output preceding line after having grep'd on string

[dominicdiddy]

Hi All,

Any ideas on how to output several lines below or above (or both) when searching for a string in a ASCII file please?

Example below,

---

Thu Nov 24 07:45:36 2005 hdgej
hdgej: New journal file opened.
hdgej: /lbtest/db/file.test.out lgdmn: version 1, sequence 5111, volume 1.

----

So if having searched for string, say "version 1", then how could I output the entire record (ie the 4 lines) surrounded by blank lines...

Any help greatly appreciated!!!

Thanks in advance,

Dominic

 

[PHV]

You may try this:
awk 'BEGIN{RS=""}/version 1/' /path/to/input

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[feherke]

Hi

Maybe some options ?

-A NUM, --after-context=NUM
Print NUM lines of trailing context after matching lines. Places a line containing -- between con­tiguous groups of matches.

-B NUM, --before-context=NUM
Print NUM lines of leading context before matching lines. Places a line containing -- between contiguous groups of matches.

-C NUM, --context=NUM
Print NUM lines of output context. Places a line containing -- between contiguous groups of matches.

Feherke.

http://rootshell.be/~feherke/

 

[dominicdiddy]

Thanks both for your help.

 

[dominicdiddy]

Afternoon all,

Once I have found the entries with the "emergency" string in them, how would I just select the ones for today and then echo a staement if found?

Any help greatly appreciated...

 

[dominicdiddy]

sorry the previous string should have been "version 1"...

 

[p5wizard]

re_today=$(date +'%a %b %d ..:..:.. %Y')
grep -p "version 1" file|egrep "^${re_today}"

should do it.


HTH,

p5wizard

 

[dominicdiddy]

Thanks for that but now getting this... Any ideas?

echo $DATE
+ echo Jan 13
Jan 13
# grep -p "emergency" /lb/db/file.ral | egrep "^${DATE}"
+ grep -p emergency /lb/db/file.ral
+ egrep ^Jan 13
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
/lbprod/db/file.ral: 0652-034 The maximum paragraph length is 10,000 bytes.
#
# grep -p "emergency" /lb/db/file.ral | egrep "^${DATE}

 

[p5wizard]

Some of your paragraphs are too long...

either ignore, if your paragraphs of interest are small enough, or use PHV's awk solution to get the paragraphs of interest:

awk 'BEGIN{RS=""}/version 1/' /path/to/input | grep "^${DATE}"

or

grep -p "version 1" /path/to/input 2>/dev/null | grep "^${DATE}"


but I would also put the weekday in your date search string if you're grepping for the date at the beginning of the line (^):

DATE="Fri Jan 13"

substitute "version 1" for "emergency" if that's what you're really after...


HTH,

p5wizard

 

[dominicdiddy]

sorry but....

# awk 'BEGIN{RS=""}/emergency/' /lb/db/file.ral | grep "^${DATE}"
+ awk BEGIN{RS=""}/emergency/ /lb/db/file.ral
+ grep ^Fri Jan 13
awk: 0602-534 Input line Mon Sep 19 19:09:49 cannot be longer than 10,239 bytes
.
The input line number is 580. The file is /lb/db/file.ral.
The source line number is 1.

why on Fri afternoon???!

 

[p5wizard]

Well it was friday 13th, right?

You might want to reduce line length for awk to process by first fold-ing the input file into smaller lines...

fold -w 1000 /path/to/inputfile | awk 'BEGIN...


HTH,

p5wizard

 

[Ghodmode]

Here's the method I prefer to use ...

tail +$((`grep -n 'emergency' /lb/db/file.ral | cut -d: -f1` - 5)) /lb/db/file.ral | head

It may be a little inefficient, but I think it does what you want. It will give ten lines of the file, starting five lines before your matched expression.

[ol]
[li][tt]tail +#[/tt] gives the end of a file starting at # lines from the beginning of the file.[/li]
[li][tt]grep -n[/tt] precedes it's output with the line number of the match and a colon.[/li]
[li][tt]cut -ddelimiter -f#[/tt] uses the colon as a delimiter and returns the first field.[/li]
[li][tt]$((expr))[/tt] is a bashism that does some math for you. I think it works in Korn, too. If that one doesn't work for you, I'm sure there are other ways to get the math done. So, my sample subtracts 5 from the line number.[/li]
[li]The second occurrence of the filename is an argument to the tail command at the beginning of the line.[/li]
[li]Pipe (|) the whole mess through head, which defaults to 10 lines.[/li]
[/ol]
So, you get 10 lines, starting at 5 lines before the match. If you want more, or different, lines, be sure to adjust both the 5 in the double-parentheses and add a -n to [tt]head[/tt].

My grep command doesn't have a [tt]-p[/tt] argument... what's it do?

--
-- Ghodmode

 

[p5wizard]

on AIX - grep -p does a paragraph search - paragraphs containing a RE

HTH,

p5wizard