Discard part of a file based on a pattern ---


 
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# 1  
Old 05-19-2014
Discard part of a file based on a pattern ---

I have the file:

Code:
s3_T0(2) Pos  "1" "2"
s1_T1(2) Pos  "1" "2"
---
0 0
1 0
0 1
1 1
---
1 2 "tau0"
1 2 "h10"

I want to patternmatch on ---
and get only the third part i.e.

Code:
1 2 "tau0"
1 2 "h10"

I wanted to start simple but even something like
Code:
awk '/---/{x="F"++i;}{print > x;}' file2

does not work.
Any suggestion appreciated.

Moderator's Comments:
Mod Comment edit by bakunin: please use CODE-tags for your code as well as file content and terminal output. Thank you.

Last edited by bakunin; 05-19-2014 at 12:17 PM..
# 2  
Old 05-19-2014
try
Code:
awk '/1 2/' file

# 3  
Old 05-19-2014
Thanks, but the rest of the file can be quite big aND not necessarily starting with 1 2.
I want the part after the second --- no matter what it is.
# 4  
Old 05-19-2014
Hello try
Code:
nawk '{A[++c] = $0} END { for ( i = 1; i <=c; i++ ) {if(A[i] ~ "---") {d++} { if ( d  == 2) print A[i+1]}}}' file

# 5  
Old 05-19-2014
Try this:
Code:
awk '/---/{p++;next} p==2' file

This User Gave Thanks to Subbeh For This Post:
# 6  
Old 05-19-2014
thanks, both solutiuons work well!
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