blocks|key|190825|text|若要匹配不包含连续重复3次以上的字符的字符串：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|190826|%5E((.)\2?(?!\2\2))%2B$|offset|length|style|CODE|190827|它的工作原理：|190828|%5E++++++++++++Start+of+string
(
++(.)++++++++Match+any+character+(not+a+new+line)+and+store+it+for+back+reference.
++++\2?++++++Optionally+match+one+more+exact+copies+of+that+character.
++++(?!++++++Make+sure+the+upcoming+character(s)+is/are+not+the+same+character.
++++++\2\2+++Repeat+'\2'+for+as+many+times+as+you+need
++++)
)%2B+++++++++++Do+ad+nauseam
$++++++++++++End+of+string|code-block|syntax|javascript|190829|因此，整个表达式中的/2数将是允许一个字符连续重复的次数，并且不会得到匹配。|190830|例如。|190831|%5E((.)\2?(?!\2\2\2))%2B$将匹配不连续重复字符超过4次的所有字符串。|unordered-list-item|190832|%5E((.)\2?(?!\2\2\2\2))%2B$将匹配不连续重复字符超过5次的所有字符串。|190833|请注意，此解决方案使用的是负面展望，但并非所有regex风格都支持它。|190834|entityMap^0|0|0|J|0|0|0|A|2|0|0|0|L|0|0|N|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|12|8|@$D|13|E|14|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|15|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|16|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|17|8|@$D|18|E|19|F|G]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1A|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|1B|8|@$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|V|3|W|5|U|7|1E|8|@$D|1F|E|1G|F|G]]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1H|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|10|$]]

To match strings not containing a character repeated more than 3 times consecutively:

<code>^((.)\2?(?!\2\2))+$</code>

How it works:

<pre><code>^ Start of string
(
 (.) Match any character (not a new line) and store it for back reference.
 \2? Optionally match one more exact copies of that character.
 (?! Make sure the upcoming character(s) is/are not the same character.
 \2\2 Repeat '\2' for as many times as you need
 )
)+ Do ad nauseam
$ End of string
</code></pre>

So, the number of <code>/2</code> in your whole expression will be the number of times you allow a character to be repeated consecutively, any more and you won't get a match.

E.g.

<ul>
<li><code>^((.)\2?(?!\2\2\2))+$</code> will match all strings that don't repeat a character more than 4 times in a row.</li>
<li><code>^((.)\2?(?!\2\2\2\2))+$</code> will match all strings that don't repeat a character more than 5 times in a row.</li>
</ul>

Please be aware this solution uses negative lookahead, but not all not all regex flavors support it.

blocks|key|2111249|text|使用带反向引用的负前瞻：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2111250|%5E(?:(.)(?!\1\1))*$|code-block|syntax|javascript|2111251|请参见使用示例的现场演示。|offset|length|2111252|(.)捕获组1中的每个字符，而前面的负面展望表明，接下来的2个字符不会重复捕获的字符。|style|CODE|2111253|entityMap|0|LINK|mutability|MUTABLE|url|http://rubular.com/r/XFQR6n8qxQ^0|0|0|8|4|0|0|0|3|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@$I|Z|J|10|1|11]]|A|$]]|$1|K|3|L|5|6|7|12|8|@$I|13|J|14|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

Use a negative lookahead with back references:

<pre><code>^(?:(.)(?!\1\1))*$
</code></pre>

See <a href="http://rubular.com/r/XFQR6n8qxQ" rel="nofollow noreferrer">live demo</a> using your examples.

<code>(.)</code> captures each character in group 1 and the negative look ahead asserts that the next 2 chars are not repeats of the captured character.

blocks|key|190898|text|我在回答这个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|190899|是否有用于匹配不超过2个重复字符的字符串的正则表达式？|offset|length|190900|被标记为这个问题的确切副本。|190901|相反，否定比赛要快得多。|190902|if+(!Regex.Match("hello+world",+@"(.)\1{2}").Success)+Console.WriteLine("No+dups");|code-block|syntax|javascript|190903|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/59275616/is-there-a-regular-expression-for-matching-a-string-that-has-no-more-than-2-repe^0|0|0|R|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@$D|Y|E|Z|1|10]]|A|$]]|$1|F|3|G|5|6|7|11|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|12|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|13|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

I'm answering this question : 

<a href="https://stackoverflow.com/questions/59275616/is-there-a-regular-expression-for-matching-a-string-that-has-no-more-than-2-repe">Is there a regular expression for matching a string that has no more than 2 repeating characters?</a>

which was marked as an exact duplicate of this question. 

<hr>

Its much quicker to negate the match instead

<pre><code>if (!Regex.Match("hello world", @"(.)\1{2}").Success) Console.WriteLine("No dups");
</code></pre>

I want to match strings that do not contain more than 3 of the same character repeated in a row. So:

<ul>
<li>abaaaa [no match]</li>
<li>abawdasd [match]</li>
<li>abbbbasda [no match]</li>
<li>bbabbabba [match]</li>
</ul>

Yes, it would be much easier and neater to do a regex match for containing the consecutive characters, and then negate that in the code afterwards. However, in this case that is not possible.

I would like to open out the question to x consecutive characters so that it can be extended to the general case to make the question and answer more useful.

Negative lookahead is supported in this case.

What is the regex to match a string not containing more than x consecutive characters

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我希望匹配不包含一行中重复的相同字符中的3个以上的字符串。所以：abaaaa没有对手abawdasd匹配abbbbasda没有匹配bbabbabba火柴是的，如果对包含连续字符的正则表达式进行匹配，然后在随后的代码中否定这一点，就会更容易、更整洁。然而，在这种情况下，这是不可能的。我想把这个问题扩展到x个连续字符，这样...

问匹配不包含x个连续字符的字符串的正则表达式是什么？
EN

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问匹配不包含x个连续字符的字符串的正则表达式是什么？EN