How to use regex to match ASTERISK in awk

Question

I'm stil pretty new to regular expression and just started learning to use awk. What I am trying to accomplish is writing a ksh script to read-in lines from text, and and for every lines that match the following:

*RECORD 0000001 [some_serial_#]

to replace $2 (i.e. 000001) with a different number. So essentially the script read in batch record dump, and replace the record number with date+record#, and write to separate file.

So this is what I'm thinking the format should be:

awk 'match($0,"/*RECORD")!=0{$2="$DATE-n++"; print $0} match($0,"/*RECORD")==0{print $0}' $BATCH > $OUTPUT

but obviously "/*RECORD" is not going to work, and I'm not sure if changing $2 and then write the whole line is the correct way to do this. So I am in need of some serious enlightenment.

Answer 1

So you want your example line to look like

*RECORD $DATE-n++ [some_serial_#]

after awk's done with it?

awk '{ if (match($0, "*RECORD") != 0) { $2="$DATE-n++"; }; print }' $BATCH > $OUTPUT

---

Based on your update, it looks like you instead expect $DATE to be an environment variable which is used in the awk expression and n is a variable in the awk script that keeps count of how many records matched the pattern. Given that, this may look more like what you want.

$ cat script.awk
BEGIN { n=0 }
{
    if (match($0, "\*RECORD") != 0) {
        n++;
        $2 = (ENVIRON["DATE"] "-" n);
    }
    print;
}

$ awk -f script.awk $BATCH > $OUTPUT

Answer 2

use equality.

D=$(date +%Y%m%d)
awk -vdate="$D" '
{
  for(i=1;i<=NF;i++){
   if ( $i == "*RECORD" ){
      $(i+1) = date"00002"
      break # break after searching for one record, otherwise, remove break
   }
  }
}1' file