CODEWITHSUNDEEP
shellcommand-linezsh
Jon Carter
Jon Carter

Reputation: 3436

How do I get the filename (not the invoking command) in a zsh script?

I'm writing a zsh script that has a couple of symbolic links to it. Normal operation is to invoke the script via one of the links.

At a certain point, I want to handle the condition where the script was called directly, instead of being called via one of the links to it. To do that, I need to take the command used to invoke the script and compare it to the name of the script file itself.

There seem to be several ways to get the invoking command ($0 and ${(%):-%N} are two examples). I do not know, however, how to determine the name of the FILE containing the actual script source. Every attempt to find out just seems to lead me to how to get the invoking command.

Here's some example code that may help to illustrate what I mean:

invoking_command=$(basename $0)

# If we're called via one symbolic link, do one thing.
if [[ $invoking_command = "link-one" ]]; then                                                                                                         
    condition="link-one"                                                                                                                               
fi                                                                                                                                                

# If we're called via the other link, do something else.                                                                                                                                                  
if [[ $invoking_command = "sapti" ]]; then                                                                                                        
    condition="link-two"                                                                                                                              
fi   

# If we're called directly, do something like, for example, 
# recommend to the user what the expected usage is.
if [[ $invoking_command = (??? WHAT GOES HERE ???) ]]
    condition="script file was run directly"
    usage && exit
fi

In the specific example I use here, I suppose I could just print usage if neither of the first conditions are true. That would handle this case, but I'm still left with the question of how to find the file name if I ever need to.

Surely this is possible. Ideas?

Upvotes: 0

Views: 431

Answers (1)

user993553
user993553

Reputation: 1077

from man zshexpn:

   a      Turn a file name into an absolute path:   prepends  the  current
          directory, if necessary, and resolves any use of `..' and `.' in
          the path.  Note that the transformation takes place even if  the
          file or any intervening directories do not exist.

   A      As  `a',  but also resolve use of symbolic links where possible.
          Note that resolution of `..' occurs before  resolution  of  sym-
          bolic  links.   This  call is equivalent to a unless your system
          has the realpath system call (modern systems do).

#!/usr/local/bin/zsh
mypath=$0:A
invoker=$0
echo $mypath
echo $invoker

Upvotes: 1

Related Questions