Regular expression to find the length of a field


 
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# 8  
Old 09-22-2010
Quote:
Originally Posted by aigles
Try and adapt the following script :
Code:
awk '

function fmtSize( fmt, typ   ,s) {
   if (! fmt) return 0;
   if (sub("^" typ "*\\(", "", fmt)) {
      s = fmt+0
   } else {
      match(fmt, "^" typ "*");
      s = RLENGTH;
   }
   return s
}

NF>=3 {
   column = $2;
   format = toupper($3);
   gsub(/^[SZ]|\.$/, "", format);
   type   = substr(format, 1, 1);
   split(format, f, /V/);
   if (type == "9")
      size = fmtSize(f[1], type) + fmtSize(f[2], type);
   else if (type == "X")
      size = fmtSize(f[1], type);
   else
      next;
   printf "Column=%s Format=%s Size=%d\n", column, $3, size;
}

' inputfile

With your sample inputfile, the output is:
Code:
Column=col1 Format=s9(10)V99. Size=12
Column=Col2 Format=s9(10)V9(3). Size=13
Column=Col3 Format=XXXX. Size=4
Column=Col4 Format=X(3). Size=3
Column=Col5 Format=S999999. Size=6
Column=Col6 Format=s999V99. Size=5
Column=Col7 Format=Z9(9). Size=9

Jean-Pierre.
Thanks Jean-Pierre. the code works perfectly fine. I kind of like your solutions especially that you post with awk. Would you please explain it as well? - Thanks again.
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