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-   -   Why $0 shows as "-bash", not the script name? (https://www.linuxquestions.org/questions/linux-newbie-8/why-%240-shows-as-bash-not-the-script-name-930266/)

johnifanx98 02-19-2012 09:48 PM
Why $0 shows as "-bash", not the script name?
 
test(){
echo $0
echo $1
echo $2
}

run "test 1 2", and it gives

-bash
1
2

I expect

test
1
2

pinga123 02-19-2012 10:06 PM
Quote:
Originally Posted by johnifanx98 (Post 4606996)
test(){
echo $0
echo $1
echo $2
}

run "test 1 2", and it gives

-bash
1
2

I expect

test
1
2
$0 is supposed to print the name of the script.
Here is the output when i run it.
Code:
test
1
2
Check the script.
Code:
test()
{
echo $0
echo $1
echo $2
}
test 1 2
Are you supplying command line arguments?

chrism01 02-20-2012 12:21 AM
How are you calling it
Code:
bash yourscript

# OR

./yourscript

AnanthaP 02-20-2012 12:30 AM
Actually the op is saying run "test 1 2".

Your $0 is run (under bash) and "test" is "shift"ed out.

OK

catkin 02-20-2012 03:08 AM
Quote:
Originally Posted by AnanthaP (Post 4607054)
Actually the op is saying run "test 1 2"
Code:
c@CW8:~$ type run
bash: type: run: not found

colucix 02-20-2012 03:32 AM
I think the most robust method is by using the appropriate bash variables. This avoid confusion between executed and sourced scripts. There is also a specific variable to retrieve function names:
Code:
#!/bin/bash
#
test () {
  echo $FUNCNAME
  echo $1
  echo $2
}

test 1 2

echo $(readlink -f "$BASH_SOURCE")
The last statement gives the full path of the script:
Code:
$ ./test.sh
test
1
2
/home/colucix/test.sh
$ . test.sh
test
1
2
/home/colucix/test.sh
$ bash test.sh
test
1
2
/home/colucix/test.sh
Hope this helps.

johnifanx98 02-20-2012 10:22 AM
Quote:
Originally Posted by colucix (Post 4607138)
I think the most robust method is by using the appropriate bash variables. This avoid confusion between executed and sourced scripts. There is also a specific variable to retrieve function names:
Code:
#!/bin/bash
#
test () {
  echo $FUNCNAME
  echo $1
  echo $2
}

test 1 2

echo $(readlink -f "$BASH_SOURCE")
The last statement gives the full path of the script:
Code:
$ ./test.sh
test
1
2
/home/colucix/test.sh
$ . test.sh
test
1
2
/home/colucix/test.sh
$ bash test.sh
test
1
2
/home/colucix/test.sh
Hope this helps.
Thanks a lot for all responses. Now I see I'm running a function not script which caused this unexpected result...


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