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awk to parse jil output

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# 1  
Old 12-03-2013
awk to parse jil output
Hi ,

I have a jil file which i am trying to parse and print the job name and the condition corresponding to it.

Below is the input file

Code:
/* -------------------- testjob1 -------------------- */

insert_job: testjob1   job_type: c
machine: unix
owner: chidori
condition: s(joba) and s(jobc)
std_out_file: /export/home/chidori/test.out
std_err_file: /export/home/chidori/test.err

   /* -------------------- testjob2 -------------------- */

   insert_job: testjob2   job_type: c
   machine: unix
   owner: chidori
   condition: s(jobc)
   std_out_file: /export/home/chidori/test.out
   std_err_file: /export/home/chidori/test.err

  /* -------------------- testjob3 -------------------- */

insert_job: testjob3   job_type: c
machine: unix
owner: chidori
condition: s(joba) or s(jobc)
std_out_file: /export/home/chidori/test.out
std_err_file: /export/home/chidori/test.err


I have figured out this one liner to do this

awk '{sub(/^ */,"");if($0~/^$){a++};if($0~/^\//){printf "%s = ", $3};if($0~/^condition:/){sub("condition: ","");print}}' sample.jil

Condition:
At times a job might not have any condition in them , in that case the script should print as "No_Condition".

I am not able to figure out how to write the code that can check if condition string does not exist then print "No_Condition". Can someone please help me out.

Also i was thinking if we can consider from the line that begins with /* to the line before the next /* as one record and do some processing. But was not sure how to implement the same.
# 2  
Old 12-03-2013
Here is one way of doing it:
Code:
awk '
        /^[ ]*\/\*/ {
                if ( NR > 1 && !(f) )
                        print "No_Condition"
                f = 0
                printf ( "%s = ", $3 )
        }
        /condition:/ {
                f = 1
                sub ( /[ ]*condition:[ ]*/, X)
                print
        }
        END {
                if (!(f))
                        print "No_Condition"
        }
' sample.jil

This User Gave Thanks to Yoda For This Post:
chidori
# 3  
Old 12-03-2013
That worked!!
Can you please explain the logic around here
# 4  
Old 12-03-2013
This is a starting point and needs some refinement, but it can guide you to a better solution:
Code:
awk 'NF==1 {next} {for (i=1; i<=NF; i++) if ($i ~ /insert|condition| or|and/) print $(i+1)} !/condition/ {print "no_condition"}' RS=-+ file
testjob1
s(joba)
s(jobc)
testjob2
s(jobc)
testjob3
no_condition

(I removed the condition line from the third record)
# 5  
Old 12-03-2013
If am using a flag variable: f and it is set when awk finds pattern /condition:/

So when awk finds next pattern /^[ ]*\/\*/ for job name, I am checking if this flag variable is set or not.

If not set, then print No_Condition. Same is done in the END rule to cover last job name.
# 6  
Old 12-03-2013
Thanks, Whats up with NR > 1 why we are going for a record greater than 1

---------- Post updated at 12:28 PM ---------- Previous update was at 12:26 PM ----------

Quote:
Originally Posted by RudiC
This is a starting point and needs some refinement, but it can guide you to a better solution:
Code:
awk 'NF==1 {next} {for (i=1; i<=NF; i++) if ($i ~ /insert|condition| or|and/) print $(i+1)} !/condition/ {print "no_condition"}' RS=-+ file
testjob1
s(joba)
s(jobc)
testjob2
s(jobc)
testjob3
no_condition

(I removed the condition line from the third record)
Hey RudiC, Can you please explain the record seperator we use here. RS=-+
# 7  
Old 12-03-2013
I used NR > 1 because for first job name, the flag variable: f is going to be initially 0 or null anyway.

So I don't want to check if flag is set or not for first job, but for all subsequent jobs.
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