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
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"
其他回答
从c++ 11开始，它可以这样做:
std::vector<std::string> splitString(const std::string& str,
const std::regex& regex)
{
return {std::sregex_token_iterator{str.begin(), str.end(), regex, -1},
std::sregex_token_iterator() };
}
// usually we have a predefined set of regular expressions: then
// let's build those only once and re-use them multiple times
static const std::regex regex1(R"some-reg-exp1", std::regex::optimize);
static const std::regex regex2(R"some-reg-exp2", std::regex::optimize);
static const std::regex regex3(R"some-reg-exp3", std::regex::optimize);
string str = "some string to split";
std::vector<std::string> tokens( splitString(str, regex1) );
注:
这是对这个答案的一个小小的改进 参见std::regex_constants::optimize使用的优化技术
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);
}
Strtok允许您传入多个字符作为分隔符。我敢打赌，如果你传入“>=”，你的示例字符串将被正确分割(即使>和=被算作单独的分隔符)。
EDIT如果您不想使用c_str()将字符串转换为char*，您可以使用substr和find_first_of进行标记化。
string token, mystring("scott>=tiger");
while(token != mystring){
token = mystring.substr(0,mystring.find_first_of(">="));
mystring = mystring.substr(mystring.find_first_of(">=") + 1);
printf("%s ",token.c_str());
}
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;
}
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"
