c++ - 如果包含字符串，如何删除整个句子

Question

如果字符串包含模式，我需要从字符串中删除整个句子。
这里我有模式“链接”或“链接”，如果它存在于字符串中，我需要删除包含它的整个句子。

std::string subject = "This is previous sentence. This can be any sentences. Link 2.1.19.3 [Example]. This is can be any other sentence. This is next sentence.";   

std::string removeRedundantString(std::string subject)
{
    std::string removeSee = subject;
    std::smatch match;  

    std::regex redundantSee("(Link.*$)");

    if (std::regex_search(subject, match, redundantSee))
    {
        removeSee = std::regex_replace(subject, redundantSee, "");
    }
}

预期输出:

This is previous sentence. This can be any sentences.This is can be any other sentence. This is next sentence.

实际输出:

This is previous sentence. This can be any sentences.

由于使用了正则表达式 "(Link.*$)"，上述实际输出即将到来删除从链接开始到字符串末尾的句子。
我无法弄清楚使用什么正则表达式来获得预期的输出。
以下是我需要测试的不同测试用例:
测试用例 1:

std::string subject = "Note this is second pattern, Ops that next the scheduler; link the amount for the full list of docs. The number of value varies from 0 to 4.";

输出: Note this is second pattern, Ops that next the scheduler;The number of value varies from 0 to 4.测试用例 2:

std::string subject = "This is another pattern. (Link Doc::78::hello::Core::mount). Since this patern includes non-numeric value.";

输出: This is another pattern.Since this patern includes non-numeric value.任何帮助，将不胜感激。

Answer 1

我会推荐

std::regex redundantSee(R"(\W*\b[Ll]ink\b(?:\d+(?:\.\d+)*|[^.])*[.?!])")

见 its online demo .注意原始字符串文字语法， R"(...)" .字符串模式可以简单地放在里面而不是 ...没有任何额外的转义。
正则表达式详细信息:

\W* - 零个或多个非单词字符

\b - 一个词边界

[Ll]ink - Link或 link字

\b - 一个词边界

(?:\d+(?:\.\d+)*|[^.])* - 零个或多个序列

\d+(?:\.\d+)* - 一个或多个数字后跟零个或多个 . 序列和一位或多位数字

| - 或

[^.] - 除 . 以外的任何字符

[.?!] - 一个 ? , .或 ! .