awk question


 
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# 1  
Old 07-23-2012
awk question

Hi experts,

Is it possiable to use awk print the line before one line 2 line for example

Code:
1111
2222
3333
4444
5555

I want to print the line bofore the line "5555" 2 line, the result should be 3333 something like grep -A (but not exactly the same, becasue grep -A 2 will print 3333 4444)
# 2  
Old 07-23-2012
Try:
Code:
awk '/555/{print A[(NR-2)%3]}{A[NR%3]=$0}' infile

more general:
Code:
awk '/555/{print A[(NR-c)%(c+1)]}{A[NR%(c+1)]=$0}' c=2 infile

# 3  
Old 07-23-2012
Quote:
Originally Posted by Scrutinizer
Try:
Code:
awk '/555/{print A[(NR-2)%3]}{A[NR%3]=$0}' infile

more general:
Code:
awk '/555/{print A[(NR-c)%(c+1)]}{A[NR%(c+1)]=$0}' c=2 infile


can you explain to me?I plan to buy one awk book, which one you recommand?Smilie

Lei
# 4  
Old 07-23-2012
Why not
Code:
grep -B2 5555 | head -1

(-B, not -A)

---

And yes, this is a GNU grep extension (thanks, Corona688).

Last edited by yazu; 07-23-2012 at 01:01 PM..
# 5  
Old 07-23-2012
Yet another way with ex...
Code:
ex +'/5555/-2 | q!' file

# 6  
Old 07-23-2012
Quote:
Originally Posted by yazu
Why not
Code:
grep -B2 5555 | head -1

(-B, not -A)
I think that's a GNU extension not found on most non-linux systems...
This User Gave Thanks to Corona688 For This Post:
# 7  
Old 07-23-2012
Quote:
Originally Posted by shamrock
Yet another way with ex...
Code:
ex +'/5555/-2 | q!' file

I get an empty line...
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