如何从字符串中删除第一个单词？

Question

假设我有

string sentence{"Hello how are you."}

我希望字符串句子有“你好吗”而没有“你好”。我该怎么做呢。

我尝试做类似的事情：

stringstream ss(sentence);
ss>> string junkWord;//to get rid of first word

但是当我这样做时：

cout<<sentence;//still prints out "Hello how are you"

很明显stringstream 不会改变实际的字符串。我也尝试过使用strtok，但它不适用于string。

Answer 1

试试下面的

#include <iostream>
#include <string>

int main() 
{
    std::string sentence{"Hello how are you."};

    std::string::size_type n = 0;
    n = sentence.find_first_not_of( " \t", n );
    n = sentence.find_first_of( " \t", n );
    sentence.erase( 0,  sentence.find_first_not_of( " \t", n ) );

    std::cout << '\"' << sentence << "\"\n";

    return 0;
}

输出是

"how are you."

Answer 2

str=str.substr(str.find_first_of(" \t")+1);

测试：

string sentence="Hello how are you.";
cout<<"Before:"<<sentence<<endl;
sentence=sentence.substr(sentence.find_first_of(" \t")+1);
cout<<"After:"<<sentence<<endl;

执行：

> ./a.out
Before:Hello how are you.
After:how are you.

假设该行不以空格开头。在这种情况下，这不起作用。

find_first_of("<list of characters>").

在我们的例子中，字符列表是空格和制表符。这将搜索任何字符列表的第一次出现并返回一个迭代器。之后添加 +1 将位置移动一个字符。然后位置指向该行的第二个单词。 Substr(pos) 将获取从位置开始到字符串最后一个字符的子字符串。

Answer 3

有无数种方法可以做到这一点。我想我会选择这个：

#include <iostream>
#include <string>

int main() {
    std::string sentence{"Hello how are you."};

    // First, find the index for the first space:
    auto first_space = sentence.find(' ');

    // The part of the string we want to keep
    // starts at the index after the space:
    auto second_word = first_space + 1;

    // If you want to write it out directly, write the part of the string
    // that starts at the second word and lasts until the end of the string:
    std::cout.write(
        sentence.data() + second_word, sentence.length() - second_word);
    std::cout << std::endl;

    // Or, if you want a string object, make a copy from the start of the
    // second word. substr copies until the end of the string when you give
    // it only one argument, like here:
    std::string rest{sentence.substr(second_word)};
    std::cout << rest << std::endl;
}

当然，除非你有很好的理由不这样做，否则你应该检查first_space != std::string::npos，这意味着没有找到空间。为了清楚起见，我的示例代码中省略了检查:)

Answer 4

您可以使用string::find() 来定位第一个空格。一旦你有了它的索引，然后从空间索引之后的索引到字符串末尾的索引中获取带有string::substr()的子字符串。

Answer 5

例如，您可以获取剩余的子字符串

string sentence{"Hello how are you."};
stringstream ss{sentence};
string junkWord;
ss >> junkWord;
cout<<sentence.substr(junkWord.length()+1); //string::substr

不过，这也取决于你想进一步做什么

Answer 6

一个班轮：

std::string subStr = sentence.substr(sentence.find_first_not_of(" \t\r\n", sentence.find_first_of(" \t\r\n", sentence.find_first_not_of(" \t\r\n"))));

工作示例：

#include <iostream>
#include <string>

void main()
{
    std::string sentence{ "Hello how are you." };

    char whiteSpaces[] = " \t\r\n";

    std::string subStr = sentence.substr(sentence.find_first_not_of(whiteSpaces, sentence.find_first_of(whiteSpaces, sentence.find_first_not_of(whiteSpaces))));

    std::cout << subStr;

    std::cin.ignore();
}

Answer 7

以下是如何使用stringstream 提取垃圾词，同时忽略前后的任何空格（使用std::ws），然后通过强大的错误处理获取句子的其余部分......

std::string sentence{"Hello how are you."};
std::stringstream ss{sentence};
std::string junkWord;
if (ss >> junkWord >> std::ws && std::getline(ss, sentence, '\0'))
    std::cout << sentence << '\n';
else
    std::cerr << "the sentence didn't contain ANY words at all\n";

看到它正在运行on ideone here....

Answer 8

#include <iostream>  // cout
#include <string>   // string
#include <sstream> // string stream
using namespace std;

int main()
{
     string testString = "Hello how are you.";

     istringstream iss(testString); // note istringstream NOT sstringstream

     char c; // this will read the delima (space in this case)
     string firstWord;

     iss>>firstWord>>c; // read the first word and end after the first ' '

    cout << "The first word in \"" << testString << "\" is \"" << firstWord << "\""<<endl;

    cout << "The rest of the words is \"" <<testString.substr(firstWord.length()+1) << "\""<<endl;
    return 0;
}

输出

“你好，你好吗”中的第一个词。是“你好” 剩下的词是“你好吗”。

live testing at ideon