Shell Programming and Scripting

I am trying to grep a section of .txt file...but once I grep the certain area of the file I would like to display all lines below it as well....how do I have to go about doing this...

example

grep "Sidney Crosby" hockey.txt

result
Sidney Crosby

Age
Goals
Assist


Can this be done with grep?

Code:

$ awk '/Sidney Crosby/,0' hockey.txt

Afaik grep will work, but you need to know beforehand how many lines you want to be displayed when the string is found e.g.

Code:

$ grep -A10 "Sidney Crosby" hockey.txt

Perfect! Thank you very much

is there an easy way to make it display only the information up to the next players name?

You're welcome.
There is definitely a way to extract that data (although I'm not sure if I myself can provide you that command, but I will try).
Can you post some sample data with at least 2 or 3 players?

I have not actaully created the text file yet but it will look something like this:

I can add an @ sign before each name if that helps at all

@Sidney Crosby

Goals - 51
Assist - 47
Pts - 98
Country: Canada


@Alex Ovechkin

Goals - 48
Assist - 30
Pts - 78
Country: Russia


@Jordan Stall

Goals - 22
Assist - 39
Pts - 61
Country: Canada

---------- Post updated at 09:08 PM ---------- Previous update was at 09:01 PM ----------

some players will have more lines than others because some will have playoff stats, olympic stats, ect and others wont

In this case it's not necessary to add "@", because the structure is pretty clear and does not vary.

Code:

 grep -A5 "Alex Ovechkin" hockey.txt

Edit: Okay, I've read your update right now regarding varying number of lines....

can i use an UNTIL loop somehow

something like :

until [ line = '^@' ] display lines

i dont know the syntax obv but is that an idea that might work?