正则表达式查找字符串中所有出现的模式

问题描述

我在查找字符串中所有出现的模式时遇到问题.

I have a problem finding all occurences of a pattern in a string.

检查此字符串:

string msg= "=?windows-1258?B?UkU6IFRyIDogUGxhbiBkZSBjb250aW51aXTpIGQnYWN0aXZpdOkgZGVz?= =?windows-1258?B?IHNlcnZldXJzIFdlYiBHb1ZveWFnZXN=?=";

我想返回2次出现(以便以后对其进行解码):

I want to return the 2 occurrences (in order to later decode them):

=?windows-1258?B?UkU6IFRyIDogUGxhbiBkZSBjb250aW51aXTpIGQnYWN0aXZpdOkgZGVz?=

和

=?windows-1258?B?IHNlcnZldXJzIFdlYiBHb1ZveWFnZXN =?="

使用以下正则表达式代码，它仅返回1次出现:完整字符串.

With the following regex code, it returns only 1 occurrence: the full string.

var charSetOccurences = new Regex(@"=\?.*\?B\?.*\?=", RegexOptions.IgnoreCase);
var charSetMatches = charSetOccurences.Matches(input);
foreach (Match match in charSetMatches)
{
    charSet = match.Groups[0].Value.Replace("=?", "").Replace("?B?", "").Replace("?b?", "");
}

你知道我想念什么吗?

推荐答案

当 regexp 解析器看到.* 字符序列时，它将匹配所有内容，直到字符串，然后逐个返回，(贪心匹配).因此，为避免此问题，您可以使用非贪婪匹配或显式定义可以出现在字符串中的字符.

When regexp parser sees the .* character sequence, it matches everything up to the end of the string and goes back, char by char, (greedy match). So, to avoid the problem, you can use a non-greedy match or explicitly define the characters that can appear at a string.

"=\?[a-zA-Z0-9?=-]*\?B\?[a-zA-Z0-9?=-]*\?="

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