[SOLVED] awk to search a pattern and print lines with pattern and one more condition

Issue

I have a input file as shown below:

SR  policy name    state             error code
1   backup01       successful         0
2   backup01       fail               13
3   backup01       fail               58
4   backup02       successful         0
5   backup02       successful         0
6   backup02       successful         0

I want to have an output where if any line for a particular policy (example backup01 has a line with "fail" state, it will only show "fail" state lines and will not show the "success" lines. Similarly where all lines of a policy (example backup02) have a state "successful" it will print all the "successful" state lines.

2   backup01       fail               13
3   backup01       fail               58
4   backup02       successful         0
5   backup02       successful         0
6   backup02       successful         0

I have tried using awk with little success, but not able to get ahead with a final solution.

awk '{if ($4 == 0) {print $0} else if($4 !=0 && $4 == 0) {print $0}}' input_file.txt

any other way using sed or other tool is also fine. ALso header from input file can also be ignored.

Solution

It can be done in two-stage processing of the same file. First pass, build up the status of the policy if it is going to be a "fail" or a "successful" and then exclude the contradicting lines in the next pass

awk '
  FNR == NR && NR > 1 { 
    t = key[$2]
    if ( $3 == "fail" || t == "fail" ) { 
      key[$2] = "fail" 
    } else {
      key[$2] = $3
    }
    next 
  } 
  key[$2] == $3
' file file

Answered By - Inian

Answer Checked By - Robin (WPSolving Admin)