Shell Programming and Scripting

Thank you radoulov.
But here is one problem

Code:

~$ bash s
s: line 3: cd: s: Not a directory
-->

You wrote:

Is your previous code working in that case?

Perhaps I'm missing something but I cannot see how can you get the file position when invoked as shell scriptname and found somwhere in your path ...

It's Friday, sorry. I should have read the above comment before posting.

It works in any case because I use the result of the command

Code:

ps -p $$ --no-heading -o args

Which outputs something like that:

Code:

path_to_bash path_to_script arguments

This pattern will work in any case because even if it's not specified, a bash will always appear first :

Code:

$ bash s
bash s
$ ./s
/bin/bash s
$ s
/bin/bash /bin/s

I thought you wanted the path to the script even when invoked as shell scriptname:

Code:

bash /real/path/to/the/script

and not:

Code:

bash s

Hello chebarbudo,

Have you tried calling calling it "by it's bare name". If the script is somewhere in the search path then $0 should return the actual path to the script. Anyway, I a still not sure I understand what you mean. I think what you mean is to have a simple script that returns the absoute path to itself?

Code:

#!/bin/bash
echo $(cd $(dirname $0);pwd)

Should always do that I think..
If that is not what you mean could you state what you are looking for exactly?

S.

What I want is to be able to know the address (full path) of the script that is currently running without knowing how the user called it.

Thank you for your proposal Scrutinizer but once again, this is a script that will fail in some cases :

Code:

# /bin/s
/bin
# s
/bin
# bash s
/root

In the last case, this command fails.
So far the only command I found that works all the time is my own insanely complicated one.