Bash – Any non-whitespace regular expression

bashregular expression

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

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

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.

Best 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.

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).

Related Question