grep仅适用于包含“标准”美国字符的行

Question

I'm trying to figure out how to grep for lines that are made up of AZ and az exclusively, that is, the "American" alphabet of letters. I would expect this to work, but it does not:

$ echo -e "Jutland\nJastrząb" | grep -x '[A-Za-z]*'
Jutland
Jastrząb

I want this to only print "Jutland", because ą is not a letter in the American alphabet. How can I achieve this?

Answer 1

You need to add LC_ALL=C before grep :

printf '%b\n' "Jutland\nJastrząb" | LC_ALL=C grep -x '[A-Za-z]*'

Jutland

You may also use -i switch to ignore case and reduce regex:

printf '%b\n' "Jutland\nJastrząb" | LC_ALL=C grep -ix '[a-z]*'

LC_ALL=C avoids locale-dependent effects otherwise your current LOCALE treats ą as [a-zA-Z] .

Answer 2

You can use perl regex:

$ echo -e "Jutland\nJastrząb" | grep -P '^[[:ascii:]]+$'
Jutland

It's experimental though:

-P, --perl-regexp
      Interpret  the  pattern as a Perl-compatible regular expression (PCRE).  This is experimental and
      grep -P may warn of unimplemented features.

EDIT

For letters only, use [A-Za-z] :

$ echo -e "L'Egyptienne\nJutland\nJastrząb" | grep -P '^[A-Za-z]+$'
Jutland