Im trying to match a string agains a regular expression inside an if statement on bash. Code below:

If [[ $var =~ ^b\S+[a-z]$ ]]; then 
echo $var
echo 'none'

Match should be a string that starts with 'b' followed by one or more non-whitespace character and ending on a letter a-z. I can match the start and end of the string but the \S is not working to match the non-whitespace characters. Thanks in advance for the help.

  • 5
    case $var in b[![:space:]]*[[:lower:]]) echo "$var";; esac
    – mikeserv
    Commented Dec 26, 2015 at 19:10
  • @mikeserv, that * would match a string containing a spacing characters. Commented Dec 26, 2015 at 20:41
  • @StéphaneChazelas - very good point. i overlooked +. thought it was just about the second char. its why i upvoted your answer.
    – mikeserv
    Commented Dec 26, 2015 at 20:42
  • 3
    if should not be capitalized
    – zwol
    Commented Dec 26, 2015 at 21:50
  • @StéphaneChazelas You are fully wright. [:graph:] is match all symbols from \x21 to \x7E
    – Costas
    Commented Dec 26, 2015 at 23:04

3 Answers 3


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:


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]$'

  [[ $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.

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"
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).

  • what do the spaces do as ERE special characters?
    – mikeserv
    Commented Dec 28, 2015 at 19:17
[[ $var =~ ^b[^[:space:]]+[abcdefghijklmnopqrstuvwxyz]$ ]]

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
$ bash -O compat31 -c "[[ x =~ '\S' ]]" && echo yes

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

$ a='\S' bash -c '[[ x =~ $a ]]' && echo 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"
  echo none


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


# match any non-whitespace char--works in bash and `grep` too
[^\r\n\t\f\v ]


Matching \S (any non-whitespace character) apparently doesn't work in regular expressions in bash or grep or similar. So, instead of using this to match one or more occurrences of any non-whitespace character:

# INSTEAD OF THESE (which do NOT work in bash or `grep`)

# match one or more non-whitespace chars
# or (same thing)

...use this:

How to match all non-whitespace characters in bash and grep

# match one or more non-whitespace chars (DOES work in bash and `grep`!)
[^\r\n\t\f\v ]+

I learned this from https://regex101.com/. Click here: https://regex101.com/r/kM041K/1, and on the right-hand side of the screen, under the "EXPLANATION" section, you will see:

\S matches any non-whitespace character (equivalent to [^\r\n\t\f\v ])

So, if in doubt about any regular expression, go to that website and see what it says.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .