Regular expression to find the length of a field


 
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# 1  
Old 09-22-2010
Regular expression to find the length of a field

Hi,

in the cobol copy books is there any regular expressions to be used in awk to fetch the length of each columns?

below mentioned are the examples.
Copy Book Sample
Code:
01 tablename.
   02 group header.
     03 col1  s9(10)V99.
     03 Col2  s9(10)V9(3).
     03 Col3  XXXX
   02 group header1.
     03 Col4  X(3).
     03 Col5  S999999.
     03 Col6  s999V99.
     03 Col7  Z9(9).

Expected output:

Code:
Col1,12
Col2,13
Col3,4
Col4,3
Col5,6
Col6,5
Col7,9

I have tried the below code looks a bit dirty can it be written more precisely and it is failing when the column is having the definition like 03 Col X. & 03 Col S999 xxxx.

Code:
awk -v brack="(" '{ ZZ[NR]=$0 }

END {
for ( i in ZZ)
{

if ( ZZ[i] ~ brack )
{
len=substr(ZZ[i], index(ZZ[i], "(" )+1, index(ZZ[i], ")") - index(ZZ[i], "(" )-1)
split(ZZ[i],Z)
print ZZ[i]","len
}

else 
{
split(ZZ[i], XX)

for ( i in XX )
{
gsub(".", "", XX[i])
if ( XX[i] ~ /[sSvV9]$ )
{
cntre=(/^[9]$/, "@@@", ZZ[i])
print XX[i]","cntre
}
else if ( XX[i] ~ /[X]$/ )

{
cntre=gsub(/[X]$/, "~~~", XX[i])
print XX[i]","cntre
}


}
}



}


}


Last edited by jim mcnamara; 09-22-2010 at 09:08 AM.. Reason: please use code tags not quote tags; removed munged color
# 2  
Old 09-22-2010
yeeks...

what's wrong with:

Code:
#  echo hello | awk '{print length($1)}'
5

?
# 3  
Old 09-22-2010
Hmmm definitely misread that. I need coffee. :-)
# 4  
Old 09-22-2010
Well, I am not trying to print the first field in the row. I want to fetch the length of the column. for instance col1(that is row3) has got the length 10 +2 = 12. and col4 has got the length of 4 (counting the number of X's)
# 5  
Old 09-22-2010
Are you saying you want to get the field sizes defined by the (messed up) COBOL source? There are no PIC elements... I am going to edit your post, it is completely unreadable.

Just for the heck of it - does this code, as written, actually compile? If it does what COBOL compiler does this?

I have code somewhere that will work with normal COBOL, the kind that has PIC 9(8)V99 or PIC XX. If it turns out that your compiler does use ANSI COBOL, and your post is in error, please let me know.

Last edited by jim mcnamara; 09-22-2010 at 09:15 AM..
# 6  
Old 09-22-2010
yes, thats right need to get the length of each column. its a bit messed up without any proper standards, some times there are no PIC elements as well. Just to make it more readable I tried to highlight the lengths in orange.

The cobol definitions are not acutally used in the cobol compiler. they are used in different platform.

here I need to modify and identify the lengths as there are 1000's of copy books so manually doing it is very tough,

Last edited by ahmedwaseem2000; 09-22-2010 at 09:18 AM..
# 7  
Old 09-22-2010
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.
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