我在c++中使用以下方法解析字符串:

using namespace std;

string parsed,input="text to be parsed";
stringstream input_stringstream(input);

if (getline(input_stringstream,parsed,' '))
{
     // do some processing.
}

使用单个字符分隔符进行解析是可以的。但是如果我想使用字符串作为分隔符呢?

例子:我想拆分:

scott>=tiger

用>=作为分隔符,这样我就可以得到斯科特和老虎。


当前回答

std::vector<std::string> parse(std::string str,std::string delim){
    std::vector<std::string> tokens;
    char *str_c = strdup(str.c_str()); 
    char* token = NULL;

    token = strtok(str_c, delim.c_str()); 
    while (token != NULL) { 
        tokens.push_back(std::string(token));  
        token = strtok(NULL, delim.c_str()); 
    }

    delete[] str_c;

    return tokens;
}

其他回答

这是一个简洁的分裂函数。我决定让背靠背分隔符返回为空字符串,但您可以很容易地检查子字符串是否为空,如果是,则不将其添加到向量。

#include <vector>
#include <string>
using namespace std;



vector<string> split(string to_split, string delimiter) {
    size_t pos = 0;
    vector<string> matches{};
    do {
        pos = to_split.find(delimiter);
        int change_end;
        if (pos == string::npos) {
            pos = to_split.length() - 1;
            change_end = 1;
        }
        else {
            change_end = 0;
        }
        matches.push_back(to_split.substr(0, pos+change_end));
        
        to_split.erase(0, pos+1);

    }
    while (!to_split.empty());
    return matches;

}

该方法使用std::string::find,而不改变原始字符串,记住前一个子字符串标记的开始和结束。

#include <iostream>
#include <string>

int main()
{
    std::string s = "scott>=tiger";
    std::string delim = ">=";

    auto start = 0U;
    auto end = s.find(delim);
    while (end != std::string::npos)
    {
        std::cout << s.substr(start, end - start) << std::endl;
        start = end + delim.length();
        end = s.find(delim, start);
    }

    std::cout << s.substr(start, end);
}

您可以使用std::string::find()函数来查找字符串分隔符的位置,然后使用std::string::substr()来获取一个令牌。

例子:

std::string s = "scott>=tiger";
std::string delimiter = ">=";
std::string token = s.substr(0, s.find(delimiter)); // token is "scott"

find(const string& str, size_t pos = 0)函数的作用是:返回字符串中str第一次出现的位置,如果没有找到则返回npos。 substr(size_t pos = 0, size_t n = npos)函数的作用是:返回对象的子字符串,从位置pos开始,长度npos。


如果你有多个分隔符,在你提取了一个标记之后,你可以删除它(包括分隔符)以继续后续的提取(如果你想保留原始字符串,只需使用s = s.s substr(pos + delimiter.length());):

s.erase(0, s.find(delimiter) + delimiter.length());

这样就可以轻松地循环获取每个令牌。

完整的示例

std::string s = "scott>=tiger>=mushroom";
std::string delimiter = ">=";

size_t pos = 0;
std::string token;
while ((pos = s.find(delimiter)) != std::string::npos) {
    token = s.substr(0, pos);
    std::cout << token << std::endl;
    s.erase(0, pos + delimiter.length());
}
std::cout << s << std::endl;

输出:

scott
tiger
mushroom

作为奖励,这里有一个分割函数和宏的代码示例,很容易使用,你可以在其中选择容器类型:

#include <iostream>
#include <vector>
#include <string>

#define split(str, delim, type) (split_fn<type<std::string>>(str, delim))
 
template <typename Container>
Container split_fn(const std::string& str, char delim = ' ') {
    Container cont{};
    std::size_t current, previous = 0;
    current = str.find(delim);
    while (current != std::string::npos) {
        cont.push_back(str.substr(previous, current - previous));
        previous = current + 1;
        current = str.find(delim, previous);
    }
    cont.push_back(str.substr(previous, current - previous));
    
    return cont;
}

int main() {
    
    auto test = std::string{"This is a great test"};
    auto res = split(test, ' ', std::vector);
    
    for(auto &i : res) {
        std::cout << i << ", "; // "this", "is", "a", "great", "test"
    }
    
    
    return 0;
}

由于这是c++分割字符串或类似的顶级Stack Overflow谷歌搜索结果,我将发布一个完整的、复制/粘贴可运行的示例,展示这两种方法。

splitString使用stringstream(在大多数情况下可能是更好更简单的选项)

splitString2使用find和substr(一种更手动的方法)

// SplitString.cpp

#include <iostream>
#include <vector>
#include <string>
#include <sstream>

// function prototypes
std::vector<std::string> splitString(const std::string& str, char delim);
std::vector<std::string> splitString2(const std::string& str, char delim);
std::string getSubstring(const std::string& str, int leftIdx, int rightIdx);


int main(void)
{
  // Test cases - all will pass
  
  std::string str = "ab,cd,ef";
  //std::string str = "abcdef";
  //std::string str = "";
  //std::string str = ",cd,ef";
  //std::string str = "ab,cd,";   // behavior of splitString and splitString2 is different for this final case only, if this case matters to you choose which one you need as applicable
  
  
  std::vector<std::string> tokens = splitString(str, ',');
  
  std::cout << "tokens: " << "\n";
  
  if (tokens.empty())
  {
    std::cout << "(tokens is empty)" << "\n";
  }
  else
  {
    for (auto& token : tokens)
    {
      if (token == "") std::cout << "(empty string)" << "\n";
      else std::cout << token << "\n";
    }
  }
    
  return 0;
}

std::vector<std::string> splitString(const std::string& str, char delim)
{
  std::vector<std::string> tokens;
  
  if (str == "") return tokens;
  
  std::string currentToken;
  
  std::stringstream ss(str);
  
  while (std::getline(ss, currentToken, delim))
  {
    tokens.push_back(currentToken);
  }
  
  return tokens;
}

std::vector<std::string> splitString2(const std::string& str, char delim)
{
  std::vector<std::string> tokens;
  
  if (str == "") return tokens;
  
  int leftIdx = 0;
  
  int delimIdx = str.find(delim);
  
  int rightIdx;
  
  while (delimIdx != std::string::npos)
  {
    rightIdx = delimIdx - 1;
    
    std::string token = getSubstring(str, leftIdx, rightIdx);
    tokens.push_back(token);
    
    // prep for next time around
    leftIdx = delimIdx + 1;
    
    delimIdx = str.find(delim, delimIdx + 1);
  }
  
  rightIdx = str.size() - 1;
  
  std::string token = getSubstring(str, leftIdx, rightIdx);
  tokens.push_back(token);
  
  return tokens;
}

std::string getSubstring(const std::string& str, int leftIdx, int rightIdx)
{
  return str.substr(leftIdx, rightIdx - leftIdx + 1);
}