在引用我的问题正则表达式EX pression到全字匹配有特殊字符不工作？，

我这说的答案

@"(?<=^|s)" + pattern + @"(?=s|$)"

这工作正常，除1例的所有情况。它失败时有空间的格局。

This works fine for all cases except 1 case. It fails when there is space in the pattern.

假设字符串你好，这是计算器的模式是本，那么它​​说没有匹配。出现这种情况是因为图案的实际字符串后，一个空的空间。

Assume string is "Hi this is stackoverflow" and pattern is "this " , then it says no matches. This happens because of an empty space after the actual string in pattern.

我们如何处理呢？理想的情况来讲，应该说一场比赛找到了！

推荐答案

试试这个

(?:(?<=^|s)(?=S)|(?<=S|^)(?=s))this (?:(?<=S)(?=s|$)|(?<=s)(?=S|$))

请参阅它这里Regexr

这也将适用于与一个空格开头的模式。

This will also work for pattern that starts with a whitespace.

基本上，我在做什么是定义一个定制的字的边界。但它不是在真实的一个 W =＆GT; W 或 W =＆GT; W 变化，其真正在 S =＆GT; S 或 S =＆GT;！ S 变化

Basically, what I am doing is to define a custom "word" boundary. But it is not true on a W=>w or a w=>W change, its true on a S=>s or a s=>S change!

下面是在C＃中的一个例子：

string str = "Hi this is stackoverflow";
string pattern = Regex.Escape("this");
MatchCollection result = Regex.Matches(str, @"(?:(?<=^|s)(?=S)|(?<=S|^)(?=s))" + pattern + @"(?:(?<=S)(?=s|$)|(?<=s)(?=S|$))", RegexOptions.IgnoreCase);

Console.WriteLine("Amount of matches: " + result.Count);
foreach (Match m in result)
{
    Console.WriteLine("Matched: " + result[0]);
}
Console.ReadLine();

更新：

这个空白的边界可以做到更普遍，从而使在图案两侧是一样的EX pression，像这样

This "Whitespace" boundary can be done more general, so that on each side of the pattern is the same expression, like this

(?:(?<=^|s)(?=S|$)|(?<=^|S)(?=s|$))

在C＃：

In c#:

MatchCollection result = Regex.Matches(str, @"(?:(?<=^|s)(?=S|$)|(?<=^|S)(?=s|$))" + pattern + @"(?:(?<=^|s)(?=S|$)|(?<=^|S)(?=s|$))", RegexOptions.IgnoreCase);