答案已经在那里，但选择答案使用擦除功能，这是非常昂贵的，想想一些非常大的字符串(以mb为单位)。因此我使用下面的函数。

vector<string> split(const string& i_str, const string& i_delim)
{
    vector<string> result;
    size_t startIndex = 0;

    for (size_t found = i_str.find(i_delim); found != string::npos; found = i_str.find(i_delim, startIndex))
    {
        result.emplace_back(i_str.begin()+startIndex, i_str.begin()+found);
        startIndex = found + i_delim.size();
    }
    if (startIndex != i_str.size())
        result.emplace_back(i_str.begin()+startIndex, i_str.end());
    return result;      
}

如果你不想修改字符串(就像Vincenzo Pii回答的那样)，并且想输出最后一个令牌，你可能想使用这种方法:

inline std::vector<std::string> splitString( const std::string &s, const std::string &delimiter ){
    std::vector<std::string> ret;
    size_t start = 0;
    size_t end = 0;
    size_t len = 0;
    std::string token;
    do{ end = s.find(delimiter,start); 
        len = end - start;
        token = s.substr(start, len);
        ret.emplace_back( token );
        start += len + delimiter.length();
        std::cout << token << std::endl;
    }while ( end != std::string::npos );
    return ret;
}

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;
}

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);
}

功能:

std::vector<std::string> WSJCppCore::split(const std::string& sWhat, const std::string& sDelim) {
    std::vector<std::string> vRet;
    size_t nPos = 0;
    size_t nLen = sWhat.length();
    size_t nDelimLen = sDelim.length();
    while (nPos < nLen) {
        std::size_t nFoundPos = sWhat.find(sDelim, nPos);
        if (nFoundPos != std::string::npos) {
            std::string sToken = sWhat.substr(nPos, nFoundPos - nPos);
            vRet.push_back(sToken);
            nPos = nFoundPos + nDelimLen;
            if (nFoundPos + nDelimLen == nLen) { // last delimiter
                vRet.push_back("");
            }
        } else {
            std::string sToken = sWhat.substr(nPos, nLen - nPos);
            vRet.push_back(sToken);
            break;
        }
    }
    return vRet;
}

单元测试:

bool UnitTestSplit::run() {
bool bTestSuccess = true;

    struct LTest {
        LTest(
            const std::string &sStr,
            const std::string &sDelim,
            const std::vector<std::string> &vExpectedVector
        ) {
            this->sStr = sStr;
            this->sDelim = sDelim;
            this->vExpectedVector = vExpectedVector;
        };
        std::string sStr;
        std::string sDelim;
        std::vector<std::string> vExpectedVector;
    };
    std::vector<LTest> tests;
    tests.push_back(LTest("1 2 3 4 5", " ", {"1", "2", "3", "4", "5"}));
    tests.push_back(LTest("|1f|2п|3%^|44354|5kdasjfdre|2", "|", {"", "1f", "2п", "3%^", "44354", "5kdasjfdre", "2"}));
    tests.push_back(LTest("|1f|2п|3%^|44354|5kdasjfdre|", "|", {"", "1f", "2п", "3%^", "44354", "5kdasjfdre", ""}));
    tests.push_back(LTest("some1 => some2 => some3", "=>", {"some1 ", " some2 ", " some3"}));
    tests.push_back(LTest("some1 => some2 => some3 =>", "=>", {"some1 ", " some2 ", " some3 ", ""}));

    for (int i = 0; i < tests.size(); i++) {
        LTest test = tests[i];
        std::string sPrefix = "test" + std::to_string(i) + "(\"" + test.sStr + "\")";
        std::vector<std::string> vSplitted = WSJCppCore::split(test.sStr, test.sDelim);
        compareN(bTestSuccess, sPrefix + ": size", vSplitted.size(), test.vExpectedVector.size());
        int nMin = std::min(vSplitted.size(), test.vExpectedVector.size());
        for (int n = 0; n < nMin; n++) {
            compareS(bTestSuccess, sPrefix + ", element: " + std::to_string(n), vSplitted[n], test.vExpectedVector[n]);
        }
    }

    return bTestSuccess;
}

下面是一个使用Boost string Algorithms库和Boost Range库将一个字符串与另一个字符串分割的示例。这个解决方案的灵感来自StringAlgo库文档，请参阅Split部分。

下面是split_with_string函数的完整程序，以及全面的测试-用godbolt试试:

#include <iostream>
#include <string>
#include <vector>
#include <boost/algorithm/string.hpp>
#include <boost/range/iterator_range.hpp>

std::vector<std::string> split_with_string(std::string_view s, std::string_view search) 
{
    if (search.empty()) return {std::string{s}};

    std::vector<boost::iterator_range<std::string_view::iterator>> found;
    boost::algorithm::ifind_all(found, s, search);
    if (found.empty()) return {};

    std::vector<std::string> parts;
    parts.reserve(found.size() + 2); // a bit more

    std::string_view::iterator part_begin = s.cbegin(), part_end;
    for (auto& split_found : found)
    {
        // do not skip empty extracts
        part_end = split_found.begin();
        parts.emplace_back(part_begin, part_end);
        part_begin = split_found.end();
    }
    if (part_end != s.end())
        parts.emplace_back(part_begin, s.end());

    return parts;
}

#define TEST(expr) std::cout << ((!(expr)) ? "FAIL" : "PASS") << ": " #expr "\t" << std::endl

int main()
{
    auto s0 = split_with_string("adsf-+qwret-+nvfkbdsj", "");
    TEST(s0.size() == 1);
    TEST(s0.front() == "adsf-+qwret-+nvfkbdsj");
    auto s1 = split_with_string("adsf-+qwret-+nvfkbdsj", "-+");
    TEST(s1.size() == 3);
    TEST(s1.front() == "adsf");
    TEST(s1.back() == "nvfkbdsj");
    auto s2 = split_with_string("-+adsf-+qwret-+nvfkbdsj-+", "-+");
    TEST(s2.size() == 5);
    TEST(s2.front() == "");
    TEST(s2.back() == "");
    auto s3 = split_with_string("-+adsf-+qwret-+nvfkbdsj", "-+");
    TEST(s3.size() == 4);
    TEST(s3.front() == "");
    TEST(s3.back() == "nvfkbdsj");
    auto s4 = split_with_string("adsf-+qwret-+nvfkbdsj-+", "-+");
    TEST(s4.size() == 4);
    TEST(s4.front() == "adsf");
    TEST(s4.back() == "");
    auto s5 = split_with_string("dbo.abc", "dbo.");
    TEST(s5.size() == 2);
    TEST(s5.front() == "");
    TEST(s5.back() == "abc");
    auto s6 = split_with_string("dbo.abc", ".");
    TEST(s6.size() == 2);
    TEST(s6.front() == "dbo");
    TEST(s6.back() == "abc");
}

测试输出:

PASS: s0.size() == 1    
PASS: s0.front() == "adsf-+qwret-+nvfkbdsj" 
PASS: s1.size() == 3    
PASS: s1.front() == "adsf"  
PASS: s1.back() == "nvfkbdsj"   
PASS: s2.size() == 5    
PASS: s2.front() == ""  
PASS: s2.back() == ""   
PASS: s3.size() == 4    
PASS: s3.front() == ""  
PASS: s3.back() == "nvfkbdsj"   
PASS: s4.size() == 4    
PASS: s4.front() == "adsf"  
PASS: s4.back() == ""   
PASS: s5.size() == 2    
PASS: s5.front() == ""  
PASS: s5.back() == "abc"    
PASS: s6.size() == 2    
PASS: s6.front() == "dbo"   
PASS: s6.back() == "abc"