如何在bash中使用正则表达式运算符=〜匹配重复的字符? [英] How to match repeated characters using regular expression operator =~ in bash?

问题描述

我想使用=〜运算符判断一个字符串是否重复了6次或更多次.

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

上面的代码不起作用.

推荐答案

BASH正则表达式风格，即 ERE 不支持正则表达式中的反向引用. ksh93和zsh对此都支持.

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

作为替代解决方案，您可以使用grep中的扩展正则表达式选项来实现:

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

---

编辑:某些ERE实现支持将反向引用作为扩展.例如，Ubuntu 14.04支持它.参见下面的代码段:

---

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

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