Get just the filename from a path in a Bash script

This question already has an answer here:

Extract filename and extension in Bash 35 answers

---

Most UNIX-like operating systems have a basename executable for a very similar purpose (and dirname for the path):

pax> a=/tmp/file.txt
pax> b=$(basename $a)
pax> echo $b
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> a=/tmp/xx/file.tar.gz
pax> xpath=${a%/*} 
pax> xbase=${a##*/}
pax> xfext=${xbase##*.}
pax> xpref=${xbase%.*}
pax> echo;echo path=${xpath};echo pref=${xpref};echo 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).

---

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

---

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

链接地址: http://www.djcxy.com/p/57080.html

上一篇: 如何在bash中检索文件名或扩展名

下一篇: 从Bash脚本中的路径获取文件名