I want to output two lines , line no 8 and line no 4 .The condition being the first occurrence of 0- or -0 and then the looking back and selecting the line with 1234.
So the two lines out put would be
Hi Akshay,
Works like a charm on the text snippet that I had attached earlier.
Somehow it is not working on the text snippet that I'm working on :-
Snippet starts :-
-------------------------------------------------
Load step no. = 1
Load increment scaled to
maximum load level
Load increment = 1.000
New load level = 1.000
Current stiffness parameter = 1.000
Solution accuracy parameter = 3.848*E-00009
Determinant of tangential matrix = 4.427*E189782
Number of Negative Pivot Element = 0
Total energy absorbtion = 8.734*E 00006
-------- I N T E R A C T I O N F U N C T I O N V A L U E S Fb(Fy) ---
ELEM ES Node1 Midspan Node2
1446 0 -0.44(-0.29) -0.73(-0.65) -1.00(-1.00)
1447 0 -0.51(-0.47) -0.55(-0.52) -0.60(-0.57)
1448 0 -0.42(-0.38) -0.46(-0.42) -0.51(-0.47)
1449 0 -0.40(-0.25) -0.71(-0.63) -0.98(-0.98)
1451 0 -0.51(-0.48) -0.55(-0.52) -0.59(-0.56)
1453 Yield at end 1
1453 Yield at end 2
1453 3 -0.07( 0.21) -0.82(-0.74) -0.21( 0.04) O---+---O
1454 0 -0.12(-0.05) -0.24(-0.18) -0.38(-0.33)
1455 Yield at end 2
I did try to change the awk command but I'm getting a null string as output.
Rgds
SShinde
---------- Post updated at 01:42 PM ---------- Previous update was at 12:34 PM ----------
Hi Akshay,
Discovered a shortcoming in the awk command .
The awk command is
awk '/1234/{p=$0;X=NR}/0\-|\-0/ && NR-X == 4{print p RS $0}' file
In the awk command, the absolute value of 4 is creating the problem.
After finding 0- or -0 , I look back till I find 1234, which may occur more than 4 lines behind the first occurrence of 0- or -0.
---------- Post updated 29-11-13 at 09:23 AM ---------- Previous update was 28-11-13 at 09:13 PM ----------
-n : only lines explicitly selected for output are written.
here, we treat three pattern: /0-/ , /-0/ and /1234/ .
For the patterns /0-/ and /-0/ , we do code block x;p;x;p;q where: x : Exchange the contents of the pattern and hold space p : Write the pattern space to standard output. q : Quit
For the pattern /1234/ we exchange the contents of the pattern and hold space.
When sed begin a new cycle, the pattern space is empty and the new line will insert in. The hold space is inchanged.
The idea is to catch the pattern /1234/ in hold space and write to output the hold space before the pattern space when the pattern /0-/ or /-0/ are detected and to quit.
Thanks Made In Germany,
I will give a it a try to interpret it.Please comment on any mistakes on my part.
Look for regex 1234, if true, then pass this to hold buffer.
Look for pattern 0-.-0, if true then exchange, hold buffer goes to pattern space and print pattern space. This prints line containing 1234
3.Then again exchange and print .This prints line containing 0-.-0 .
4.Then quit. This ensures that ONLY the first instance of the occurrence gets printed.
Rgds
SShinde
---------- Post updated at 04:14 AM ---------- Previous update was at 04:07 AM ----------
Just another thought,
Supposing I were looking to print the FIRST THREE instances of the pattern.
How would the sed command look like?
Rgds
SShinde
You got it, well explained!
--
sed is limited - no variables that can be used as counters.
If you have a "do it n-times" requirement then it's time for awk or perl.
BTW sed has a n-times modifier for its s command, and its buffers have multi-line capability, so there might be even an artistic solution...
Artistic solution as here (in red, all pattern find (five) , but print only FIRST FOUR print - see red value in sed command -)
But here, this syntax is not for all sed version:
Phew!