Bash – Using Regex in Bash and Regex101

bashregular expression

Using https://regex101.com/ I built a regular expression to return the first occurrence of an IP address in a string.

RegExp:

(?:\d{1,3}\.)+(?:\d{1,3})

RegExp including delimiters:

/(?:\d{1,3}\.)+(?:\d{1,3})/

With the following test string:

eu-west                       140.243.64.99

It returns a full match of:

140.243.64.99

No matter what I try with anchors etc, the following bash script will not work with the regular expression generated.

temp="eu-west                       140.243.64.99            "
regexp="(?:\d{1,3}\.)+(?:\d{1,3})"
if [[ $temp =~ $regexp ]]; then
  echo "found a match"
else
  echo "No IP address returned"
fi

Best Answer

\d is a nonstandard way for saying "any digit". I think it comes from Perl, and a lot of other languages and utilities support Perl-compatible REs (PCRE), too. (and e.g. GNU grep 2.27 in Debian stretch supports the similar \w for word characters even in normal mode.)

Bash doesn't support \d, though, so you need to explicitly use [0-9] or [[:digit:]]. Same for the non-capturing group (?:..), use just (..) instead.

This should print match:

temp="eu-west                       140.243.64.99            "
regexp="([0-9]{1,3}\.)+([0-9]{1,3})"
[[ $temp =~ $regexp ]] && echo match

Related Solutions

Bash – Lazy regex using Bash

There is a reason why the internet is packed with alternative approaches. I can't really think of any situation where you would be forced to use bash for this. Why not use one of the tools designed for the job?

Anyway, as far as I know there is no way of doing non-greedy matches using the =~ operator. That's because it does not use bash's internal regex engine but your system's C one as defined in man 3 regex. This is explained in man bash:

   An additional binary operator, =~, is available, with the  same  prece‐
   dence  as  ==  and !=.  When it is used, the string to the right of the
   operator is considered  an  extended  regular  expression  and  matched
   accordingly  (as  in  regex(3)).

You can, however, do more or less what you want (bearing in mind that this is really not a good way of parsing HTML files) with a slightly different regex:

string='<span class="circle"> </span>foo</span></span>'
regex='<span class="circle"> </span>([^<]+)</span>'
[[ $string =~ $regex ]]; 
echo "${BASH_REMATCH[1]}"

The above will return foo as expected.

Correct Regex Not Working in Grep – Troubleshooting Guide

You seem to have defined the right regex, but not set the sufficient flags in command-line for grep to understand it. Because by default grep supports BRE and with -E flag it does ERE. What you have (look-aheads) are available only in the PCRE regex flavor which is supported only in GNU grep with its -P flag.

Assuming you need to extract only the matching string after prefix you need to add an extra flag -o to let know grep that print only the matching portion as

grep -oP '(?<=prefix).*$' <<< prefixSTRING

There is also a version of grep that supports PCRE libraries by default - pcregrep in which you can just do

pcregrep -o '(?<=prefix).*$' <<< prefixSTRING

Detailed explanation on various regex flavors are explained in this wonderful Giles' answer and tools that implement each of them

Best Answer

Related Solutions

Bash – Lazy regex using Bash

Correct Regex Not Working in Grep – Troubleshooting Guide

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