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Following Symlinks to Actual Script


 
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# 1  
Old 09-08-2009
Following Symlinks to Actual Script

I need to have my script know what directory it's in, even if it's run from a symlink located elsewhere. Here's what I've come up with, for the benefit of anyone with a similar need, but I'm also interested to know if there's a more elegant solution. I'd rather not get into awk-land, but I couldn't get the quoting and escaping right with sed.

Code:
#!/bin/ksh
prog="$0"
echo "prog='$prog'"
while [ -L "$prog" ]; do
  prog=$(stat --format %N "$prog" | awk '{s=gensub("^.*-> ","","g");print substr(s,2,length(s)-2);}')
  echo "prog='$prog'"
done
dir=`dirname "$prog"`
echo "dir='$dir'"

# 2  
Old 09-08-2009
This perhaps?
Code:
#!/bin/ksh
dir=$(pwd -P)

or do you mean the directory the scripts itself is located in?
Code:
#!/bin/ksh
dir=$(cd ${0%/*};pwd -P)


Last edited by Scrutinizer; 09-08-2009 at 05:07 PM..
# 3  
Old 09-08-2009
Quote:
Originally Posted by jeffclough
I need to have my script know what directory it's in, even if it's run from a symlink located elsewhere.

There should be no reason to need to know that. Rethink your program design.
Quote:
Here's what I've come up with, for the benefit of anyone with a similar need, but I'm also interested to know if there's a more elegant solution. I'd rather not get into awk-land, but I couldn't get the quoting and escaping right with sed.

Code:
#!/bin/ksh
prog="$0"
echo "prog='$prog'"
while [ -L "$prog" ]; do
  prog=$(stat --format %N "$prog" | awk '{s=gensub("^.*-> ","","g");print substr(s,2,length(s)-2);}')
  echo "prog='$prog'"
done
dir=`dirname "$prog"`
echo "dir='$dir'"


You don't need awk:

Code:
prog=$(stat --format %N "$prog")
prog=${prog#* -> ?}
prog=${prog%?}


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