unconstant delimiter

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1

I have file with unconstatnt delimiter for each field which are non-printable characters like tab and space

file
 
6271        manchester          (tab) 11/09/09      200        accepted 
6272        manchester          (tab) 11/09/09      200        accepted

I want only first 3 fields as it were with same delimiters

can any one help me out

thanks

2 
$ awk '{print $1,$2,$3}' infile
3

This will give the same tab and space

 
$ nawk '{print substr($0,0,index($0,"/")+2)}' test
6271        manchester          (tab) 11/09
6272        manchester          (tab) 11/09
4

suppose I dont have any / in my 3 field

or

I want first 4 fields from file

thanks

5 
 
$ perl -lane '$a=$_; $a =~ s/(.*\d\d\s).*([a-z])/$1/; print $a;' test
6271        manchester          (tab) 11/09/09      200  
6272        manchester          (tab) 11/09/09      200
6

can we acheive using sed or awk

because I dont perl

thanks

7 
 
$ nawk '{print substr($0,1,index($0,$4)+length($4))}' test
6271        manchester          (tab) 11/09/09 
6272        manchester          (tab) 11/09/09

---------- Post updated at 04:11 PM ---------- Previous update was at 04:09 PM ----------

$ nawk '{print substr($0,1,index($0,$5)+length($5))}' test
6271        manchester          (tab) 11/09/09      200 
6272        manchester          (tab) 11/09/09      200
8

With GNU awk 4 you can easily reconstruct the original variable field separator(s):

 split(string, array [, fieldsep [, seps ] ])
Divide string into pieces separated by fieldsep and store the pieces in array and
the separator strings in the seps array.

For example:

awk '{
  split($0, t, FS, s)
  for (i = 1; ++i <= l;)
    printf "%s", (t (i < l ? s : RS))
  }' l=4 infile

% awk '{
  split($0, t, FS, s)
  for (i = 0; ++i <= l;)
    printf "%s", (t (i < l ? s : RS))
  }' l=5 infile
6271        manchester          (tab) 11/09/09      200
6272        manchester          (tab) 11/09/09      200
% awk '{
  split($0, t, FS, s)
  for (i = 0; ++i <= l;)
    printf "%s", (t (i < l ? s : RS))
  }' l=4 infile
6271        manchester          (tab) 11/09/09
6272        manchester          (tab) 11/09/09
% awk '{
  split($0, t, FS, s)
  for (i = 0; ++i <= l;)
    printf "%s", (t (i < l ? s : RS))
  }' l=2 infile
6271        manchester
6272        manchester