How to match repeated characters using regular expression operator =~ in bash?

Question

I want to know if a string has repeated letter 6 times or more, using the =~ operator.

a="aaaaaaazxc2"
if [[ $a =~ ([a-z])\1{5,} ]];
then
     echo "repeated characters"
fi

The code above does not work.

Answer 1

BASH regex flavor i.e. ERE doesn't support backreference in regex. ksh93 and zsh support it though.

As an alternate solution, you can do it using extended regex option in grep:

a="aaaaaaazxc2"
grep -qE '([a-zA-Z])\1{5}' <<< "$a" && echo "repeated characters"

repeated characters

---

EDIT: Some ERE implementations support backreference as an extension. For example Ubuntu 14.04 supports it. See snippet below:

$> echo $BASH_VERSION
4.3.11(1)-release

$> a="aaaaaaazxc2"
$> re='([a-z])\1{5}'
$> [[ $a =~ $re ]] && echo "repeated characters"
repeated characters

Answer 2

[[ $var =~ $regex ]] parses a regular expression in POSIX ERE syntax.

See the POSIX regex standard, emphasis added:

BACKREF - Applicable only to basic regular expressions. The character string consisting of a character followed by a single-digit numeral, '1' to '9'.

Backreferences are not formally specified by the POSIX standard for ERE; thus, they are not guaranteed to be available (subject to platform-specific libc extensions) in bash's native regex syntax, thus mandating the use of external tools (awk, grep, etc).