Executing Shell Script from Current Directory Without '"./Filename"

Executing Shell Script from current directory without './filename

You can execute it without ./ by using:

sh testfile

Or

sh /path/to/file/testfile

Edit
If you want to execute the program directly with a command, what you can do is to define an alias:

alias execute_testfile="sh /path/to/file/testfile"

And then, you will execute the program whenever you write

execute_testfile

or whatever name you define.

To make this alias persistent, do include the alias ... line in your ~/.profile or ~/.bash_profile files.

Get current directory or folder name (without the full path)

No need for basename, and especially no need for a subshell running pwd (which adds an extra, and expensive, fork operation); the shell can do this internally using parameter expansion:

result=${PWD##*/}          # to assign to a variable
result=${result:-/}        # to correct for the case where PWD=/

printf '%s\n' "${PWD##*/}" # to print to stdout
                           # ...more robust than echo for unusual names
                           #    (consider a directory named -e or -n)

printf '%q\n' "${PWD##*/}" # to print to stdout, quoted for use as shell input
                           # ...useful to make hidden characters readable.

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Note that if you're applying this technique in other circumstances (not PWD, but some other variable holding a directory name), you might need to trim any trailing slashes. The below uses bash's extglob support to work even with multiple trailing slashes:

dirname=/path/to/somewhere//
shopt -s extglob           # enable +(...) glob syntax
result=${dirname%%+(/)}    # trim however many trailing slashes exist
result=${result##*/}       # remove everything before the last / that still remains
result=${result:-/}        # correct for dirname=/ case
printf '%s\n' "$result"

Alternatively, without extglob:

dirname="/path/to/somewhere//"
result="${dirname%"${dirname##*[!/]}"}" # extglob-free multi-trailing-/ trim
result="${result##*/}"                  # remove everything before the last /
result=${result:-/}                     # correct for dirname=/ case

How do I get the directory where a Bash script is located from within the script itself?

#!/usr/bin/env bash

SCRIPT_DIR=$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )

is a useful one-liner which will give you the full directory name of the script no matter where it is being called from.

It will work as long as the last component of the path used to find the script is not a symlink (directory links are OK). If you also want to resolve any links to the script itself, you need a multi-line solution:

#!/usr/bin/env bash

SOURCE=${BASH_SOURCE[0]}
while [ -L "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )
  SOURCE=$(readlink "$SOURCE")
  [[ $SOURCE != /* ]] && SOURCE=$DIR/$SOURCE # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
done
DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )

This last one will work with any combination of aliases, source, bash -c, symlinks, etc.

Beware: if you cd to a different directory before running this snippet, the result may be incorrect!

Also, watch out for $CDPATH gotchas, and stderr output side effects if the user has smartly overridden cd to redirect output to stderr instead (including escape sequences, such as when calling update_terminal_cwd >&2 on Mac). Adding >/dev/null 2>&1 at the end of your cd command will take care of both possibilities.

To understand how it works, try running this more verbose form:

#!/usr/bin/env bash

SOURCE=${BASH_SOURCE[0]}
while [ -L "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  TARGET=$(readlink "$SOURCE")
  if [[ $TARGET == /* ]]; then
    echo "SOURCE '$SOURCE' is an absolute symlink to '$TARGET'"
    SOURCE=$TARGET
  else
    DIR=$( dirname "$SOURCE" )
    echo "SOURCE '$SOURCE' is a relative symlink to '$TARGET' (relative to '$DIR')"
    SOURCE=$DIR/$TARGET # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
  fi
done
echo "SOURCE is '$SOURCE'"
RDIR=$( dirname "$SOURCE" )
DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )
if [ "$DIR" != "$RDIR" ]; then
  echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"

And it will print something like:

SOURCE './scriptdir.sh' is a relative symlink to 'sym2/scriptdir.sh' (relative to '.')
SOURCE is './sym2/scriptdir.sh'
DIR './sym2' resolves to '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
DIR is '/home/ubuntu/dotfiles/fo fo/real/real1/real2'

To show only file name without the entire directory path

ls whateveryouwant | xargs -n 1 basename

Does that work for you?

Otherwise you can (cd /the/directory && ls) (yes, parentheses intended)

Reliable way for a Bash script to get the full path to itself

Here's what I've come up with (edit: plus some tweaks provided by sfstewman, levigroker, Kyle Strand, and Rob Kennedy), that seems to mostly fit my "better" criteria:

SCRIPTPATH="$( cd -- "$(dirname "$0")" >/dev/null 2>&1 ; pwd -P )"

That SCRIPTPATH line seems particularly roundabout, but we need it rather than SCRIPTPATH=`pwd` in order to properly handle spaces and symlinks.

The inclusion of output redirection (>/dev/null 2>&1) handles the rare(?) case where cd might produce output that would interfere with the surrounding $( ... ) capture. (Such as cd being overridden to also ls a directory after switching to it.)

Note also that esoteric situations, such as executing a script that isn't coming from a file in an accessible file system at all (which is perfectly possible), is not catered to there (or in any of the other answers I've seen).

The -- after cd and before "$0" are in case the directory starts with a -.

How to extract directory path from file path?

dirname and basename are the tools you're looking for for extracting path components:

$ VAR='/home/pax/file.c'
$ DIR="$(dirname "${VAR}")" ; FILE="$(basename "${VAR}")"
$ echo "[${DIR}] [${FILE}]"
[/home/pax] [file.c]

They're not internal bash commands but they are part of the POSIX standard - see dirname and basename. Hence, they're probably available on, or can be obtained for, most platforms that are capable of running bash.

Linux Bash Script: How to get file without path?

basename path/to/file.b.c should give you file.b.c

However re-reading the question, I think a temporary cd to the path and then an ls may be better:

(cd /path/to/file; ls -l *.*.*)

How can I remove the extension of a filename in a shell script?

You should be using the command substitution syntax $(command) when you want to execute a command in script/command.

So your line would be

name=$(echo "$filename" | cut -f 1 -d '.')

Code explanation:

1. echo get the value of the variable $filename and send it to standard output
2. We then grab the output and pipe it to the cut command
3. The cut will use the . as delimiter (also known as separator) for cutting the string into segments and by -f we select which segment we want to have in output
4. Then the $() command substitution will get the output and return its value
5. The returned value will be assigned to the variable named name

Note that this gives the portion of the variable up to the first period .:

$ filename=hello.world
$ echo "$filename" | cut -f 1 -d '.'
hello
$ filename=hello.hello.hello
$ echo "$filename" | cut -f 1 -d '.'
hello
$ filename=hello
$ echo "$filename" | cut -f 1 -d '.'
hello

How to obtain the absolute path of a file via Shell (BASH/ZSH/SH)?

Use realpath

$ realpath example.txt
/home/username/example.txt

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