Replace using awk on fixed width file.

Shell Programming and Scripting
1

All,

I used to use following command to replace specific location in a fixed width file.

Recently looks like my command stopped working as intended. We are on AIX unix.

awk 'function repl(s,f,t,v)
{ return substr(s,1,f-1) sprintf("%-*s", t-f+1, v) substr(s,t+1) }
NR<=10 { a=repl($0,474,1,"Y")
  a=repl(a,476,1,"N")
  a=repl(a,478,10,"2011-01-01")
  a=repl(a,489,10,"2011-12-31")
  print a
} 
NR>10 && NR<=20 { a=repl($0,474,1,"N")
  a=repl(a,476,1,"Y")
  a=repl(a,478,10,"2015-01-01")
  a=repl(a,489,10,"2015-12-31")
  print a
}
NR>20 ' filename  > filename.new

Currently the command is replacing incorrectly and the output seems to be wierd layout than the original file.

Appreciate your help on fixing this.

2

Could you also post an anonymized sample of your input and the intended output...

3

May I question that your script has ever worked, no matter what the system was? t-f+1 with f=478 and t=1 will yield a negative number not suitable as a width parameter for sprintf . Even if you remove the - sign from the format string, the result is inacceptable, adding about 2000 spaces to each of your lines.
Try

return substr(s,1,f-1) sprintf("%-*s", t, v) substr(s,f+t) }

and report back on the results.

4

It looks like your repl function was written to expect FROM and TO values so you should be calling it like this:

a=repl(a,478,487,"2011-01-01")

Or update repl() to use FROM and LENGTH as RudiC mentioned:

function repl(s,f,L,v)
{ return substr(s,1,f-1) sprintf("%-*s", L, v) substr(s,f+L) }

$ printf "%c" {1..9} {0..9} | awk '
function repl(s,f,L,v) { return substr(s,1,f-1) sprintf("%-*s", L, v) substr(s,f+L) }
{print repl($0,8,3,"AB") }'
1234567AB 123456789