How do I manipulate $PATH elements in shell scripts?

Question

Is there a idiomatic way of removing elements from PATH-like shell variables?

That is I want to take

PATH=/home/joe/bin:/usr/local/bin:/usr/bin:/bin:         


        

Answer 1

This is easy using awk.

Replace

{
  for(i=1;i<=NF;i++) 
      if($i == REM) 
          if(REP)
              print REP; 
          else
              continue;
      else 
          print $i; 
}

Start it using

function path_repl {
    echo $PATH | awk -F: -f rem.awk REM="$1" REP="$2" | paste -sd:
}

$ echo $PATH
/bin:/usr/bin:/home/js/usr/bin
$ path_repl /bin /baz
/baz:/usr/bin:/home/js/usr/bin
$ path_repl /bin
/usr/bin:/home/js/usr/bin

Append

Inserts at the given position. By default, it appends at the end.

{ 
    if(IDX < 1) IDX = NF + IDX + 1
    for(i = 1; i <= NF; i++) {
        if(IDX == i) 
            print REP 
        print $i
    }
    if(IDX == NF + 1)
        print REP
}

Start it using

function path_app {
    echo $PATH | awk -F: -f app.awk REP="$1" IDX="$2" | paste -sd:
}

$ echo $PATH
/bin:/usr/bin:/home/js/usr/bin
$ path_app /baz 0
/bin:/usr/bin:/home/js/usr/bin:/baz
$ path_app /baz -1
/bin:/usr/bin:/baz:/home/js/usr/bin
$ path_app /baz 1
/baz:/bin:/usr/bin:/home/js/usr/bin

Remove duplicates

This one keeps the first occurences.

{ 
    for(i = 1; i <= NF; i++) {
        if(!used[$i]) {
            print $i
            used[$i] = 1
        }
    }
}

Start it like this:

echo $PATH | awk -F: -f rem_dup.awk | paste -sd:

Validate whether all elements exist

The following will print an error message for all entries that are not existing in the filesystem, and return a nonzero value.

echo -n $PATH | xargs -d: stat -c %n

To simply check whether all elements are paths and get a return code, you can also use test:

echo -n $PATH | xargs -d: -n1 test -d