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
template<typename C, typename T>
auto insert_in_container(C& c, T&& t) -> decltype(c.push_back(std::forward<T>(t)), void()) {
c.push_back(std::forward<T>(t));
}
template<typename C, typename T>
auto insert_in_container(C& c, T&& t) -> decltype(c.insert(std::forward<T>(t)), void()) {
c.insert(std::forward<T>(t));
}
template<typename Container>
Container splitR(const std::string& input, const std::string& delims) {
Container out;
size_t delims_len = delims.size();
auto begIdx = 0u;
auto endIdx = input.find(delims, begIdx);
if (endIdx == std::string::npos && input.size() != 0u) {
insert_in_container(out, input);
}
else {
size_t w = 0;
while (endIdx != std::string::npos) {
w = endIdx - begIdx;
if (w != 0) insert_in_container(out, input.substr(begIdx, w));
begIdx = endIdx + delims_len;
endIdx = input.find(delims, begIdx);
}
w = input.length() - begIdx;
if (w != 0) insert_in_container(out, input.substr(begIdx, w));
}
return out;
}
其他回答
一个更简单的解决方案是-
可以使用strtok在多字符分隔符的基础上进行分隔。 记住使用strdup，这样原始字符串就不会发生变化。
#include <stdio.h>
#include <string.h>
const char* str = "scott>=tiger";
char *token = strtok(strdup(str), ">=");
while (token != NULL)
{
printf("%s\n", token);
token = strtok(NULL, ">=");
}
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;
}
我查看了答案，没有看到一个基于迭代器的方法可以被送入范围循环，所以我做了一个。
这使用了c++ 17 string_views，所以它不应该分配字符串的副本。
struct StringSplit
{
struct Iterator
{
size_t tokenStart_ = 0;
size_t tokenEnd_ = 0;
std::string str_;
std::string_view view_;
std::string delimiter_;
bool done_ = false;
Iterator()
{
// End iterator.
done_ = true;
}
Iterator(std::string str, std::string delimiter)
: str_{std::move(str)}, view_{str_}, delimiter_{
std::move(delimiter)}
{
tokenEnd_ = view_.find(delimiter_, tokenStart_);
}
std::string_view operator*()
{
return view_.substr(tokenStart_, tokenEnd_ - tokenStart_);
}
Iterator &operator++()
{
if (tokenEnd_ == std::string::npos)
{
done_ = true;
return *this;
}
tokenStart_ = tokenEnd_ + delimiter_.size();
tokenEnd_ = view_.find(delimiter_, tokenStart_);
return *this;
}
bool operator!=(Iterator &other)
{
// We only check if both points to the end.
if (done_ && other.done_)
{
return false;
}
return true;
}
};
Iterator beginIter_;
StringSplit(std::string str, std::string delim)
: beginIter_{std::move(str), std::move(delim)}
{
}
Iterator begin()
{
return beginIter_;
}
Iterator end()
{
return Iterator{};
}
};
示例用法如下:
int main()
{
for (auto token : StringSplit{"<>foo<>bar<><>bar<><>baz<><>", "<>"})
{
std::cout << "TOKEN: '" << token << "'" << std::endl;
}
}
打印:
TOKEN: ''
TOKEN: 'foo'
TOKEN: 'bar'
TOKEN: ''
TOKEN: 'bar'
TOKEN: ''
TOKEN: 'baz'
TOKEN: ''
TOKEN: ''
它正确地处理字符串开头和结尾的空项。
用c++ 20实现的一种方法:
#include <iostream>
#include <ranges>
#include <string_view>
int main()
{
std::string hello = "text to be parsed";
auto split = hello
| std::ranges::views::split(' ')
| std::ranges::views::transform([](auto&& str) { return std::string_view(&*str.begin(), std::ranges::distance(str)); });
for (auto&& word : split)
{
std::cout << word << std::endl;
}
}
看到的: https://stackoverflow.com/a/48403210/10771848 https://en.cppreference.com/w/cpp/ranges/split_view
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);
}
