Extract the first 2 directories in a path

Question

I need to extract the first 2 directories in a path that is in the variable $ORACLE_HOME

ORACLE_HOME=/oradba/app/oracle/product/11.2.0.4/testdb

I need the value: /oradba/app/

Answer 1

1. Using cut (tacking on the final /):

   ORACLE_HOME=$(printf "%s" "$ORACLE_HOME" | cut -d/ -f1-3)/
   

2. Using parameter expansion twice; the first strips away the first two elements of the directory, then the second strips that remainder away from the original variable:

   suffix=${ORACLE_HOME#/*/*/}
   ORACLE_HOME=${ORACLE_HOME%"$suffix"}
   

Answer 2

Two alternatives. The latter is more convoluted, but safer as it doesn't require you to know how many directories are in the variable:

$ echo ${ORACLE_HOME%/*/*/*/*}/
/oradba/app/
$ echo $ORACLE_HOME | awk -F/ 'BEGIN {OFS="/"} {print $1,$2,$3,""}'
/oradba/app/

Answer 3

Try:

echo $ORACLE_HOME | cut -d/ -f-3

For example:

$ ORACLE_HOME=/oradba/app/oracle/product/11.2.0.4/testdb
$ echo $ORACLE_HOME | cut -d/ -f-3
/oradba/app
$