c＃中以下要求的正则表达式 [英] Regular Expression for below requirement in c#

问题描述

必须包含至少6个字母字符和至少2个数字。用户ID和密码不得包含无效字符，最大长度为12个字符。



我需要从我的数据库中获取正则表达式并验证。我不想在我的c＃代码中检查它。我很抱歉让人感到困惑。

Must contain at least 6 alpha characters and at least 2 numbers. User ID and Passwords must not contain invalid characters and have a maximum length of 12 characters.

I need to get the regular expression from my database and verify. I dont want to check it in my c# code. I apologize for confusing.

推荐答案

你的数字与字母自由混合，正则表达式非常不合适。如果你想要允许所有的Unicode字母，不仅要为这样的混合编写正则表达式非常困难，任何字母的概念几乎不可能表达。使用C＃，很容易做到。如果你只想要英文字符，那么使用C＃也更容易。



如下：拳头，检查最大长度。然后遍历字符串并计算3个类中的字符：字母，数字和另一个。在循环之前，将两个计数器变量初始化为零：一个用于字母，一个用于数字。在一个循环中，如果出现非字母和非数字，立即返回false（失败），否则递增数字字母的计数器。



你可以使用功能

Char.IsLetter Method（Char）（系统） [ ^ ]，

Char .IsDigit Method（Char）（System） [ ^ ]，

或更好， Char.IsLetterOrDigit方法（字符）（系统） [ ^ ]。



如果你想允许英文cha仅限racters，只检查角色是否在A和z之间;对于其他语言，类似的检查可能不那么简单。



这就是全部。



只需一个注意：建议的标准对于密码来说真的很糟糕。您可以在许多地方阅读有关良好密码限制的建议。

-SA

It your numbers are freely mixed with letters, Regular Expression is quite unsuitable. Not only it's extremely hard to write the regular expression for such a mix, the idea of "any letter" is nearly impossible to express, if you want to allow all Unicode letters. With C#, it's easy to do. If you want, say, English characters only, it's also easier to do with C#.

Here is how: Fist, check up maximum length. Then traverse the string and count characters in 3 classes: letters, digits and the other. Before the cycle, initialize two counter variables to zero: one for letters and one for digits. In a loop, if non-letter and non-digit appears, return false (failure) immediately, otherwise increment the counter of letters of digits.

You can use the functions
Char.IsLetter Method (Char) (System)[^],
Char.IsDigit Method (Char) (System)[^],
or, even better, Char.IsLetterOrDigit Method (Char) (System)[^].

If you want to allow English characters only, just checkup that the character is between "A" and "z"; for other languages, the similar check can be not as simple.

That's all.

Just one note: the suggested criterion is really bad for a password. You can read recommendations on good password constraints in many places.

—SA

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