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awk to get string

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# 1  
Old 06-22-2017
awk to get string
Hello Gurus,
I am facing one issue to get a portion of string from /etc/oraInst.loc
I ran the below command
Code:
cat /etc/oraInst.loc | grep VIS | cut -d "=" -f2

the result is as below :
Code:
/u01/oracle/VIS/oraInventory

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

How to achieve this.

Please advice.

Thanks-
P


Moderator's Comments:
Mod Comment Please use CODE tags correctly as required by forum rules!
Last edited by RudiC; 06-22-2017 at 06:39 PM.. Reason: Changed CODE tags.
# 2  
Old 06-22-2017
Hi pokhraj_d,

Obviously
Code:
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  
Old 06-22-2017
Try
Code:
awk -F= '/VIS/ {n=split ($2, T, "/"); sub ("/" T[n] "$", "", $2); print $2}' /etc/oraInst.loc
/u01/oracle/VIS

Last edited by RudiC; 06-23-2017 at 02:55 AM.. Reason: Added field separator.
# 4  
Old 06-23-2017
Try:
Code:
while IFS== read key val rest
do
  case $val in
    (*VIS*) echo "${val%/*}"
  esac
done < /etc/oraInst.loc

# 5  
Old 06-23-2017
sed version:
Code:
sed -n 's#^.*=\(.*VIS\).*$#\1#p' /etc/oraInst.loc

Andrew
# 6  
Old 06-23-2017
Hello Gurus,
Thank you very much... It worked..

Thanks-
P
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