awk question


 
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# 1  
Old 12-21-2010
awk question

Hi
I was just wondering ... what is the most simple and/or efficient way you suggest in awk, to "shift" fields to left ?

(let's say we are not interested by the N first field but we want to print all N+1 and more fields , until the end of line)

Of course i want to get rid off any potential leading FS (those at beginning of line)

Last edited by ctsgnb; 12-21-2010 at 05:55 AM..
# 2  
Old 12-21-2010
One way to remove the first 3 fields:
Code:
awk '{sub(".*" $3 FS,"")}1' file

This User Gave Thanks to Franklin52 For This Post:
# 3  
Old 12-21-2010
awk lack that shift command Smilie
# 4  
Old 12-21-2010
Quote:
Originally Posted by ctsgnb
awk lack that shift command Smilie
Alas, no shift operator in AWK Smilie
# 5  
Old 12-21-2010
Yep, cruel lack of that shift thing
I was dreaming about a "kind of" work around like
Code:
$0=system (set -- $0 ; shift 3 )

but this will of course also fail ... grrrrr Smilie
# 6  
Old 12-21-2010
Yep , I think sub is the best option. Sub alternative:
Code:
awk 'sub(".*"$4,$4)'


Last edited by Scrutinizer; 12-21-2010 at 11:29 AM..
# 7  
Old 01-27-2011
awk question here is a suggestion

A bit longer but should shift and clear FS in beginning of line (start at field 2)
Code:
#!/bin/ksh
cat <file name>  | awk '{
  ct=split($0,a)
  ftime=1
  for(x=2;x<ct;x++){
    if(ftime == 1){
      out_val=a[x]
      ftime=0
    }else{
      out_val=out_val " " a[x]
    }
  }
  printf("%s\n",out_val)
}'


Last edited by Scott; 01-27-2011 at 10:43 AM.. Reason: Added code tags; indentation
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