mjhjmtu

I have a config file and its content looks something like below:

 Jobname|Type|Silo|Description
 #comment1
 #comment2
 job1|me|silo1|test_job1
 job1|me|silo1|test_job2
 job1|prod|silo1|test_job3

Now I need the conditional content of the file, say the content with TYPE =me.
For this I am using a grep with me:

 job_detail=$((cat config_file | grep me | awk 'print $4'))

In this case, I am getting the first line also , as JOBNAME is getting matching character me.
I escaped comments with -v options. I can't comment the first line of the config file, as it is used by other unknown processes.

Is there a way i can grep the whole word match? would be better if there is a way to grep whole word with a particular character as condition.

A way to divide the line with '|' and then grep?

try

awk -F| -v select="$var" '$2 == select print $4;' config_file

where

• 
  $var contains the field you want to select


• -F| tell awk to use | as separator, | (pipr) must be escaped.


• -v select="$var" transfer $var to awk variable (select)


• 
  $2 == select select line whose second arg is "$var" or select.


• 
  print $4; print fourth field.

man grep will show you the -w flag:

-w, --word-regexp Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character. Similarly, it must be either at the end of the line or followed by a non-word constituent character. Word-constituent characters are letters, digits, and the underscore. 

Or, stick | egrep -v Jobname early in your pipline.

A variation of Archemar's solution which assumes that the me that you want to search for is $LOGNAME, i.e. the username of the current user:

awk -F '|' '$2 == ENVIRON["LOGNAME"] print $4 ' <config_file