Awk: for i in 1,2,n-1,n

Shell Programming and Scripting
1

hello,

a fast question:

how can i loop over four numbers in awk like in bash:

n ist a number >3

n=4
for i in 1 2 $((n-1)) $n
do .. done

i tried:

awk: for (i in 1 2 n-1 n)
with all possible brackets (){} and punctuations , ;
but this doesn't work...

2

Put the values in an array and to iterate in order use a for loop with an index variable.

Regards,
Alister

1 Like
3

try a loop like:

echo "" | awk '{x=4 ; while (i++ <x) print i}'
4

You could try somehting like this:

awk 'BEGIN{n=4; s="1,2,"n-1; split(s,a,","); for (i in a) print a}'
1 Like
5

EDIT : The solution below does not fit the question, sorry about that :X

Is this OK?

~/unix.com$ awk 'BEGIN{n=4; for(i=1; i<=n; i++)print i}'
1
2
3
4
~/unix.com$ awk 'BEGIN{n=4; while(i++<n)print i}'
1
2
3
4
6

Maybe not quite what you need, but at least pointing in the desired direction:

awk '{for (i=n-2; i<=n+1; i++) print i%n+1} ' n=15 file
14
15
1
2

A little sort might help...

1 Like
7

@alister: works, but not very elegant

@tukuyomi, rdrtx1: n can be >>> 4
i need the first two (1,2) and the last two (n-1,n)...

@rudiC:
elegant, modulo! the order is no problem

@Subbeh: could work, I will try...

8

try also:

awk '{for (i=1; i<=n; i++) if (i<=2 || i>n-2) print i} ' n=1000 infile