如何从c ++中的字符串中剥离所有非字母数字字符？ [英] How to strip all non alphanumeric characters from a string in c++?

问题描述

我在写一个软件，它需要我处理从libcurl的网页中获取的数据。当我得到数据，由于某种原因，它有额外的换行符。我需要找出一种方法，只允许字母，数字和空格。并删除一切，包括换行符。有什么简单的方法吗？感谢。

I am writing a piece of software, and It require me to handle data I get from a webpage with libcurl. When I get the data, for some reason it has extra line breaks in it. I need to figure out a way to only allow letters, numbers, and spaces. And remove everything else, including line breaks. Is there any easy way to do this? Thanks.

推荐答案

编写一个需要 char code> true 如果要删除该字符，或 false ，如果要保留：

Write a function that takes a char and returns true if you want to remove that character or false if you want to keep it:

bool my_predicate(char c);

然后使用 std :: remove_if 算法从字符串中删除不需要的字符：

Then use the std::remove_if algorithm to remove the unwanted characters from the string:

std::string s = "my data";
s.erase(std::remove_if(s.begin(), s.end(), my_predicate), s.end());

根据您的要求，您可以使用标准库谓词之一，例如 std :: isalnum ，而不是写你自己的谓词（你说你需要匹配字母数字字符和空格，所以也许这不完全适合你需要的）。

Depending on your requirements, you may be able to use one of the Standard Library predicates, like std::isalnum, instead of writing your own predicate (you said you needed to match alphanumeric characters and spaces, so perhaps this doesn't exactly fit what you need).

如果你想使用标准库 std :: isalnum 函数，你需要一个cast来消除<$在C标准库头< cctype> （这是您要使用的）中的c $ c> std :: isalnum 在C ++标准库头中的 std :: isalnum < locale> （这不是你想要的使用，除非您要执行特定于语言环境的字符串处理）：

If you want to use the Standard Library std::isalnum function, you will need a cast to disambiguate between the std::isalnum function in the C Standard Library header <cctype> (which is the one you want to use) and the std::isalnum in the C++ Standard Library header <locale> (which is not the one you want to use, unless you want to perform locale-specific string processing):

s.erase(std::remove_if(s.begin(), s.end(), (int(*)(int))std::isalnum), s.end());

这对任何序列容器同样有效（包括 std :: string ， std :: vector 和 std :: deque ）。这个成语通常被称为擦除/移除成语。 std :: remove_if 算法也适用于普通数组。 std :: remove_if 只对序列进行一次遍历，因此它具有线性时间复杂性。

This works equally well with any of the sequence containers (including std::string, std::vector and std::deque). This idiom is commonly referred to as the "erase/remove" idiom. The std::remove_if algorithm will also work with ordinary arrays. The std::remove_if makes only a single pass over the sequence, so it has linear time complexity.

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