grep and pull number of lines above...

[ryanc2]

I need to pull the 2 lines before a string of "^" charactors.

For example:

Date
Version
^^^^^^^^^^^^^^^^^^^

Is it possible to grep for the ^ and retrieve the previous 2 lines?

Thanks in advance.

 

[Salem]

Well if you have GNU grep, then you can use this option

`-B NUM'
`--before-context=NUM'
     Print NUM lines of leading context before matching lines.

 

[ryanc2]

I'm working in SCO UNIX. I found that option searching some other forums, but it won't work for me.

 

[Salem]

Well you can do it with an AWK script

#!/bin/awk -f
# find the match
/^\^/ {
  print l1
  print l2
}

# record the previous two lines
{
    l1 = l2
    l2 = $0
}
# end

Save this as say 'myscript' (choose your own name), then type this command line

awk -f myscript file.txt

file.txt is the file you want to search

 

[Ygor]

Or just...

awk '/^\^/ {print a[(NR-2)%2] "\n" a[(NR-1)%2]} {a[NR%2]=$0}' file.txt