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Selective printing based on matched record

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# 1  
Old 04-07-2017
Selective printing based on matched record
Hi,

I have file with thousands of lines somewhat similar to the below 6 lines

Code:
06MXXXXXXXXXXXXXXXX 0328 003529         J27300022      MICROSOFT   *MSN         
06<1      000000001344392   JPN151-85830 MSBILL.INFO      F                     
06<A17087454000328651551    MSBILL.INFO  JPN0000000000000000000000000060300     
06<BZ20C0WS0L208                              5627                              
06=0000000000000000001344MCCAW5PIO0328      4816    0                           
06=51709000100020 09000--

I want to search for the line starting with a '06 then an M or V' Or' 07 then an M or V' with any characters till 003529 (which is at the 26-31st position). Once this match is found, it should print out the next 5 lines, including the matched line, so in effect a total of 6 lines.

I was using a somewhat similar awk command to search for lines which have 003529 at the specific position
, but couldn't figure out how to write the regex. Also couldn't figure out how I would print the next 5 (total 6 including the match) lines.
Code:
awk '{if (substr($0,26,6)=="003529")print $0}' file

what I had thought was if I could get the matched line $0 as NR==1, then I could print out NR1 to NR==6 when there was a match.

Please need some expert advice on this
# 2  
Old 04-07-2017
Hello dsid,

Could you please try following and let me know if this helps.
Code:
awk '/^06[MV].*003529/'  Input_file

NOTE: You had mentioned like after chars M or V, 1 more time M or V should come but couldn't see this into your output, let us know if this is having more conditions etc.

Thanks,
R. Singh
# 3  
Old 04-07-2017
Try also
Code:
grep -A5 "^0[67][MV].\{22\}003529" file

OR

Code:
awk '/^0[67][MV]......................003529/ {L = NR + 6} NR < L' file

This User Gave Thanks to RudiC For This Post:
dsid
# 4  
Old 04-07-2017
Quote:
Originally Posted by RavinderSingh13
Hello dsid,

Could you please try following and let me know if this helps.
Code:
awk '/^06[MV].*003529/'  Input_file

NOTE: You had mentioned like after chars M or V, 1 more time M or V should come but couldn't see this into your output, let us know if this is having more conditions etc.

Thanks,
R. Singh

I had tried your command with a little tweak

Code:
awk '/^0[67][MV].*003529/'

and the output I got was what it was supposed to give. However, it isn't exactly suitable for me as I am looking for only those lines which have 003529 in the 26-31 position.

Also there is either a 06M or 06V or 07M or 07V in the line that I need to search. When the regex matches, it should print the next 5 lines, including $0, in total 6 lines.

Hope I am clear now.

---------- Post updated at 10:28 AM ---------- Previous update was at 10:19 AM ----------

Quote:
Originally Posted by RudiC
Try also
Code:
grep -A5 "^0[67][MV].\{22\}003529" file

OR

Code:
awk '/^0[67][MV]......................003529/ {L = NR + 6} NR < L' file

Hi RudiC,

Can you please explain to me what does the part in RED mean?

Also is there a way where we can specify that 003529 comes between position 26,31. Reason Im asking is if there is a need for me to specify characters which are say between the position 100-106 then it kind of becomes difficult for me to mention the dots.

Code:
grep -A5 "^0[67][MV].\{22\}003529" file

is really fast compared to awk
# 5  
Old 04-07-2017
NR is the actual line No., for instance 17. You want six additional lines printed, so L becomes 23, and awk will print (default action if a pattern is TRUE) as long as NR is less than 23.

My awk versions don't seem to accept what is called a "bound" (c.f. man regex); you need to try it on your own, and YMMV.
Or, compose the pattern from a leading regex match, and a substr () match afterwards.
# 6  
Old 04-07-2017
Quote:
Originally Posted by dsid
I had tried your command with a little tweak
Code:
awk '/^0[67][MV].*003529/'

and the output I got was what it was supposed to give. However, it isn't exactly suitable for me as I am looking for only those lines which have 003529 in the 26-31 position.
Also there is either a 06M or 06V or 07M or 07V in the line that I need to search. When the regex matches, it should print the next 5 lines, including $0, in total 6 lines.
Hope I am clear now.
Hello dsid,

Could you please try following and let me know if this helps you.
Code:
awk '/^0[67][MV]/ && substr($0,26,1)==0 && substr($0,27,1)==0 && substr($0,28,1)==3 && substr($0,29,1)==5 && substr($0,30,1)==2 && substr($0,31,1)==9'  Input_file

EDIT: In case you want to print the next 6 lines after the matching line, you could do the following then.
Code:
awk '/^0[67][MV]/ && substr($0,26,1)==0 && substr($0,27,1)==0 && substr($0,28,1)==3 && substr($0,29,1)==5 && substr($0,30,1)==2 && substr($0,31,1)==9{;print;A=6;next} A>0{print;A--}'  Input_file

Thanks,
R. Singh
Last edited by RavinderSingh13; 04-07-2017 at 07:10 AM..
# 7  
Old 04-07-2017
Quote:
Originally Posted by RavinderSingh13
Hello dsid,

Could you please try following and let me know if this helps you.
Code:
awk '/^0[67][MV]/ && substr($0,26,1)==0 && substr($0,27,1)==0 && substr($0,28,1)==3 && substr($0,29,1)==5 && substr($0,30,1)==2 && substr($0,31,1)==9'  Input_file

EDIT: In case you want to print the next 6 lines after the matching line, you could do the following then.
Code:
awk '/^0[67][MV]/ && substr($0,26,1)==0 && substr($0,27,1)==0 && substr($0,28,1)==3 && substr($0,29,1)==5 && substr($0,30,1)==2 && substr($0,31,1)==9{;print;A=6;next} A>0{print;A--}'  Input_file

Thanks,
R. Singh
both of your commands work, however, I had to tweak your second command and put a '5' instead of 6 cos it was printing that extra line at the end. can you please explain the whole command and why are individual substring() needed.


Quote:
Originally Posted by RudiC
NR is the actual line No., for instance 17. You want six additional lines printed, so L becomes 23, and awk will print (default action if a pattern is TRUE) as long as NR is less than 23.

My awk versions don't seem to accept what is called a "bound" (c.f. man regex); you need to try it on your own, and YMMV.
Or, compose the pattern from a leading regex match, and a substr () match afterwards.
After a lot of head scratching, I finally managed to find a way out using sed
Code:
sed -n '/0[67][MV].\{22\}003529/,+5p' 2017087222049-26-S-T-M-0_SW.inc

I couldn't figure out a way to mention specific sets using regex in awk (highlighted in RED). The only thing I could figure out was
Code:
awk '/\y003529\y/ {L = NR + 6} NR < L'

but that wasn't doing the trick either
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