printing multiple spaces with awk

Hallo,

I have a listing with following form:

a  b c d  xx y   zz.txt
f r  f     g    yz x  xx zzz.doc

In article <7966a6cc.0402260540.79f08419@posting.google.com>,
Brigitte Selch <Brigitte_Selch@mn.man.de> wrote:
>Hallo,
>
>       I have a listing with following form:
>
>       a  b c d  xx y   zz.txt
>       f r  f     g    yz x  xx zzz.doc
>
>       .
>       .
>       .
>
>       Now ich want to eliminate the first 4 columns and print the all
>       following columns as they are (with multiple blanks, because they ar
e
>       filenames).
>       In my example:
>
>       xx y   zz.txt
>       yz x  xx zzz.doc

I'm not sure I understand the underlying problem - what exactly is this
data, and, if they are filenames, why do you need to preserve the
inter-field spacing - but anyway, this is a well-known "feature" of AWK (*),
and here is one solution (**).

{
for (i=0; i<4; i++)
sub("[ \t]*"$1"[ \t]*","")
print
}

(*) It has its plusses and minusses.

(**) Modulu issues of reg ex magic chars in $1.

Hi Brigitte,

index($0,$5)            returns the index of $5 in the string $0.
substr($0,index($0,$5)) returns the rest of $0, starting at the
beginning of $5.

awk '{print substr($0,index($0,$5));}'

HTH,
Stefan

Brigitte Selch schrieb:
>        I have a listing with following form:
>
>        a  b c d  xx y   zz.txt
>        f r  f     g    yz x  xx zzz.doc
[...]
>        Now ich want to eliminate the first 4 columns and print the all
>        following columns as they are (with multiple blanks, because they a
re
>        filenames).
>        In my example:
>
>        xx y   zz.txt
>        yz x  xx zzz.doc

Test:

awk '{print substr($0,index($0,$5));}' << EOF
>        a  b c d  xx y   zz.txt
>        f r  f     g    yz x  xx zzz.doc
> EOF
xx y   zz.txt
yz x  xx zzz.doc

Hello,

Wow, it works.

Thank you very much,
Brigitte