mjhjmtu

Looking to use any command to produce the following outcome. Edited for greater clarity. I'm trying to remove any ^abc that is not a line before ^xyz.

stdout:

abc def
abc ghi
abc jkl
xyz mno
xyz pqr
abc def
abc ghi
abc jkl
xyz mno
xyz pqr

desired result:

abc jkl
xyz mno
xyz pqr
abc jkl
xyz mno
xyz pqr

You could maintain a two-line buffer (using a N D loop) and print the first line before deleting it only if there's a xyz in the buffer:

$ cat stdout | sed ':a; $!N; /xyz/P; D; ba'
abc jkl
xyz mno
xyz pqr
abc jkl
xyz mno
xyz pqr

Here is an awk one-liner.

awk '/^abc/ prev = $0; prevabc = "true"; /^xyz/ if (prevabc == "true") print prev; prevabc = "false" ; print' file.txt

Explanation

• 
  /^abc/ prev = $0; prevabc = "true";: If the line starts with abc, don't print it, but instead store the value of the line in the variable prev, and indicated that this line started with abc by storing true in variable prevabc.


• 
  /^xyz/ if (prevabc == "true") print prev; prevabc = "false" ; print: If the line starts with xyz, the do the following.
  
  
  • 
    if (prevabc == "true") print prev; prevabc = "false" : If the previous line started with abc (because prevabc == "true"), then print the previous line prev and reset prevabc to false.
  
  
  • 
    print: Since this line started with xyz, then print this line too.