find command not giving any output

Question

Please anyone know why this command is not giving me any output? allthe variables are correct but no result.

find . -name "$cdr_type*$DAYZ*.unl*" |
  xargs zcat |
  awk -F "|" '{if($14==$tdate && $22==$misdn) print $0}'|
  head

When I echo the command like this

echo "find . -name $cdr_type*$DAYZ*.unl* | xargs zcat | awk -F | '{if($14==$tdate && $22==$misdn) print $0}'|head" 

here is my output:

find . -name mgr*20160928*.unl* |
  xargs zcat |
  awk -F | '{if(4==20160928093911 && 2==2348094398953)}'|
  head

My observation is its showing $4 and $2 instead of $14 and $22

Answer 1

$14 when expanded by the shell in most Bourne-like shells is the same as ${1}4, and if $1 is empty or unset, that gives 4 as expected. Here you want to pass a litteral $14 to awk so that awk treats it as its 14th field.

Normally within single quotes, that $1 variable would not be expanded but you're not telling us how you echo that command.

In any case, here, it looks like you want:

find . -name "$cdr_type*$DAYZ*.unl*" -exec zcat {} + |
  awk -F "|" -v tdate="$tdate" -v misdn="$misdn" '
   $14 == tdate && $22 == misdn' |
 head

That is pass the content of the $tdate and $misdn shell variables as awk variables.