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Getting number of lines of nth occurrency

 
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# 1  
Old 12-09-2009
Getting number of lines of nth occurrency
Hi all,
I would like to extract the line number of the n-th occurrency of a given string in a file.
e.g.

xxx
yyy
xxx
zzz
xxx

the second occurrency of xxx is at line 3.

What is the fastest way to do it in bash?
Thank you,
# 2  
Old 12-09-2009
Try:

Code:
awk '/xxx/ && (++x==2){ print NR}'  file

# 3  
Old 12-09-2009
Hi Dennis,

i think 5 must be replaced by 2 .


awk '/xxx/ && (++x==2){ print NR}' filename

Thanks,
Penchal
# 4  
Old 12-09-2009
Thanks,
that works.
what if xxx and 2 are variables and I want to assign the value to a variable?

I tried
Code:
STRING=xxx
ID=2

NUMBER=`awk '/$STRING/ && (++x==$ID){ print NR}' file`
echo $NUMBER

but it doesn't work
# 5  
Old 12-09-2009
A bit prettier using variables:
Code:
$ awk -vSTR=xxx -vNUM=2 '$0 ~ STR { occ++ } occ == NUM {print NR; exit }' test.txt
3
$ awk -vSTR=xxx -vNUM=3 '$0 ~ STR { occ++ } occ == NUM {print NR; exit }' test.txt
5

That way you won't have to play around with double and single quotes. Just replace xxx and the number by any variable you like.
# 6  
Old 12-09-2009
Try:

Code:
var1=xxx
var2=2
awk "/$var1/ && (++x==$var2){ print NR}"  file

# 7  
Old 12-09-2009
Thank you.
I have the problem that also the file name is variable

Code:
var1=xxx
var2=2
file=file_name
NUMBER=`awk "/$var1/ && (++x==$var2){ print NR}"  $file`
echo NUMBER

will give
Code:
23199 file_name

so filename becomes part of number!
How to solve it?
Last edited by f_o_555; 12-09-2009 at 05:34 AM..
 
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