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.
You must use regex ^ to denote start of string:
$ awk '$2 ~ /^[[:upper:]]/' file ID A56 DS AGE 56 You could use awk as @cuonglm suggested, or
GNU grep
grep -P '^[^\s]+\s+[A-Z]' file Perl
perl -lane 'print if $F[1]=~/^[A-Z]/' file GNU sed
sed -rn '/^[^\s]+\s+[A-Z]/p' file 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 
grep, GNU sed and for the last one, recent versions of ksh93 zsh or bash and that the file doesn't contain backslash characters. Except for the perl one what [A-Z] matches depends on the locale and doesn't make much sense except in the C locale. -P is a GNU extension? OK. Why does the [A-Z] not make sense? Presumably, the OP would want whatever is defined as a capital letter in their locale right? I added -r for the backslashes. -P is a GNU extension though it's found as well in some BSDs that have forked or rewritten their GNU grep. See there for [A-Z]. Also note that \s is [[:space:]], not [[:blank:]]. printf then, just in case.
awk '$2~/^[A-Z]/' file