awk to get string

Shell Programming and Scripting
1

Hello Gurus,
I am facing one issue to get a portion of string from /etc/oraInst.loc
I ran the below command

cat /etc/oraInst.loc | grep VIS | cut -d "=" -f2

the result is as below :

/u01/oracle/VIS/oraInventory

But I want only upto /u01/oracle/VIS .

How to achieve this.

Please advice.

Thanks-
P

2

Hi pokhraj_d,

Obviously

cat /etc/oraInst.loc | grep VIS | cut -d "=" -f2 | sed 's#/oraInventory##'

would work but if you need a more generic answer, you need to provide the content of /etc/oraInst.loc.

Regards
Santiago

3

Try

awk -F= '/VIS/ {n=split ($2, T, "/"); sub ("/" T[n] "$", "", $2); print $2}' /etc/oraInst.loc
/u01/oracle/VIS
4

Try:

while IFS== read key val rest
do
  case $val in
    (*VIS*) echo "${val%/*}"
  esac
done < /etc/oraInst.loc
5

sed version:

sed -n 's#^.*=\(.*VIS\).*$#\1#p' /etc/oraInst.loc

Andrew

6

Hello Gurus,
Thank you very much... It worked..

Thanks-
P