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
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);
}
其他回答
这是一个完整的方法，它在任何分隔符上分割字符串，并返回分割后的字符串的向量。
这是改编自ryanbwork的答案。然而，他的检查:if(token != mystring)给出错误的结果，如果你的字符串中有重复的元素。这是我对那个问题的解决方案。
vector<string> Split(string mystring, string delimiter)
{
vector<string> subStringList;
string token;
while (true)
{
size_t findfirst = mystring.find_first_of(delimiter);
if (findfirst == string::npos) //find_first_of returns npos if it couldn't find the delimiter anymore
{
subStringList.push_back(mystring); //push back the final piece of mystring
return subStringList;
}
token = mystring.substr(0, mystring.find_first_of(delimiter));
mystring = mystring.substr(mystring.find_first_of(delimiter) + 1);
subStringList.push_back(token);
}
return subStringList;
}
我会使用boost::tokenizer。下面的文档解释了如何创建适当的标记器函数:http://www.boost.org/doc/libs/1_52_0/libs/tokenizer/tokenizerfunction.htm
这里有一个对你的案子有用。
struct my_tokenizer_func
{
template<typename It>
bool operator()(It& next, It end, std::string & tok)
{
if (next == end)
return false;
char const * del = ">=";
auto pos = std::search(next, end, del, del + 2);
tok.assign(next, pos);
next = pos;
if (next != end)
std::advance(next, 2);
return true;
}
void reset() {}
};
int main()
{
std::string to_be_parsed = "1) one>=2) two>=3) three>=4) four";
for (auto i : boost::tokenizer<my_tokenizer_func>(to_be_parsed))
std::cout << i << '\n';
}
以下是我的看法。它处理边缘情况，并接受一个可选参数从结果中删除空条目。
bool endsWith(const std::string& s, const std::string& suffix)
{
return s.size() >= suffix.size() &&
s.substr(s.size() - suffix.size()) == suffix;
}
std::vector<std::string> split(const std::string& s, const std::string& delimiter, const bool removeEmptyEntries = false)
{
std::vector<std::string> tokens;
for (size_t start = 0, end; start < s.length(); start = end + delimiter.length())
{
size_t position = s.find(delimiter, start);
end = position != std::string::npos ? position : s.length();
std::string token = s.substr(start, end - start);
if (!removeEmptyEntries || !token.empty())
{
tokens.push_back(token);
}
}
if (!removeEmptyEntries &&
(s.empty() || endsWith(s, delimiter)))
{
tokens.push_back("");
}
return tokens;
}
例子
split("a-b-c", "-"); // [3]("a","b","c")
split("a--c", "-"); // [3]("a","","c")
split("-b-", "-"); // [3]("","b","")
split("--c--", "-"); // [5]("","","c","","")
split("--c--", "-", true); // [1]("c")
split("a", "-"); // [1]("a")
split("", "-"); // [1]("")
split("", "-", true); // [0]()
这个方法使用字符串find和字符串substr
vector<string> split(const string& str,const string delim){
vector<string> vtokens;
size_t start = 0;
size_t end = 0;
while((end = str.find(delim,start))!=string::npos){
vtokens.push_back(str.substr(start,end-start));
start = end +1;
}
vtokens.push_back(str.substr(start));
return vtokens;
}
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
