[Solved] Removing Leading Zeros within awk file [duplicate]

Try this: sub(/^0+/, "", $2)?

$ awk 'BEGIN{b="000000000000787301"; sub(/^0+/, "", b); print b;}'
787301

or typecast it by adding zero:

$ awk 'BEGIN{b="000000000000787301"; print b+0;}'
787301

update A working example based on comments:

$ echo 'E2EDP19001 02002000000000000797578' | awk '{$2 = substr($2,6); sub(/^0+/, "", $2); print $2;}'
797578

or, preserve $2:

$ echo 'E2EDP19001 02002000000000000797578' | awk '{a = substr($2,6); sub(/^0+/, "", a); print a;}'
797578

up-update Based on revised post, here’s a working sample, just like I proposed many, many revisions ago:

sub()

$ echo 'E2EDP19001                                                   02002000000000000797578' \
   | awk 'BEGIN {E2EDP19seen = 1 } 
       {if (substr($0, 1, 7) == "E2EDP19" && E2EDP19seen == 1) {
       out = substr($2, 6); sub(/^0+/, "", out); print out } }'
797578

typecast

$ echo 'E2EDP19001                                                   02002000000000000797578' \
  | awk 'BEGIN {E2EDP19seen = 1 } 
      {if (substr($0, 1, 7) == "E2EDP19" && E2EDP19seen == 1) {
      print substr($2, 6) + 0 } }'
797578