extract the filename from a path in a Bash

I have a file which has a lot of links and I need bash script to extract all the names of the files which end with .pdf format ?

3 Answers

You can extract the names with

grep -o '[^/]*\.pdf' example

• [^/] matches any single character that is not /
• [^/]*\.pdf is a (possibly empty) sequence of non-/ characters, followed by the characters .pdf (the backslash before the period makes it literal - otherwise . in a regex matches any character)
• the grep -o flag outputs each matching portion, one match per line

To de-duplicate the, pipe through sort and uniq, or sort -u

grep -o '[^/]*\.pdf' example | sort -u

3

• Here, cut will split the string with space and f1 means first field
• Now rev will reverse the string and split the string by '/' using cut with -f1 which wiil give us the last part of the url however in reverse order. So we need to reverse it again!

cat filename | cut -d' ' -f1 | rev | cut -d'/' -f1 | rev

2

basename /usr/bin/poop.txt
would give you
poop.txt

i generated a testy.txt file from /usr/bin that contains all of its path/files and edited some of the names to end in .pdf.

so basically testy.txt looks like this:
/usr/bin/aa-enabled
/usr/bin/aconnect.pdf
/usr/bin/alsaucm
/usr/bin/xargs.pdf
/usr/bin/xcursogen
/usr/bin/znew.pdf

You can use basename to extract just the names in the file after grepping pdf.

basename -a $(grep "\\.pdf" testy.file)

aconnect.pdf
xargs.pdf
znew.pdf

the -a allows the multiple arguments

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