.net 字母数字的正则表达式，但至少有一个字符

regex for alphanumeric, but at least one character

提问by mrblah

In my asp.net page, I have an input box that has to have the following validation on it:

在我的 asp.net 页面中，我有一个输入框，必须对其进行以下验证：

Must be alphanumeric, with at least 1 character (i.e. can't be ALL numbers).

必须是字母数字，至少有 1 个字符（即不能是所有数字）。

回答by cletus

^\d*[a-zA-Z][a-zA-Z0-9]*$

Basically this means:

基本上这意味着：

• Zero or more ASCII digits;
• One alphabetic ASCII character;
• Zero or more alphanumeric ASCII characters.

• 零个或多个 ASCII 数字；
• 1 个字母 ASCII 字符；
• 零个或多个字母数字 ASCII 字符。

Try a few tests and you'll see this'll pass any alphanumeric ASCII string where at least one non-numeric ASCII character is required.

尝试一些测试，您会看到这将传递任何需要至少一个非数字 ASCII 字符的字母数字 ASCII 字符串。

The key to this is the \d*at the front. Without it the regex gets much more awkward to do.

关键是\d*在前面。没有它，正则表达式会变得更加尴尬。

回答by Philippe Leybaert

Most answers to this question are correct, but there's an alternative, that (in some cases) offers more flexibility if you want to change the rules later on:

这个问题的大多数答案都是正确的，但有一个替代方案，如果您想稍后更改规则，则（在某些情况下）提供更大的灵活性：

^(?=.*[a-zA-Z].*)([a-zA-Z0-9]+)$

This will match any sequence of alphanumerical characters, but only if the first group also matches the whole sequence. It's a little-known trick in regular expressions that allows you to handle some very difficult validation problems.

这将匹配任何字母数字字符序列，但前提是第一组也匹配整个序列。这是正则表达式中一个鲜为人知的技巧，它允许您处理一些非常困难的验证问题。

For example, say you need to add another constraint: the string should be between 6 and 12 characters long. The obvious solutions posted here wouldn't work, but using the look-ahead trick, the regex simply becomes:

例如，假设您需要添加另一个约束：字符串的长度应介于 6 到 12 个字符之间。此处发布的明显解决方案不起作用，但使用前瞻技巧，正则表达式简单地变为：

^(?=.*[a-zA-Z].*)([a-zA-Z0-9]{6,12})$

回答by John Kugelman

^[\p{L}\p{N}]*\p{L}[\p{L}\p{N}]*$

Explanation:

解释：

• [\p{L}\p{N}]*matches zero or more Unicode letters or numbers
• \p{L}matches one letter
• [\p{L}\p{N}]*matches zero or more Unicode letters or numbers
• ^and $anchor the string, ensuring the regex matches the entirestring. You may be able to omit these, depending on which regex matching function you call.

• [\p{L}\p{N}]*匹配零个或多个 Unicode 字母或数字
• \p{L}匹配一个字母
• [\p{L}\p{N}]*匹配零个或多个 Unicode 字母或数字
• ^并$锚定字符串，确保正则表达式匹配整个字符串。您可以省略这些，具体取决于您调用的正则表达式匹配函数。

Result: you can have any alphanumeric string except there's got to be a letter in there somewhere.

结果：您可以有任何字母数字字符串，除非某处必须有一个字母。

\p{L}is similar to [A-Za-z]except it will include all letters from all alphabets, with or without accents and diacritical marks. It is much more inclusive, using a larger set of Unicode characters. If you don't want that flexibility substitute [A-Za-z]. A similar remark applies to \p{N}which could be replaced by [0-9]if you want to keep it simple. See the MSDN page on character classesfor more information.

\p{L}类似于，[A-Za-z]除了它将包括所有字母表中的所有字母，带有或不带有重音符号和变音符号。它更具包容性，使用更大的 Unicode 字符集。如果您不想要那种灵活性替代品[A-Za-z]。如果您想保持简单\p{N}，可以将其替换为类似的评论[0-9]。有关更多信息，请参阅有关字符类的MSDN 页面。

The less fancy non-Unicode version would be

不那么花哨的非 Unicode 版本是

^[A-Za-z0-9]*[A-Za-z][A-Za-z0-9]*$

回答by Dexter

^[0-9]*[A-Za-z][0-9A-Za-z]*$

is the regex that will do what you're after. The ^ and $ match the start and end of the word to prevent other characters. You could replace the [0-9A-z] block with \w, but i prefer to more verbose form because it's easier to extend with other characters if you want.

是正则表达式，它将做你所追求的。^ 和 $ 匹配单词的开头和结尾以防止其他字符。您可以用 \w 替换 [0-9A-z] 块，但我更喜欢更详细的形式，因为如果需要，使用其他字符扩展更容易。

Add a regular expression validator to your asp.net page as per the tutorial on MSDN: http://msdn.microsoft.com/en-us/library/ms998267.aspx.

按照 MSDN 上的教程向您的 asp.net 页面添加正则表达式验证器：http: //msdn.microsoft.com/en-us/library/ms998267.aspx。

回答by Welbog

^\w*[\p{L}]\w*$

This one's not that hard. The regular expression reads: match a line starting with any number of word characters (letters, numbers, punctuation (which you might not want)), that contains one letter character (that's the [\p{L}]part in the middle), followed by any number of word characters again.

这个没那么难。正则表达式读取：匹配以任意数量的单词字符（字母、数字、标点符号（您可能不想要））开头的一行，其中包含一个字母字符（即[\p{L}]中间部分），后跟任意数量的单词又是字符。

If you want to exclude punctuation, you'll need a heftier expression:

如果要排除标点符号，则需要一个更复杂的表达式：

^[\p{L}\p{N}]*[\p{L}][\p{L}\p{N}]*$

And if you don't care about Unicode you can use a boring expression:

如果你不关心 Unicode，你可以使用一个无聊的表达式：

^[A-Za-z0-9]*[A-Za-z][A-Za-z0-9]*$

回答by eKek0

^[0-9]*[a-zA-Z][a-zA-Z0-9]*$

Can be

可

• any number ended with a character,
• or an alphanumeric expression started with a character
• or an alphanumeric expression started with a number, followed by a character and ended with an alphanumeric subexpression

• 任何以字符结尾的数字，
• 或以字符开头的字母数字表达式
• 或以数字开头的字母数字表达式，后跟一个字符并以字母数字子表达式结尾