Shell Programming and Scripting

Hello,

I am using a if condition within my awk script and I have a problem with the search pattern.

I would like awk to find all the fields starting with 123. For now I have something like that:

Code:

 
awk '{for(i=1;i<9;i+=2)if($i=="123"...)print $i}' test

but that is not working as "..." is not an authorized pattern. All my fields are 6 digits numbers so I want to look for "123XXX".

I hope you can help me...

Thank you

Code:

% print 123 321 321 123 | 
  awk '{ for (i=1; i<=NF; i++) if ($i ~ /^123/) print "field:", i, $i }'
field: 1 123
field: 4 123

Edit: Sorry,
just realized that the example was too bad
Anyway, the code is correct:

Code:

% print 123456 321234 321432 12354313 | 
  awk '{ for (i=1; i<=NF; i++) if ($i ~ /^123/) print "field:", i, $i }'
field: 1 123456
field: 4 12354313

Thank you for your answer... I never used the ~ and ~! to compare expression! I was always struggling with == !!