How to Know the Script File Name in a Bash Script

How do I know the script file name in a Bash script?

me=`basename "$0"`

For reading through a symlink¹, which is usually not what you want (you usually don't want to confuse the user this way), try:

me="$(basename "$(test -L "$0" && readlink "$0" || echo "$0")")"

IMO, that'll produce confusing output. "I ran foo.sh, but it's saying I'm running bar.sh!? Must be a bug!" Besides, one of the purposes of having differently-named symlinks is to provide different functionality based on the name it's called as (think gzip and gunzip on some platforms).

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¹ That is, to resolve symlinks such that when the user executes foo.sh which is actually a symlink to bar.sh, you wish to use the resolved name bar.sh rather than foo.sh.

bash script to check file name begins with expected string

In bash, you can get the first 7 characters of a shell variable with:

${sourceFile:0:7}

and the last four with:

${sourceFile:${#sourceFile}-4}

Armed with that knowledge, simply use those expressions where you would normally use the variable itself, something like the following script:

arg=$1
shopt -s nocasematch
i7f4="${arg:0:7}${arg:${#arg}-4}"
if [[ "${i7f4}" = "adusers.txt" ]] ; then
    echo Okay
else
    echo Bad
fi

You can see it in action with the following transcript:

pax> check.sh hello
Bad

pax> check.sh addUsers.txt
Bad

pax> check.sh adUsers.txt
Okay

pax> check.sh adUsers_new.txt
Okay

pax> check.sh aDuSeRs_stragngeCase.pdf.gx..txt
Okay

How to find filename in file content with bash script?

on given file ('+')

+ executes the command on all found files. The {} is replaced by the list of files in current directory.

Debug with -exec echo grep -il {} +.

You want:

-exec grep -Fil {} {} ';'

To search the filename {} as a pattern in the file named {}. The ';' terminates the command.

I also added -F to interpret pattern literally.

How to find out name of script called (sourced) by another script in bash?

Change file2.sh to:

#!/bin/bash
echo "from file2: ${BASH_SOURCE[0]}"

Note that BASH_SOURCE is an array variable. See the Bash man pages for more information.

How to get the Directory name and file name of a bash script by bash?

#!/bin/bash

echo "/$(basename "$(dirname "$0")")/$(basename "$0")"
echo
echo

read -r

Output:

/Dirname/Filname.Extension

Get only file name in variable in a for loop

You need two lines; chained operators aren't allowed.

f=${f##*/}  # Strip the directory
f=${f%%.*}  # Strip the extensions

Or, you can use the basename command to strip the directory and one extension (assuming you know what it is) in one line.

f=$(basename "$f" .txt)

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