in reply to Get Parent Directory
This may help you understand what the code trying to do.
# Set the path variable. $path = "\\default\\main\\Anand\\toipcs\\Tutorials\\internet"; # Make a copy of it $curPath = $path; # Extract (what I assume to be) your user/home directory ("Anand"). # However this line won't work because the substitution operator is m +issing the 's' $curPath =~ !\default\main\(.*)\Tutorials\internet!$1!; # ^ There should be an 's' here # Make a copy of the extracted directory name ("Anand") and print it +out $parentdir = $curPath; print "parent Directory : $curPath\n"; # This line doesn't doesn't make any sense at all??. # It's obviously trying to extract 2 different parts of the path fro +om something # But # a) The is no variable to operate on! It would need to look somethi +ng like: # ($Pdir, $Sdir) = $path =~ m!(.*)\\(.*)!; !!!NOTE the double +d backslash (\\)! # b) Quite what the intent of the '$1,$2' at the end was meant to d +o. ($Pdir,$Sdir) =~ m!(.*)\(.*)$!$1,$2; # Nothing will be printed as the line above did nothing! print "parent branch : $Pdir \t sub branch $Sdir";
However, I think you need to try and explain what it is that you want the code to do, then it may be possible to help you further.
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Get Parent Directory : redefined
by Anonymous Monk on Oct 22, 2002 at 12:05 UTC | |
by BrowserUk (Patriarch) on Oct 22, 2002 at 12:42 UTC | |
by Ananda (Pilgrim) on Oct 23, 2002 at 04:06 UTC |