selective deletion of lines between two patterns

Question

I have a file which contains following content,

Hello world Unix!!
Its bright world
Current time is HH:mm:ss
Next action plan is item #3
End of task.
Thank you.


Hello world Linux!!
All actions completed.
End of Activity.
Thanks.

I would like to selectively delete all the lines that contain keyword, "world" and also all the subsequent lines until a line with specific set of keywords (Example, "Activity" or "task") are encountered.

Desired output :

End of task.
Thank you.
End of Activity.
Thanks. 

Answer 1

Try:

$ awk  -v f=1 '/world/{f=0} /Activity|task/{f=1} f' File
End of task.
Thank you.


End of Activity.
Thanks.

How it works:

1. -v f=1

   Create an awk variable f and set it to 1.

2. /world/{f=0}

   If the current line contains world, set variable f to zero.

3. /Activity|task/{f=1}

   If the line contains either Activity or task, then set f to 1.

   Note that world above and Activity or task here are treated as regular expressions. Also, they are case-sensitive.

4. f

   If f is nonzero, print the line.

To also delete blank lines

$ awk  -v f=1 '/world/{f=0} /Activity|task/{f=1} f && /./' File
End of task.
Thank you.
End of Activity.
Thanks.

f && /./ tells awk to print the line only if f is non-zero and the line contains at least one character (of any kind).

Answer 2

Not as pretty as the awk solution, but since the OP tagged the question with sed too, here goes:

#with blank lines
sed '/world/,/task\|Activity/{//!d};/world/d' file

#without blank lines
sed '/world/,/task\|Activity/{//!d};/world\|^$/d' file

Note the use of //, which means the previous regular expression used :)