Unable to get script name as output

starqazi · 06-15-2015, 08:46 AM

Hi All,

I m trying to get script name of executed script.
The command in the script is:

prog1=$(echo $0 | sed 's/.*\///g;s/\.sh//g')
echo $prog1

It is only returning "bash".

What is wrong with this command

catkin · 06-15-2015, 10:02 AM

It works for me, assuming your intention is to strip off any leading directory names and any .sh extension.

It would be safer to put double quotes around the $0

Using $0 is not robust though, as detailed at http://mywiki.wooledge.org/BashFAQ/028

How did you test it? Meaning what command did you use. And what was the result?

suicidaleggroll · 06-15-2015, 11:13 AM

Works for me too.

Code:

$ echo $0 | sed 's/.*\///g;s/\.sh//g'
bash
$ cat script.sh
#!/bin/bash

echo $0 | sed 's/.*\///g;s/\.sh//g'
$ ./script.sh
script

Habitual · 06-15-2015, 12:35 PM

Code:

vi LQRocks.sh

and add

Code:

#!/bin/bash
echo `basename $0`

gives you the script name when run from a script. NOT a shell.

Code:

./LQRocks.sh

LQRocks.sh

Change the script name...reflects in the script:

Code:

mv LQRocks.sh YesItDoes.sh

Code:

./YesItDoes.sh

YesItDoes.sh

Enjoy the Goodness.

starqazi · 06-16-2015, 02:13 AM

Thank All who replied my post. I found your posts really help and with your help I resolved the issue.

Thanks & Regards,
Qazi

pan64 · 06-16-2015, 04:14 AM

echo ${0##*/}

Lsatenstein · 06-16-2015, 11:32 AM

test.sh

Code:

prog1=$(echo $0 | sed 's/.*\///g;s/\.sh//g')
echo $prog1

If you run
./test.sh
test

the response was test
If however, you source the file, the response is bash

Code:

.  ./test.sh
bash

With sourcing, the result of $0 is the name of the calling script

in place of echo and sed

test.sh

Code:

prog1=$(basename $0 .sh)    
echo "$prog1 is my script name"

Please note: $0 changes when the script is sourced.
If you are going to source your script, you would need test.sh as an argument to another script.

test2.sh

Code:

#!/bin/bash
prog1=$(basename $1 .sh )

Now ./test2.sh test.sh
yields test
and
. ./test2.sh test.sh
yields test.

Why do we want to do this anyway. Well here is a situation
Suppose I have a group of directories titled project1, project2, project3, ...
and my "project1.sh", "project2.sh", "project3.sh" ... script have

Code:

#!/bin/bash
prog1=$(basename $0 .sh)
cd $prog1
ls -l | grep -v total
#

If I do project1.sh
I will see the contents of project1

If I want to also switch to project1
I would just like to do
. project1.sh

to switch to that directory and after the listing, to become my current directory.