Any Non-whitespace Regular Expression

Answer :

In non-GNU systems what follows explain why \S fail:

The \S is part of a PCRE (Perl Compatible Regular Expressions). It is not part of the BRE (Basic Regular Expressions) or the ERE (Extended Regular Expressions) used in shells.

The bash operator =~ inside double bracket test [[ use ERE.

The only characters with special meaning in ERE (as opposed to any normal character) are .[\()*+?{|^$. There are no S as special. You need to construct the regex from more basic elements:

regex='^b[^[:space:]]+[a-z]$' 

Where the bracket expression [^[:space:]] is the equivalent to the \S PCRE expressions :

The default \s characters are now HT (9), LF (10), VT (11), FF (12), CR (13), and space (32).

The test would be:

var='big'            regex='^b[^[:space:]]+[a-z]$'  [[ $var =~ $regex ]] && echo "$var" || echo 'none' 

However, the code above will match bißß for example. As the range [a-z] will include other characters than abcdefghijklmnopqrstuvwxyz if the selected locale is (UNICODE). To avoid such issue, use:

var='bißß'            regex='^b[^[:space:]]+[a-z]$'  ( LC_ALL=C;   [[ $var =~ $regex ]]; echo "$var" || echo 'none' ) 

Please be aware that the code will match characters only in the list: abcdefghijklmnopqrstuvwxyz in the last character position, but still will match many other in the middle: e.g. bég.

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Still, this use of LC_ALL=C will affect the other regex range: [[:space:]] will match spaces only of the C locale.

To solve all the issues, we need to keep each regex separate:

reg1=[[:space:]]   reg2='^b.*[a-z]$'           out=none  if                 [[ $var =~ $reg1 ]]  ; then out=none elif   ( LC_ALL=C; [[ $var =~ $reg2 ]] ); then out="$var" fi printf '%6.8s\t|' "$out" 

Which reads as:

• If the input (var) has no spaces (in the present locale) then
• check that it start with a b and ends in a-z (in the C locale).

Note that both tests are done on the positive ranges (as opposed to a "not"-range). The reason is that negating a couple of characters opens up a lot more possible matches. The UNICODE v8 has 120,737 characters already assigned. If a range negates 17 characters, then it is accepting 120720 other possible characters, which may include many non-printable control characters.

It should be a good idea to limit the character range that the middle characters could have (yes, those will not be spaces, but may be anything else).

[[ $var =~ ^b[^[:space:]]+[abcdefghijklmnopqrstuvwxyz]$ ]] 

What [a-z] matches depends on the locale and generally is not (only) one of abcdefghijklmnopqrstuvwxyz.

perl's \S (horizontal and vertical spaces) now also recognised by some other regexp engines is [^[:space:]] in POSIX and bash's EREs.

bash uses the system's regexp library to match those regular expressions, but even on systems (like recent GNU ones) where the regexps have a \S operator, that won't work because in:

[[ x = \S ]] 

bash calls regcomp("S") and with:

[[ x = '\S' ]] 

bash calls regcomp("\\S") (two backslashes).

However, with bash-3.1 or if you turn bash-3.1 compatibility on with shopt -s compat31, then:

[[ x = '\S' ]] 

will work (will match a non-spacing character) on systems where EREs support \S.

$ bash -c "[[ x =~ '\S' ]]" || echo no no $ bash -O compat31 -c "[[ x =~ '\S' ]]" && echo yes yes 

Another option would be to put the regexp in a variable:

$ a='\S' bash -c '[[ x =~ $a ]]' && echo yes yes 

Again, that only works on systems that support that perl-like \S in their regexps.

The POSIX equivalent to that bash-specific code, would be:

if expr " $var" : \         ' b[^[:space:]]\{1,\}[abcdefghijklmnopqrstuvwxyz]$' \    > /dev/null; then   printf '%s\n' "$var" else   echo none fi 

Or:

case $var in   ([!b]* | *[!abcdefghijklmnopqrstuvwxyz] | *[[:space:]]* | "" | ? | ??)     echo none;;   (*) printf '%s\n' "$var" esac