Is there a way to completely delete fields in awk, so that extra delimiters do not print?

This is an oldie but goodie.

As Jonathan points out, you can't delete fields in the middle, but you can replace their contents with the contents of other fields. And you can make a reusable function to handle the deletion for you.

$ cat test.awk
function rmcol(col,     i) {
  for (i=col; i<NF; i++) {
    $i = $(i+1)
  }
  NF--
}

{
  rmcol(3)
}

1

$ printf 'one two three four\ntest red green blue\n' | awk -f test.awk
one two four
test red blue

You can't delete fields in the middle, but you can delete fields at the end, by decrementing NF.

So you can shift all the later fields down to overwrite $2 and $3 then decrement NF by two, which erases the last two fields:

$ echo 1 2 3 4 5 6 7 | awk '{for(i=2; i<NF-1; ++i) $i=$(i+2); NF-=2; print $0}'
1 4 5 6 7

If you are just looking to remove columns, you can use cut:

$ cut -f 1,4- file.txt

To emulate cut:

$ awk -F "\t" '{ for (i=1; i<=NF; i++) if (i != 2 && i != 3) { if (i == NF) printf $i"\n"; else printf $i"\t" } }' file.txt

Similarly:

$ awk -F "\t" '{ delim =""; for (i=1; i<=NF; i++) if (i != 2 && i != 3) { printf delim $i; delim = "\t"; } printf "\n" }' file.txt

HTH