awk specfic text and print 2 lines above

Question

I have file where I need 2 lines above my match result using awk. I have put together the below and so far so good, however I want to use match case for search:

#cat data.out
{ unload file name = dupli00913.unl number of rows = 251 }

create table vodac.duplicall2
{ unload file name = dupli01608.unl number of rows = 0 }

create table vodac.duplicall

The command I am using to achieve what I want is below:

cat data.out | awk "/create table vodac.duplicall/{for(i=1;i<=x;)print a[i++];print} {for(i=1;i<x;i++)a[i]=a[i+1];a[x]=\$0;}"  x=2 | head -1 | awk '{print $6}' 

The above gives me the below:

dupli00913.unl

The output command gives me what I need, but must be for "duplicall" table not "duplicall2". How can I achieve this in my command to match case specific string?

Answer 1

With your shown samples, could you please try following. Written and tested in GNU awk.

tac Input_file | 
awk '
  /create table vodac\.duplicall$/{
    found=1
    next
  }
  found && ++count==2{
    print
    exit
}'

Explanation: Adding detailed explanation for above.

tac Input_file |                        ##Using tac Input_file to print lines in reverse order(from last line to first line).
awk '                                   ##Sending tac command output to awk program here.
  /create table vodac\.duplicall$/{     ##Checking condition if line contains create table vodac\.duplicall here.
    found=1                             ##If above condition is TRUE then set found to 1 here.
    next                                ##next will skip all further statements from here.
  }
  found && ++count==2{                  ##Checking condition if found is SET and count is 2 then do following.
    print                               ##Printing current line here.
    exit                                ##exiting from program from here.
}'