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grep to get line numbers

 
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# 1  
Old 04-14-2011
grep to get line numbers
I know if i use grep -n that the output will have the lines numbered but is there a way to grep the actually line number.

so like this
Code:
grep -n "one" /usr/dict/numbers
1:one
21:twenty-one
31:thirty-one
41:forty-one
51:fifty-one
61:sixty-one
71:seventy-one
81:eighty-one
91:ninety-one

but using grep to actually work with the line numbers. So if i wanted to use grep for only even line numbers, or line numbers that are any number followed by 1.

---------- Post updated at 06:50 PM ---------- Previous update was at 06:48 PM ----------

Can i use delimiters or fields with grep/egrep?

So like when using

Code:
cut -d ":" -f1

# 2  
Old 04-14-2011
grep does one thing: match lines. It's not a language, so it can't match things conditionally.

awk is a language though, with features that let you build small scripts that do a lot.

It seems cryptic but there's a simple secret to understanding it: The code inside the single big { } block is run repeatedly, once every single line, and sets variables $0(entire line), $1 (first token), $2, ... $NR as it splits the line apart on spaces. The difference between N and $N, is that N gives you the contents of the variable N, and $N gives you the contents of token number N(if N was 4, you get token 4)

Code:
# Print with line numbers.
# It sets the entire line, $0, to the special var NR (line number),
# followed by :, followed by the rest of the original line.
# The 1 at the end is a statement telling it whether to print or not.
# 1 is always true, so always prints.
awk '{ $0=NR ":" $0; } 1' inputfile

# /one/ is only true when the input string contains one, so
# this numbers the lines but only prints lines containing 'one'.
awk '{ $0=NR ":" $0; } /one/' inputfile

# The same as above, with another condition, odd lines only
awk '{ $0=NR ":" $0; } /one/ && (NR%2)' inputfile

# Same as above except even lines only
awk '{ $0=NR ":" $0; } /one/ && !(NR%2)' inputfile

 
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