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
对于字符串(或单个字符)分隔符，这应该非常有效。不要忘记包含#include <sstream>。
std::string input = "Alfa=,+Bravo=,+Charlie=,+Delta";
std::string delimiter = "=,+";
std::istringstream ss(input);
std::string token;
std::string::iterator it;
while(std::getline(ss, token, *(it = delimiter.begin()))) {
std::cout << token << std::endl; // Token is extracted using '='
it++;
// Skip the rest of delimiter if exists ",+"
while(it != delimiter.end() and ss.peek() == *(it)) {
it++; ss.get();
}
}
第一个while循环使用字符串分隔符的第一个字符提取一个标记。第二个while循环跳过分隔符的其余部分，停在下一个标记的开头。
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, ">=");
}
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
这是一个简洁的分裂函数。我决定让背靠背分隔符返回为空字符串，但您可以很容易地检查子字符串是否为空，如果是，则不将其添加到向量。
#include <vector>
#include <string>
using namespace std;
vector<string> split(string to_split, string delimiter) {
size_t pos = 0;
vector<string> matches{};
do {
pos = to_split.find(delimiter);
int change_end;
if (pos == string::npos) {
pos = to_split.length() - 1;
change_end = 1;
}
else {
change_end = 0;
}
matches.push_back(to_split.substr(0, pos+change_end));
to_split.erase(0, pos+1);
}
while (!to_split.empty());
return matches;
}
这与其他答案相似，但它使用了string_view。这些是原始字符串的视图。类似于c++20的例子。虽然这将是一个c++17的例子。(编辑以跳过空匹配)
#include <algorithm>
#include <iostream>
#include <string_view>
#include <vector>
std::vector<std::string_view> split(std::string_view buffer,
const std::string_view delimeter = " ") {
std::vector<std::string_view> ret{};
std::decay_t<decltype(std::string_view::npos)> pos{};
while ((pos = buffer.find(delimeter)) != std::string_view::npos) {
const auto match = buffer.substr(0, pos);
if (!match.empty()) ret.push_back(match);
buffer = buffer.substr(pos + delimeter.size());
}
if (!buffer.empty()) ret.push_back(buffer);
return ret;
}
int main() {
const auto split_values = split("1 2 3 4 5 6 7 8 9 10 ");
std::for_each(split_values.begin(), split_values.end(),
[](const auto& str) { std::cout << str << '\n'; });
return split_values.size();
}
