How do I extract the last directory of a pwd output? I don't want to use any knowledge of how many levels there are in the directory structure. If I wanted to use that, I could do something like:
> pwd
/home/kiki/dev/my_project
> pwd | cut -d'/' -f5
my_project
But I want to use a command that works regardless of where I am in the directory structure. I assume there is a simple command to do this using awk or sed.
Are you looking for basename or dirname?
Something like
basename "`pwd`"
should be what you want to know.
If you insist on using sed, you could also use
pwd | sed 's#.*/##'
If you want to do it completely within a bash script without running any external binaries, ${PWD##*/} should work.
Using awk:
pwd | awk -F/ '{print $NF}'
Should work for you:
pwd | rev | cut -f1 -d'/' - | rev
Reference:
https://stackoverflow.com/a/31728689/663058
Related
I am trying to write a generic script which can have different file name inputs.
This is just a small part of my bash script.
for example, lets say folder 444-55 has 2 files
qq.filter.vcf
ee.filter.vcf
I want my output to be -
qq
ee
I tried this and it worked -
ls /data2/delivery/Stack_overflow/1111_2222_3333_23/secondary/444-55/*.filter.vcf | sort | cut -f1 -d "." | xargs -n 1 basename
But lets say I have a folder like this -
/data2/delivery/Stack_overflow/de.1111_2222_3333_23/secondary/444-55/*.filter.vcf
My script's output would then be
de
de
How can I make it generic?
Thank you so much for your help.
Something like this in a script will "cut" it:
for i in /data2/delivery/Stack_overflow/1111_2222_3333_23/secondary/444-55/*.filter.vcf
do
basename "$i" | cut -f1 -d.
done | sort
advantages:
it does not parse the output of ls, which is frowned upon
it cuts after having applied the basename treatment, and the cut ignores the full path.
it also sorts last so it's guaranteed to be sorted according to the prefix
Just move the basename call earlier in the pipeline:
printf "%s\n" /data2/delivery/Stack_overflow/1111_2222_3333_23/secondary/444-55/*.filter.vcf |
xargs -n 1 basename |
sort |
cut -f1 -d.
I have to run many python script which differ just with one parameter. I name them as runv1.py, runv2.py, runv20.py. I have the original script, say runv1.py. Then I make all copies that I need by
cat runv1.py | tee runv{2..20..1}.py
So I have runv1.py,.., runv20.py. But still the parameter v=1 in all of them.
Q: how can I also replace v parameter to read it from the file name? so e.g in runv4.py then v=4. I would like to know if there is any one-line shell command or combination of commands. Thank you!
PS: direct editing each file is not a proper solution when there are too many files.
Below for loop will serve your purpose I think
for i in `ls | grep "runv[0-9][0-9]*.py"`
do
l=`echo $i | tr -d [a-z.]`
sed -i 's/v/'"$l"'/g' runv$l.py
done
Below command was to pass the parameter to script extracted from the filename itself
ls | grep "runv[0-9][0-9]*.py" | tr -d [a-z.] | awk '{print "./runv"$0".py "$0}' | xargs sh
in the end instead of sh you can use python or bash or ksh.
Is there a shell script that runs on a mac to generate a word list from a text file, listing the unique words? Even better if it could sort by frequency....
sorry forgot to mention, yeah i prefer a bash one as i'm using mac now...
oh, my file is in french... (basically i'm reading a novel and learning french, so i try to generate a word list help myself). hope this is not a problem?
If I understood you correctly, you need something like that:
cat <filename> | sed -e 's/ /\n/g' | sort | uniq -c
This command will do
cat file.txt | tr "\"' " '\n' | sort -u
Here sort -u will not work on Macintosh machines. In that case use sort | uniq -c instead. (Thanks to Hank Gay)
cat file.txt | tr "\"' " '\n' | sort | uniq -c
Just answer my question to dot down the final version i'm using:
tr -cs "[:alpha:]" "\n" < FileIn.txt | sort | uniq -c | awk '{print $2","$1}' >> FileOut.csv
some notes:
tr can be used directly to do replacement.
since i'm interested creating a word list for my french vocabulary, i used [:alpha:]
awk is used to insert a comma, so that the output is a csv file, easier for me to upload...
thanks again for everyone helping me.
sorry i didn't put it clearly at the beginning that i'm using a mac and expect a bash script.
cheers.
ls /home/user/new/*.txt prints all txt files in that directory. However it prints the output as follows:
[me#comp]$ ls /home/user/new/*.txt
/home/user/new/file1.txt /home/user/new/file2.txt /home/user/new/file3.txt
and so on.
I want to run the ls command not from the /home/user/new/ directory thus I have to give the full directory name, yet I want the output to be only as
[me#comp]$ ls /home/user/new/*.txt
file1.txt file2.txt file3.txt
I don't want the entire path. Only filename is needed. This issues has to be solved using ls command, as its output is meant for another program.
ls whateveryouwant | xargs -n 1 basename
Does that work for you?
Otherwise you can (cd /the/directory && ls) (yes, parentheses intended)
No need for Xargs and all , ls is more than enough.
ls -1 *.txt
displays row wise
There are several ways you can achieve this. One would be something like:
for filepath in /path/to/dir/*
do
filename=$(basename $filepath)
... whatever you want to do with the file here
done
Use the basename command:
basename /home/user/new/*.txt
(cd dir && ls)
will only output filenames in dir. Use ls -1 if you want one per line.
(Changed ; to && as per Sactiw's comment).
you could add an sed script to your commandline:
ls /home/user/new/*.txt | sed -r 's/^.+\///'
A fancy way to solve it is by using twice "rev" and "cut":
find ./ -name "*.txt" | rev | cut -d '/' -f1 | rev
The selected answer did not work for me, as I had spaces, quotes and other strange characters in my filenames. To quote the input for basename, you should use:
ls /path/to/my/directory | xargs -n1 -I{} basename "{}"
This is guaranteed to work, regardless of what the files are called.
I prefer the base name which is already answered by fge.
Another way is :
ls /home/user/new/*.txt|awk -F"/" '{print $NF}'
one more ugly way is :
ls /home/user/new/*.txt| perl -pe 's/\//\n/g'|tail -1
just hoping to be helpful to someone as old problems seem to come back every now and again and I always find good tips here.
My problem was to list in a text file all the names of the "*.txt" files in a certain directory without path and without extension from a Datastage 7.5 sequence.
The solution we used is:
ls /home/user/new/*.txt | xargs -n 1 basename | cut -d '.' -f1 > name_list.txt
There are lots of way we can do that and simply you can try following.
ls /home/user/new | tr '\n' '\n' | grep .txt
Another method:
cd /home/user/new && ls *.txt
Here is another way:
ls -1 /home/user/new/*.txt|rev|cut -d'/' -f1|rev
You could also pipe to grep and pull everything after the last forward slash. It looks goofy, but I think a defensive grep should be fine unless (like some kind of maniac) you have forward slashes within your filenames.
ls folderpathwithcriteria | grep -P -o -e "[^/]*$"
When you want to list names in a path but they have different file extensions.
me#server:/var/backups$ ls -1 *.zip && ls -1 *.gz
Having trouble with grepping and cutting at the same time
I have a file test.txt.
Inside the file is this syntax
File: blah.txt Location: /home/john/Documents/play/blah.txt
File: testing.txt Location /home/john
My command is ./delete -r (filename), say filename is blah.txt.
How would i search test.txt for blah.txt and cut the /home/john/Documents/play/blah.txt out and put it in a variable
grep -P "^File: blah\.txt Location: .+" test.txt | cut -d: -f3
Prefer always to involse as less as possible external command for your task.
You can achive what you want using single awk command:
awk '/^File: blah.txt/ { print $4 }' test.txt
Try this one ;)
filename=$(grep 'blah.txt' test.txt | grep -oP 'Location:.*' | grep -oP '[^ ]+$')
./delete $filename