结合这两个正则表达式为一体 [英] Combining these two Regular Expressions into one

问题描述

我的在C＃如下：

 公共静态布尔IsAlphaAndNumeric（字符串s）
 {
返回Regex.IsMatch（S，@[A-ZA-Z] +）
和;&安培; Regex.IsMatch（S，@\d +）; 
} 
  

我要检查，如果参数取值至少包含一个字母字符和一个数字，我写了上面的方法来做到这一点。

但是，有没有办法我可以结合两个正则表达式（[A-ZA-Z] +和\d + ）合并为一个？

解决方案

  @^（？=。* [A-ZA -Z]）（？=。* \d）
 
 ^＃从字符串
的开头（？=。* [A-ZA-Z]）＃期待对于后面有一个字母任意数目的字符，不提前指针
（？=。* \d）＃期待任何数量的字符后跟数字）的
  

使用两个积极向前看符号的以保证它找到一个字母，之前的强校一个数字。您添加 ^ 来只尝试期待一次，从字符串的开始。否则，正则表达式引擎将尝试匹配的字符串中的每一个点。

I have the following in C#:

public static bool IsAlphaAndNumeric(string s)
{
    return Regex.IsMatch(s, @"[a-zA-Z]+") 
        && Regex.IsMatch(s, @"\d+");
}

I want to check if parameter s contains at least one alphabetical character and one digit and I wrote the above method to do so.

But is there a way I can combine the two regular expressions ("[a-zA-Z]+" and "\d+") into one ?

解决方案

@"^(?=.*[a-zA-Z])(?=.*\d)"

 ^  # From the begining of the string
 (?=.*[a-zA-Z]) # look forward for any number of chars followed by a letter, don't advance pointer
 (?=.*\d) # look forward for any number of chars followed by a digit)

Uses two positive lookaheads to ensure it finds one letter, and one number before succeding. You add the ^ to only try looking forward once, from the start of the string. Otherwise, the regexp engine would try to match at every point in the string.

这篇关于结合这两个正则表达式为一体的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！