Shell Programming and Scripting

So if I have an awk statement that is basically just looking at the NF at if its more than 2, then print out the first 2 words, and all the rest on another line. I know that $1 and $2 are the first two fields, but how would I symbolise telling it to print all the other fields regardless of how many there are on a newline?

Just some quick pseudocode as im away from my PC

Code:

if (NF>2){
print $1 FS $2 "\n" 
}

now after the \n I want to know how i can tell it to print out all the other fields, bearing in mind that the number of fields it checks on each line can vary.
Many thanks

Code:

awk 'NF > 2 {
  for (i = 0; ++i <= NF;)
    printf "%s", ($i (i == 2 ? RS : FS))
  print x
  }' infile

Hi.

You could use a for-loop:

Code:

for( i = 3; i <= NF; i++ ) ...

Or, perhaps:

Code:

$ echo 1 2 3 4 5 6 | awk '{if( NF > 2) { $3 = RS $3}}1'
1 2
3 4 5 6

If I got you correctly,Try:

Code:

awk 'NF>2 { print $1" "$2; $1=$2=""; print; }' file

Nice

Hi Rad,
I am new to awk,
Pls tell me what does mean by 1 at end in your code

Code:

echo 1 2 3 4 5 6 | /usr/xpg4/bin/awk '{if( NF > 2) { $3 = RS $3}}1'

This is scottn's code, not mine.
The 1 simply means true, which in the pattern/expression part causes the execution of the default action, which is print the current record.