Get just the filename from a path in a Bash script [duplicate] - bash

Question

This question already has answers here:
Extract filename and extension in Bash
(38 answers)
Closed 6 years ago.
How would I get just the filename without the extension and no path?
The following gives me no extension, but I still have the path attached:
source_file_filename_no_ext=${source_file%.*}

Answer 1

Many UNIX-like operating systems have a basename executable for a very similar purpose (and dirname for the path):
pax> full_name=/tmp/file.txt
pax> base_name=$(basename ${full_name})
pax> echo ${base_name}
file.txt
That unfortunately just gives you the file name, including the extension, so you'd need to find a way to strip that off as well.
So, given you have to do that anyway, you may as well find a method that can strip off the path and the extension.
One way to do that (and this is a bash-only solution, needing no other executables):
pax> full_name=/tmp/xx/file.tar.gz
pax> xpath=${full_name%/*}
pax> xbase=${full_name##*/}
pax> xfext=${xbase##*.}
pax> xpref=${xbase%.*}
pax> echo "path='${xpath}', pref='${xpref}', ext='${xfext}'"
path='/tmp/xx', pref='file.tar', ext='gz'
That little snippet sets xpath (the file path), xpref (the file prefix, what you were specifically asking for) and xfext (the file extension).

Answer 2

basename and dirname solutions are more convenient. Those are alternative commands:
FILE_PATH="/opt/datastores/sda2/test.old.img"
echo "$FILE_PATH" | sed "s/.*\///"
This returns test.old.img like basename.
This is salt filename without extension:
echo "$FILE_PATH" | sed -r "s/.+\/(.+)\..+/\1/"
It returns test.old.
And following statement gives the full path like dirname command.
echo "$FILE_PATH" | sed -r "s/(.+)\/.+/\1/"
It returns /opt/datastores/sda2

Answer 3

Here is an easy way to get the file name from a path:
echo "$PATH" | rev | cut -d"/" -f1 | rev
To remove the extension you can use, assuming the file name has only ONE dot (the extension dot):
cut -d"." -f1

Answer 4

$ file=${$(basename $file_path)%.*}

Answer 5

Some more alternative options because regexes (regi ?) are awesome!
Here is a Simple regex to do the job:
regex="[^/]*$"
Example (grep):
FP="/hello/world/my/file/path/hello_my_filename.log"
echo $FP | grep -oP "$regex"
#Or using standard input
grep -oP "$regex" <<< $FP
Example (awk):
echo $FP | awk '{match($1, "$regex",a)}END{print a[0]}
#Or using stardard input
awk '{match($1, "$regex",a)}END{print a[0]} <<< $FP
If you need a more complicated regex:
For example your path is wrapped in a string.
StrFP="my string is awesome file: /hello/world/my/file/path/hello_my_filename.log sweet path bro."
#this regex matches a string not containing / and ends with a period
#then at least one word character
#so its useful if you have an extension
regex="[^/]*\.\w{1,}"
#usage
grep -oP "$regex" <<< $StrFP
#alternatively you can get a little more complicated and use lookarounds
#this regex matches a part of a string that starts with / that does not contain a /
##then uses the lazy operator ? to match any character at any amount (as little as possible hence the lazy)
##that is followed by a space
##this allows use to match just a file name in a string with a file path if it has an exntension or not
##also if the path doesnt have file it will match the last directory in the file path
##however this will break if the file path has a space in it.
regex="(?<=/)[^/]*?(?=\s)"
#to fix the above problem you can use sed to remove spaces from the file path only
## as a side note unfortunately sed has limited regex capibility and it must be written out in long hand.
NewStrFP=$(echo $StrFP | sed 's:\(/[a-z]*\)\( \)\([a-z]*/\):\1\3:g')
grep -oP "$regex" <<< $NewStrFP
Total solution with Regexes:
This function can give you the filename with or without extension of a linux filepath even if the filename has multiple "."s in it.
It can also handle spaces in the filepath and if the file path is embedded or wrapped in a string.
#you may notice that the sed replace has gotten really crazy looking
#I just added all of the allowed characters in a linux file path
function Get-FileName(){
local FileString="$1"
local NoExtension="$2"
local FileString=$(echo $FileString | sed 's:\(/[a-zA-Z0-9\<\>\|\\\:\)\(\&\;\,\?\*]*\)\( \)\([a-zA-Z0-9\<\>\|\\\:\)\(\&\;\,\?\*]*/\):\1\3:g')
local regex="(?<=/)[^/]*?(?=\s)"
local FileName=$(echo $FileString | grep -oP "$regex")
if [[ "$NoExtension" != "" ]]; then
sed 's:\.[^\.]*$::g' <<< $FileName
else
echo "$FileName"
fi
}
## call the function with extension
Get-FileName "my string is awesome file: /hel lo/world/my/file test/path/hello_my_filename.log sweet path bro."
##call function without extension
Get-FileName "my string is awesome file: /hel lo/world/my/file test/path/hello_my_filename.log sweet path bro." "1"
If you have to mess with a windows path you can start with this one:
[^\\]*$

Answer 6

$ source_file_filename_no_ext=${source_file%.*}
$ echo ${source_file_filename_no_ext##*/}