shell script to find file name from its path

Question

Hello i want a simple shell script that find the name of the file from a given path of the file. like

$path = "/var/www/html/test.php";

then i want to get value "test" in some variable. Also only .php files are present.I am using bash shell. Thanks

score 62 · Accepted Answer · answered Jan 10 '11 at 10:05

62

Try:

path="/var/www/html/test.php"
name=$(basename "$path" ".php")
echo "$name"

The quotes are only there to prevent problems when $path contains spaces.

answered Jan 10 '11 at 10:05

Aaron Digulla

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we need to write the extention ?? ".php", what if we don't know the extention ? is there a way ? – Hasni Mar 11 '23 at 13:30
@Hasni Try this question: https://stackoverflow.com/questions/125281/how-do-i-remove-the-file-suffix-and-path-portion-from-a-path-string-in-bash – Aaron Digulla Mar 13 '23 at 19:50

score 27 · Answer 2 · answered Jan 10 '11 at 11:03

27

Use the built in UNIX command:

basename "/var/www/html/test.php"

answered Jan 10 '11 at 11:03

Hogsmill

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score 4 · Answer 3 · answered Jun 17 '13 at 16:36

4

Use the basename() function. It is buily in UNIX function

answered Jun 17 '13 at 16:36

Chetan Wadhwa

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1
14

score 0 · Answer 4 · answered May 03 '12 at 15:18

0

string="/var/www/html/test.php"

oIFS="$IFS"; IFS='/' 
set -A str $string
IFS="$oIFS"

echo "strings count = ${#str[@]}"
len=${#str[@]}
pos=`expr $len - 1`
echo "file : ${str[$pos]}";

Output-

 strings count = 4
 file : test.php

answered May 03 '12 at 15:18

Anjan Biswas

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shell script to find file name from its path

4 Answers4