Printing 10 lines above and below the search string: help needed

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Top Forums Shell Programming and Scripting Printing 10 lines above and below the search string: help needed
# 1  
Old 02-02-2010
Debian Printing 10 lines above and below the search string: help needed


The below code will search a particular string(say false in this case) and return me 10 lines above and below the search string in a file.

awk 'c-->0;$0~s{if(b)for(c=b+1;c>1;c--)print r[(NR-c+1)%b];print("***********************************");print;c=a;}b{r[NR%b]=$
0}' b=10 a=10 s="false"


Further to enhance I want to ignore if the same search string appears in those 10 lines. (Currently this is not happening. For Ex: If the search string exists in 5th line it again starts printing 10lines above and below causing some mess.)

It would be really great if someone could help me on this.

P.S: Being my first post apologies for any mistakes in my post and suggestions welcome.

# 2  
Old 02-02-2010
Does your version of grep support
grep -A 10 -B 10 'mypattern'   filename

That does exaclty what you want.
# 3  
Old 02-04-2010

Thanks a lot but I m afraid that when i execute the above command it says "-A" not a valid option.

Please suggest me a different option to try with.
# 4  
Old 02-04-2010
in solaris use below

/usr/sfw/bin/ggrep -A 10 -B 10 'mypattern'   filename

# 5  
Old 02-04-2010
Try this one:
awk '
/false/ {
  for(i=0;i<10;i++) {
    print a[(NR+i)%10]
' file

# 6  
Old 03-01-2010
Dear Friend,

You can use the following code also

  egrep -A 10 -B 10 "pattern" file


-A is used to print the 10 lines after the matched patten
-B is used to print the 10 lines before the matched pattern
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