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
你也可以使用regex:
std::vector<std::string> split(const std::string str, const std::string regex_str)
{
std::regex regexz(regex_str);
std::vector<std::string> list(std::sregex_token_iterator(str.begin(), str.end(), regexz, -1),
std::sregex_token_iterator());
return list;
}
这相当于:
std::vector<std::string> split(const std::string str, const std::string regex_str)
{
std::sregex_token_iterator token_iter(str.begin(), str.end(), regexz, -1);
std::sregex_token_iterator end;
std::vector<std::string> list;
while (token_iter != end)
{
list.emplace_back(*token_iter++);
}
return list;
}
像这样使用它:
#include <iostream>
#include <string>
#include <regex>
std::vector<std::string> split(const std::string str,
const std::string regex_str) {
std::regex regexz(regex_str);
return {std::sregex_token_iterator(str.begin(), str.end(), regexz, -1),
std::sregex_token_iterator()};
}
int main()
{
std::string input_str = "lets split this";
std::string regex_str = " ";
auto tokens = split(input_str, regex_str);
for (auto& item: tokens)
{
std::cout<<item <<std::endl;
}
}
在网上玩!
您可以简单地使用子字符串、字符等，或者使用实际的正则表达式来进行分割。 它也是简洁的c++ 11!
其他回答
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
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;
}
以防将来，有人想跳出Vincenzo Pii答案的盒子函数
#include <vector>
#include <string>
std::vector<std::string> SplitString(
std::string str,
std::string delimeter)
{
std::vector<std::string> splittedStrings = {};
size_t pos = 0;
while ((pos = str.find(delimeter)) != std::string::npos)
{
std::string token = str.substr(0, pos);
if (token.length() > 0)
splittedStrings.push_back(token);
str.erase(0, pos + delimeter.length());
}
if (str.length() > 0)
splittedStrings.push_back(str);
return splittedStrings;
}
我还修复了一些错误，以便如果字符串的开头或结尾有分隔符，函数将不会返回空字符串
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());
}
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
