Remove '/' from end of variable - bash

Question

I'm looking to remove the '/' from the end of a variable so I can check if a file is a symlink to a directory. I've tried just about every method I could think of or find online, is there something I'm missing?

If I check if a file is a symlink with the '/' on the end, it treats it as a directory. This can be checked by running:

if [ -L symtest/ ] ; then echo "symlink"; fi

where symtest is a symlink to a directory. The above outputs nothing.

When I remove the '/', it works fine and outputs "symlink":

if [ -L symtest ] ; then echo "symlink"; fi

My question is, is there a way to remove the '/' from the name when it's passed as a variable to a function?

The function would look something like:

function is_it_a_symlink() {
    if [ -L $1 ] ; then
        echo "This file is a symlink!"
    else
        echo "This file is not a symlink!"
    fi
}

Thanks in advance.

Note that POSIX's Pathname Resolution rules require the behaviour you observe. — Jonathan Leffler
– Jonathan Leffler, Commented Mar 10, 2017 at 1:40
@Alexander, please avoid linking the ABS as a resource -- it has a longstanding reputation of showcasing bad practices. (Using the gratuitous function keyword -- which breaks compatibility with POSIX sh but adds no value over the standardized function declaration syntax -- is onesuch). — Charles Duffy
– Charles Duffy, Commented Mar 10, 2017 at 1:46
@Alexander, ...actually, there are more serious bad practices in that particular page you linked as well. directorys=$@ loses boundary information -- one can't distinguish between ./yourscript "foo bar" "baz qux" and ./yourscript foo bar baz qux after using it; the linkchk function is missing quotes (needs to be "$1"/* and linkchk "$element"), and more. — Charles Duffy
– Charles Duffy, Commented Mar 10, 2017 at 1:48
@Alexander, ...a better link for test operators is wiki.bash-hackers.org/commands/classictest; another reliable resource is the Wooledge wiki, ie. mywiki.wooledge.org/BashGuide/TestsAndConditionals. There's also the POSIX specification for test at pubs.opengroup.org/onlinepubs/9699919799/utilities/test.html — Charles Duffy
– Charles Duffy, Commented Mar 10, 2017 at 1:49

heemayl · Accepted Answer · 2017-03-10 01:41:57Z

3

POSIX-ly, using parameter expansion to get rid of the last /:

${parameter%/}

So, your case:

[ -L "${1%/}" ]

External tool, sed:

sed 's_/$__' <<<"$1"

So:

[ -L "$(sed 's_/$__' <<<"$1")" ]

Similarly, awk:

awk '{sub("/$", "")} 1' <<<"$1"

So:

[ -L "$(awk '{sub("/$", "")} 1' <<<"$1")" ]

edited Mar 10, 2017 at 1:41

answered Mar 10, 2017 at 1:35

heemayl

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3 Comments

Philip Gibbons Over a year ago

That works perfectly! I had actually tried that earlier just without the quotation marks. Thanks!

chepner Over a year ago

To be safe, you should probably remove all trailing /; don't assume there is just one. ${1%%/}

Charles Duffy Over a year ago

@chepner, ...that works? I thought one had to use an extglob to match a sequence there. Actually, it looks to me like one does; thus, ${1%%+(/)}, after shopt -s extglob.

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