避免 cpp 中的空令牌

本文介绍了避免 cpp 中的空令牌的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

限时送ChatGPT账号..

我有一个字符串:

s = "server ('m1.labs.terada')ta') username ('user5') password('user)5') dbname ('default')";

我正在提取参数名称:例如服务器、用户名...、数据库名称.

I am extracting the argument names: such as server, username..., dbname.

为此，我使用以下正则表达式:

To do this I am using the following regex:

regex re("\\(\'[!-~]+\'\\)");
sregex_token_iterator i(s.begin(), s.end(), re, -1);
sregex_token_iterator j;
unsigned count = 0;
while(i != j)
{
    string str1 = *i;
    cout <<"token = "<<str1<< endl;
    i++;
    count++;
}
cout << "There were " << count << " tokens found." << endl

为此，我得到的输出是:

For this the output I am getting is :

token = server 
token =  username 
token =  password
token =  dbname 
token =  
There were 5 tokens found.

我应该如何避免形成第 5 个令牌.我错过了什么吗?

How shall I avoid the 5th token formed. Am I missing out on anything?

推荐答案

原来在字符串的末尾有一些非单词字符.您可以将它们与 \W* (零个或多个非单词字符)匹配.由于您的标记仅由单词字符组成，因此您可以使用 \W* 模式安全地包装您的模式:

It turns out there are some non-word chars at the end of the string. You may match them with \W* (zero or more non-word chars). Since your tokens are only composed of word chars, you may safely wrap your pattern with \W* pattern:

regex re("\\W*\\(\'[!-~]+\'\\)\\W*");

查看 C++ 在线演示

结果:

token = server
token = username
token = password
token = dbname
There were 4 tokens found.

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