-04-07 18:10:58Z

1

If you're sure that $PARTIAL is an actual path:

result="${FULLPATH#$PARTIAL}"
result="${result#/}"

answered Apr 7, 2010 at 18:10

Ignacio Vazquez-Abrams

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janks · Accepted Answer · 2010-04-07 18:12:32Z

0

# echo $FULLPATH | sed 's#'"$PARTIAL"'##'

answered Apr 7, 2010 at 18:12

janks

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