blocks|key|3005560|text|作为对澄清的响应，似乎不是严格要求单个正则表达式。在这种情况下，我建议您使用几个正则表达式或函数。我的猜测是，性能不是一个要求，因为通常这些类型的检查是在响应用户输入时完成的。用户输入验证可能需要100ms，而且看起来仍然是即时的，而且您可以在100ms内运行大量代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3005561|例如，我个人会在单独的测试中对您的每个条件进行检查。首先，检查是否有空格。第二，检查至少一个字母。接下来，检查至少一个数字。最后，查找包含三个或三个以上重复字符的任何跨度。|3005562|您的代码将更容易理解，稍后修改规则也将更容易(经验表明，这几乎肯定会发生)。|3005563|例如：|3005564|function+do_validation(string)+{
++++return+(has_no_space(string)+&&
++++++++++++has_no_special_char(string)+&&
++++++++++++has_alpha(string)+&&
++++++++++++has_digit(string)+&&
++++++++++++!+(has_repeating(string)))|code-block|syntax|javascript|3005565|我个人认为上面的几个数量级比一个复杂的正则表达式更容易阅读。此外，添加或删除规则不会使您必须重新实现复杂的正则表达式(因此，需要重新测试所有可能的组合)。|3005566|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|T|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|U|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

In response to a clarification, it seems that a single regular expression isn't strictly required. In that case I suggest you use several regular expressions or functions. My guess is, performance isn't a requirement, since usually these sorts of checks are done in response to user input. User input validation can take 100ms and still appear to be instant, and you can run a lot of code in 100ms. 

For example, I personally would do a check for each of your conditions in a separate test. First, check for spaces. Second, check for at least one letter. Next, check for at least one number. Finally, look for any spans of three or more repeated characters.

Your code will be much easier to understand, and it will be much easier to modify the rules later (which, experience has shown, is almost certainly going to happen). 

For example:

<pre><code>function do_validation(string) {
 return (has_no_space(string) &amp;&amp;
 has_no_special_char(string) &amp;&amp;
 has_alpha(string) &amp;&amp;
 has_digit(string) &amp;&amp;
 ! (has_repeating(string)))
</code></pre>

I personally consider the above to be orders of magnitude easier to read than one complex regular expression. Plus, adding or removing a rule doesn't make you have to reimplement a complex regular expression (and thus, be required to re-test all possible combinations).

blocks|key|2793138|text|要匹配连续重复3次或更多次的同一字符，请尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2793139|([a-zA-Z0-9])\1{2,}|code-block|syntax|javascript|2793140|样本匹配(同时测试here和here)：AABBAA+(不匹配)，AABBBAAA+(匹配BBB和AAA)，ABABABABABABABA+(不匹配)，ABCCCCCCCCCC+(匹配CCCCCCCCCC)。|offset|length|style|CODE|2793141|entityMap|0|LINK|mutability|MUTABLE|url|http://www.myregextester.com/|1|http://www.regexpal.com/^0|0|0|K|6|X|8|19|3|1D|3|1I|F|24|C|2K|A|9|4|0|E|4|1|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]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]|$I|17|J|18|K|L]|$I|19|J|1A|K|L]|$I|1B|J|1C|K|L]]|9|@$I|1D|J|1E|1|1F]|$I|1G|J|1H|1|1I]]|A|$]]|$1|M|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]|U|$5|P|Q|R|A|$S|V]]]]

For matching the same character repeated 3 or more times consecutively, try:

<pre><code>([a-zA-Z0-9])\1{2,}
</code></pre>

Sample matches (tested both <a href="http://www.myregextester.com/" rel="noreferrer">here</a> and <a href="http://www.regexpal.com/" rel="noreferrer">here</a>): <code>AABBAA</code> (no matches), <code>AABBBAAA</code> (matches <code>BBB</code> and <code>AAA</code>), <code>ABABABABABABABA</code> (no matches), <code>ABCCCCCCCCCC</code> (matches <code>CCCCCCCCCC</code>).

blocks|key|181808|text|这个正则表达式可以做到这一点：%5E(?!.*([A-Za-z0-9])\1{2})(?=.*[a-z])(?=.*\d)[A-Za-z0-9]%2B$|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|181809|细目如下：|181810|(?!.*([A-Za-z0-9])\1{2})确保所有字符在一行中的重复次数不超过两次。|181811|(?=.*[a-z])至少需要一个小写字母|181812|(?=.*\d)至少需要一个数字|181813|[A-Za-z0-9]%2B只允许字母和数字|181814|编辑:从负前视中删除了无关的.*|BOLD|181815|entityMap^0|F|1L|0|0|0|O|0|0|B|0|0|8|0|0|C|0|0|3|E|2|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1A|8|@$9|1B|A|1C|B|R]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|T|$]]

This regex will do it:
<code>^(?!.*([A-Za-z0-9])\1{2})(?=.*[a-z])(?=.*\d)[A-Za-z0-9]+$</code>

Here's the breakdown:

<code>(?!.*([A-Za-z0-9])\1{2})</code> makes sure that none of the chars repeat more than twice in a row.

<code>(?=.*[a-z])</code> requires at least one lowercase letter

<code>(?=.*\d)</code> requires at least one digit

<code>[A-Za-z0-9]+</code> allows only letters and digits

EDIT :
removed an extraneous <code>.*</code> from the negative lookahead

blocks|key|3005767|text|您可以在两个正则表达式中完成此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3005768|3005769|/%5E(?=.*[a-z])(?=.*[0-9])[a-z0-9]%2B$/i这将确保至少有一个数字和一个字母，同时只接受字母和数字(没有空格或特殊的characters)|ordered-list-item|offset|length|style|CODE|3005770|/([a-z0-9])\1{2,}/i如果匹配，则有一个重复的字符。这意味着您应该抛出false.|3005771|3005772|解释：|3005773|第一个正则表达式：|BOLD|3005774|3005775|%5E：匹配line|unordered-list-item|3005776|(?=.*[a-z])的开头:检查是否至少有一个letter|3005777|(?=.*[0-9])：检查是否至少有一个digit|3005778|[a-z0-9]%2B：如果检查为真，则只匹配数字/字母一个或多个times|3005779|$：匹配line|3005780|i的结尾:修饰符，匹配不区分大小写的|3005781|3005782|第二个正则表达式：|3005783|3005784|([a-z0-9])：匹配和分组数字或letter|3005785|\1{2,}：匹配组1两个或更多修饰符:修饰符，不区分大小写匹配|3005786|3005787|entityMap^0|0|0|0|10|0|0|J|17|5|0|0|0|0|9|0|0|0|1|0|0|B|0|0|B|0|0|9|0|0|1|0|0|1|0|0|0|9|0|0|0|A|0|0|6|0|0^^$0|@$1|2|3|4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|1I|8|@$F|1J|G|1K|H|I]]|9|@]|A|$]]|$1|J|3|K|5|E|7|1L|8|@$F|1M|G|1N|H|I]|$F|1O|G|1P|H|I]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1R|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1S|8|@$F|1T|G|1U|H|Q]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1V|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|1W|8|@$F|1X|G|1Y|H|I]]|9|@]|A|$]]|$1|V|3|W|5|U|7|1Z|8|@$F|20|G|21|H|I]]|9|@]|A|$]]|$1|X|3|Y|5|U|7|22|8|@$F|23|G|24|H|I]]|9|@]|A|$]]|$1|Z|3|10|5|U|7|25|8|@$F|26|G|27|H|I]]|9|@]|A|$]]|$1|11|3|12|5|U|7|28|8|@$F|29|G|2A|H|I]]|9|@]|A|$]]|$1|13|3|14|5|U|7|2B|8|@$F|2C|G|2D|H|I]]|9|@]|A|$]]|$1|15|3|-4|5|6|7|2E|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|2F|8|@$F|2G|G|2H|H|Q]]|9|@]|A|$]]|$1|18|3|-4|5|6|7|2I|8|@]|9|@]|A|$]]|$1|19|3|1A|5|U|7|2J|8|@$F|2K|G|2L|H|I]]|9|@]|A|$]]|$1|1B|3|1C|5|U|7|2M|8|@$F|2N|G|2O|H|I]]|9|@]|A|$]]|$1|1D|3|-4|5|6|7|2P|8|@]|9|@]|A|$]]|$1|1E|3|-4|5|6|7|2Q|8|@]|9|@]|A|$]]]|1F|$]]

You may do this in 2 regexes:
<ol>
<li><code>/^(?=.*[a-z])(?=.*[0-9])[a-z0-9]+$/i</code> This will assure that there is at least 1 digit and 1 letter while accepting only letters and digits (no space or special characters)</li>
<li><code>/([a-z0-9])\1{2,}/i</code> If this one is matched, then there is a repeated character. Which means you should throw <code>false</code>.</li>
</ol>
<h1>Explanation:</h1>
First regex:
<ul>
<li><code>^</code> : match begin of line</li>
<li><code>(?=.*[a-z])</code> : check if there is at least one letter</li>
<li><code>(?=.*[0-9])</code> : check if there is at least one digit</li>
<li><code>[a-z0-9]+</code> : if the checks were true, then match only digits/letters one or more times</li>
<li><code>$</code> : match end of line</li>
<li><code>i</code> : modifier, match case insensitive</li>
</ul>
Second regex:
<ul>
<li><code>([a-z0-9])</code> : match and group a digit or a letter</li>
<li><code>\1{2,}</code> : match group 1 two or more times</li>
<li><code>i</code> : modifier, match case insensitive</li>
</ul>

blocks|key|3005817|text|这个对你有效吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3005818|/(\b(?:([A-Za-z0-9])(?!\2{2}))%2B\b)/|code-block|syntax|javascript|3005819|试试看：|3005820|var+regex+=+new+RegExp(/(\b(?:([A-Za-z0-9])(?!\2{2}))%2B\b)/)
var+tests+=+['A1D3E',+'AAAA',+'AABAA',+'abccddeeff',+'abbbc',+'1234']

for(test+in+tests)+{
+++console.log(tests[test]+%2B+'+-+'+%2B+Boolean(tests[test].match(regex)))
}|3005821|将输出：|3005822|A1D3E+-+true
AAAA+-+false
AABAA+-+true
abccddeeff+-+true
abbbc+-+false
1234+-+true|3005823|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Does this one work for you?

<pre><code>/(\b(?:([A-Za-z0-9])(?!\2{2}))+\b)/
</code></pre>

Try it out:

<pre><code>var regex = new RegExp(/(\b(?:([A-Za-z0-9])(?!\2{2}))+\b)/)
var tests = ['A1D3E', 'AAAA', 'AABAA', 'abccddeeff', 'abbbc', '1234']

for(test in tests) {
 console.log(tests[test] + ' - ' + Boolean(tests[test].match(regex)))
}
</code></pre>

Will output:

<pre><code>A1D3E - true
AAAA - false
AABAA - true
abccddeeff - true
abbbc - false
1234 - true
</code></pre>

I have a requirement to handle a regular expression for no more than two of the same letters/digits in an XSL file.

<ul>
<li>no space</li>
<li>does not support special chars</li>
<li>support (a-z,A-Z,0-9)</li>
<li>require one of a-z</li>
<li>require one of 0-9</li>
<li>no more than 2 same letter/digits (i.e., <code>BBB</code> will fail, <code>BB</code> is accepted)</li>
</ul>

What I have so far 

<pre><code>(?:[^a-zA-Z0-9]{1,2})
</code></pre>

Regular expression for no more than two repeated letters/digits

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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我需要在XSL文件中处理不超过两个相同字母/数字的正则表达式。no space不支持特殊字符支持(a-z，A-Z,0-9)require a-z之一需要0-9之一不超过2个相同的字母/数字(即D12</code>将失败，BB被接受)到目前为止，我所拥有的(?:[^a-zA-Z0-9]{1,2})

问不超过两个重复字母/数字的正则表达式
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