blocks|key|430980|text|你需要副选修课。我将进一步使用范围来保持事物的可读性：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|430981|%5E([2-9]%7C10)$|code-block|syntax|javascript|430982|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You need parantheses for that. I would further use ranges to keep things readable:

<pre><code>^([2-9]|10)$
</code></pre>

blocks|key|432159|text|显然，在这种情况下，您不应该使用RegExp，而应该使用数值计算：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|432160|var+num+=+parseInt(aNumber,+10);
if+(num+>=+2+&&+num+<=+10)+{
++++alert("Match+found!");
}|code-block|syntax|javascript|432161|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is clearly a case where you shouldn't use RegExp but numerical evaluation:

<pre><code>var num = parseInt(aNumber, 10);
if (num &gt;= 2 &amp;&amp; num &lt;= 10) {
 alert("Match found!");
}
</code></pre>

blocks|key|430931|text|在交替前后使用括号，因为级联比交替具有更高的优先级：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|430932|%5E(2%7C3%7C4%7C5%7C6%7C7%7C8%7C9%7C10)$|code-block|syntax|javascript|430933|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Use parentheses around the alternations, since concatenation has higher precedence than alternation:

<pre><code>^(2|3|4|5|6|7|8|9|10)$
</code></pre>

blocks|key|869331|text|一个完整的javascript函数，可以匹配2、9或10。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|869332|<script+type="text/javascript">
++++function+validateValue(+theValue+)
++++{
++++++++if(+theValue.match(+/%5E([2-9]{1}%7C10)$/+)+)
++++++++{
++++++++++++window.alert(+'Valid'+);
++++++++}
++++++++else
++++++++{
++++++++++++window.alert(+'invalid'+);
++++++++}
++++++++return+false;
++++}
</script>|code-block|syntax|javascript|869333|regex很容易扩展到包含所有规则：|869334|/%5E(-1%7C[2-9]%7C10%7C99)$/+//+will+match+-1,+2-10,+99|869335|编辑:我忘了提到你将不得不替换你自己的真实/错误的情境逻辑。edit2:将缺失的括号添加到第二正则表达式中。|869336|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

A complete javascript function to match either 2 though 9, or 10

<pre><code>&lt;script type="text/javascript"&gt;
 function validateValue( theValue )
 {
 if( theValue.match( /^([2-9]{1}|10)$/ ) )
 {
 window.alert( 'Valid' );
 }
 else
 {
 window.alert( 'invalid' );
 }
 return false;
 }
&lt;/script&gt;
</code></pre>

The regex is easily expanded to incorporate all of your rules: 

<pre><code>/^(-1|[2-9]|10|99)$/ // will match -1, 2-10, 99
</code></pre>

edit: I forgot to mention you'll have to substitute your own true/false situational logic.
edit2: added missing parenthesis to 2nd regex.

blocks|key|869256|text|原始表达式失败的原因是匹配%5E2、3、4、5、6、7、8、9、10$。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|869257|试试看这个regex：%5E(2件事，3件事，4件事，5件事，6件事，6件事，7件事，8件事，8件事，9件事，10件事)$|869258|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

The reason your original expression fails is because you're matching ^2, 3, 4, 5, 6, 7, 8, 9, 10$.

Try this regex: ^(2|3|4|5|6|7|8|9|10)$

blocks|key|869375|text|这个在线实用工具非常擅长为数字范围生成re：http://utilitymill.com/utility/Regex_For_Range。对于包含2-10的范围，它给出了这样的表达式：\b0*([2-9]%7C10)\b。如果您想省略前面的0，请去掉"0*“部分。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|869376|entityMap|0|LINK|mutability|MUTABLE|url|http://utilitymill.com/utility/Regex_For_Range^0|2K|G|M|1A|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

This online utility is very good at generating re's for numeric ranges: <a href="http://utilitymill.com/utility/Regex_For_Range" rel="nofollow noreferrer">http://utilitymill.com/utility/Regex_For_Range</a>. For the range 2-10 inclusive, it gives this expression: <code>\b0*([2-9]|10)\b</code>. If you want to omit leading 0's, take out the "0*" part.

I have a field on a form that takes the following values: -1, 2-10, 99

I have a business rule that's concerned with answers 2-10.

I'm trying to write a regular expression that will match 2-10 but not 99, and I'm having trouble.

The original expression:

<pre><code>^2|3|4|5|6|7|8|9|10$
</code></pre>

Fails because 99 is matched (technically, twice). Also, the Line boundries are something I've never been comfortable with. I oberve different behavior from them in expresso than I do in other places (e.g. .net). In this particular instance, the regex is being run in javascript. Anyway, expresso seems to ignore them (and if I put the values in brackets: 

<pre><code>^[2|3|4|5|6|7|8|9|10]$

^[2-9]$
</code></pre>

either "all spelled out" or as a range, expresso never returns any matches if I specify the opening line/string closing line/string characters (and yes, I was trying to match the 10 separately in the second case there).

I know, I know. If you use a regex to solve a problem, then you have two problems (and presumably they'll start inviting friends over, thing 1 and thing 2 style). I don't have to use one here; I may switch to a case statement. But it seems like I should be able to use a regex here, and it seems a reasonable thing to do. I'm still pretty green when it comes to the regex;

Regex to match 2-10, but not 99

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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窗体上有一个字段，其值如下：-1，2-10，99我有一条关于2-10答案的商业规则。我正在尝试编写一个与2-10匹配的正则表达式，而不是99，而且我遇到了麻烦。原文如下：^2|3|4|5|6|7|8|9|10$失败，因为99是匹配的(技术上，两次)。而且，线赏金是我从来都不舒服的东西。我在表现力上和其他地方(例如.net)有不同的行为。在这个特定的例子中，regex正在javascript中运行。

问与2-10匹配，但不是99
EN

回答 6

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问与2-10匹配，但不是99EN

回答 6

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问与2-10匹配，但不是99
EN