如何计算字符串中特定字符集的数量

Basically, I want to count sets of specific characters in a string. In other words I need to count all Letters and Numbers, and nothing else. But I cant seem to find the right (regex) syntax. Here's what i have ...

基本上，我想计算一个字符串中的特定字符集。换句话说，我需要统计所有的字母和数字，而不是其他任何东西。但我似乎无法找到正确的（正则表达式）语法。这是我的...

public double AlphaNumericCount(string s)
{
    double count = Regex.Matches(s, "[A-Z].[a-z].[0-9]").Count;
        return count;
}

I've been looking around, but cant seem to find anything that allows more than one set of characters. Again, I'm not sure on the syntax maybe it should be "[A-Z]/[a-z]/[0-9]" or something. Anywho, go easy on me - its my first day using Regex.

我一直在环顾四周，但似乎找不到任何允许多个角色的东西。同样，我不确定语法可能应该是“[A-Z] / [a-z] / [0-9]”或其他什么。 Anywho，对我来说很容易 - 这是我使用正则表达式的第一天。

Thanks.

谢谢。

4 个解决方案

#1

Regular Expression Cheat Sheet

正则表达式备忘单

Expresso Regular Expression tool

Expresso正则表达式工具

[A-Z].[a-z].[0-9] will match any capital letter ([A-Z]), followed by any character (.), followed by any lower case letter ([a-z]), followed by any character (.), followed by any number ([0-9]).

[AZ]。[az]。[0-9]将匹配任何大写字母（[AZ]），后跟任何字符（。），后跟任何小写字母（[az]），后跟任何字符（。），后跟任意数字（[0-9]）。

What you want to match on any letter or number is [A-Za-z0-9].

您希望在任何字母或数字上匹配的是[A-Za-z0-9]。

#2

If the use of regular expressions is not required, there is an easy, alternate solution:

如果不需要使用正则表达式，那么有一个简单的替代解决方案：

return s.ToCharArray().Count(c => Char.IsNumber(c) || Char.IsLetter(c));

#3

Try this one:

试试这个：

^[a-zA-Z0-9]+$

see: regexlib.com

见：regexlib.com

#4

If you want to avoid regular expressions, you can simply iterate through the characters in the string and check if they're a letter or digit using Char.IsLetterOrDigit.

如果要避免使用正则表达式，可以简单地遍历字符串中的字符，并使用Char.IsLetterOrDigit检查它们是字母还是数字。

public int AlphaNumericCount(string s)
{
    int count = 0;

    for(int i = 0; i < s.Length; i++)
    {
       if(Char.IsLetterOrDigit(s[i])) 
          count++;
    }

    return count;
}

#1