10-15-2012
How to get top level parent directory
Hi All,
I have a directory like this: /u01/app/oracle/11gSE1/11gR203
How do i get the top level directory /u01 from this? Tried dirname and basename but dint help. I can this using echo $ORACLE_HOME | awk -F"/" '{print "/"$2}'. But I am trying to find out if there is a better way of doing it other than using awk. Thanks in advance.
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IGAWK(1) Utility Commands IGAWK(1)
NAME
igawk - gawk with include files
SYNOPSIS
igawk [ all gawk options ] -f program-file [ -- ] file ...
igawk [ all gawk options ] [ -- ] program-text file ...
DESCRIPTION
Igawk is a simple shell script that adds the ability to have ``include files'' to gawk(1).
AWK programs for igawk are the same as for gawk, except that, in addition, you may have lines like
@include getopt.awk
in your program to include the file getopt.awk from either the current directory or one of the other directories in the search path.
OPTIONS
See gawk(1) for a full description of the AWK language and the options that gawk supports.
EXAMPLES
cat << EOF > test.awk
@include getopt.awk
BEGIN {
while (getopt(ARGC, ARGV, "am:q") != -1)
...
}
EOF
igawk -f test.awk
SEE ALSO
gawk(1)
Effective AWK Programming, Edition 1.0, published by the Free Software Foundation, 1995.
AUTHOR
Arnold Robbins (arnold@skeeve.com).
Free Software Foundation Nov 3 1999 IGAWK(1)