print all but select fields in awk

Question

I have a large file with hundreds of columns that I want to remove only the third and fourth columns from and print the rest to a file. My initial idea was to make an awk script like awk '{print $1, $2, for (i=$5; i <= NF; i++) print $i }' file > outfile. However, this code does not work.

I then tried:

awk '{for(i = 1; i<=NF; i++)
if(i == 3 || i == 4) continue
else
print($i)}' file > outfile

But this just printed everything out in one field. It would be possible to split this up into two scripts and combine them with unix paste but this seems like something that should be able to be done in one line.

Answer 1

Your first try was pretty close. Modifying it to use printf and including the field separators worked for me:

awk '{printf $1FS$2; for (i=5; i <= NF; i++) printf FS$i; print NL }'

Answer 2

What about something like:

cat SOURCEFILE | cut -f1-2,5- >> DESTFILE

It prints the first two columns, skips the 3rd and 4rth, and then prints from 5 onwards to the end.

Answer 3

How about just setting the third and fourth columns to an empty string:

echo 1 2 3 4 5 6 7 8 9 10 |
awk -F" " '{ $3="";  $4=""; print}'