Print lines if given column starts with a capital letter

Question

I have a file like this:

ID  A56
DS  /A56
DS  AGE 56

And I'd like to print the whole line only if the second column starts with a capital letter.

Expected output:

ID  A56
DS  AGE 56

What I've tried so far:
awk '$2 ~ /[A-Z]/ {print $0}' file
Prints everything: capital letters are found within the second column.

awk '$2 /[A-Z]/' file
Gets a syntax error.

Answer 1

You must use regex ^ to denote start of string:

$ awk '$2 ~ /^[[:upper:]]/' file
ID  A56
DS  AGE 56

Answer 2

You could use awk as @cuonglm suggested, or

1. GNU grep

   grep -P '^[^\s]+\s+[A-Z]' file 
   

2. Perl

   perl -lane 'print if $F[1]=~/^[A-Z]/' file
   

3. GNU sed

   sed -rn '/^[^\s]+\s+[A-Z]/p' file 
   

4. shell (assumes a recent version of ksh93, zsh or bash)

   while read -r a b; do 
       [[ $b =~ ^[A-Z] ]] && printf "%s %s\n" "$a" "$b"; 
   done < file