Displaying lines above grep-ed line

Hi,

How do i display 15 lines above the line that i grep for? Lets say i grep for "searchstring" in a flat file, i want to display 15 lines before the "searchstring" line. is ther a way to do this? Any help is appreciated. Thx in advance.

There is a program called cgrep (for "context grep") that can do this. Here is a link. Go to that page and look for the cgrep links.

Also here is a perl implementation, and a shell script.

I've never used any of these, but I think they can do pretty much what you're looking for.

Hi Perderabo,

I wish i could use the program, but i cant. This server i am working on is a production server and i can install any new programs on it .Is ther any other way i can do this?

I was also wondering if i could display to the standard output cretain lines based on their line numbers. For example i could grep a line with its line number using the -n option then i could loop 15 lines before that and display those line. So lets say i grep the line at line number 42, i would -1 the line number and ouput the line at line 41, then -1 again and display line 40....and so on. Is there a way to display string based on specific line numbers? Thnx.

you have a couple of probls
- almost unix tools, works sequentailly, so processing line x
they have no idea about line x-1
- what when before 'searchstring' there are NOT 15 lines ?
use awk, perl or c

read the file in an array
if you found 'searchstring' at line-number >14
print elements of array[pos-15,pos]
restart the job

I finally managed to solve the line no problem. I grep-ed the line with the -n option and wrote the output to a file and grepped the line nos and did a while loop to read up and down of the grep-ed line. A bit slow but what the hell....it works

did u try grep -A (after) or grep -B (before) ?

a tricky but fast(er) way:
1) get line number (grep)
2) calculate -= START, if negative skipp it (bc)
3) generate a sed script (echo)
4) if successfull, exec it (sed)

Hi guggach,

for some strange reason, it dosent display anything if there is only one occurance of the word being grep-ed. And if the word is sandwiched between othe chars, it is not being displayed either.

hi mtrxreloaded,

I cant seem to find the -A or -B option for grep. Is that what you meant?

anyway.....this is my code
<code>
cat /tmp/vinod/linenos | while read LINE
do
lineno=$LINE
count=0
actline=0
lineend=`expr $lineno + 3`
linestr=`expr $lineno - 1`
actline=$lineno
echo >> results.txt
echo >> results.txt
while [ "$count" != "3" ]
do
actline=`expr $actline - 1`
grep $actline /tmp/vinod/searchtext >> /tmp/vinod/results.txt
count=`expr $count + 1`
if [ "$count" = "3" ];
then
grep $lineno /tmp/vinod/searchtext >> /tmp/vinod/results.txt
count=0
actline=$lineno
while [ "$count" != "3" ]
do
actline=`expr $actline + 1`
grep $actline /tmp/vinod/searchtext >> /tmp/vinod/results.txt
count=`expr $count + 1`
done
fi
echo >> /tmp/vinod/results.txt
echo >> /tmp/vinod/results.txt
done

done

</code>

-A -B is maybe gnu-grep.
you have solid shell misunderstandings

do you know the difference:
echo >> /tmp/vinod/results.txt
echo >> /tmp/vinod/results.txt

and:
(
echo
echo
) >/tmp/vinod/results.txt

put this in afile:

hallo
gugus
aaa
123
456
789
aaa
222
333
444
aaa
lollo
aaa

and try the script, it must print:

123
456
789
aaa
222
333
444
aaa

every thing else is cr...

GREP --help

Context control:
-B, --before-context=NUM print NUM lines of leading context
-A, --after-context=NUM print NUM lines of trailing context

exp: grep -A 4 "string"
- prints 4 lines after the matching string
exp: grep -B 4 "string"
- prints 4 lines before the matching string

One line command to display 5 lines above and 5 lines below the grep-ed line:

grep -n "<pattern>" testfile.txt | awk -F":" '{print $1-5 "," $1+5 "p"}' | xargs -t -i sed -n {} testfile.txt