Shell Programming and Scripting

In the past I needed a help with the problem how to search for pattern after the occurence of another pattern which is described in this thread:

https://www.unix.com/shell-programmin...-pattern1.html

Now I would need something quite similar, only the pattern which is to be searched must be preceded by occurence of more patterns (in given order), not just by one.

For example:

Code:

any lines
pattern1
any lines
pattern3
...
anylines
pattern1
any lines
pattern2
any lines
pattern3
any lines
pattern4
line1
line2
the line I need to get
any lines

I need to get the third line after the occurence of the pattern4 but only if pattern1, pattern2 , pattern3 has occured before in given order!.

Would you help me please?

Thanks

Moderator's Comments:

Video tutorial on how to use code tags in The UNIX and Linux Forums.

I suppose you meant the output should be line2?

Try this...

Code:

sed -n 'H;/pattern4/{x;/pattern1.*pattern2.*pattern3/{n;n;p}}' input_file

--ahamed

I mean the output should be "the line I need to get". The example wasn't originaly quoted correctly (meantime I have corrected it). But if I understand your proposal well, only one more ';n' should be added to 'print' section of your sed script.

Anyhow, I don't understand your script well - would you explain please the meaning of 'H;', 'x;' and why pattern2 & pattern3 are preceded by asterisk ('*')?

Thanks

yes you are right, add one more n

Code:

sed -n 'H;/pattern4/{x;/pattern1.*pattern2.*pattern3/{n;n;n;p}}' input_file

--ahamed

---------- Post updated at 08:44 PM ---------- Previous update was at 07:06 AM ----------

H : Hold the incoming data into hold buffer.
When you find pattern4, then get the held data into current space (using x - exchange) and search if pattern1, pattern2 and pattern3 are present and in the same order.
If present, read 3 more lines(using n) and print it.

--ahamed