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
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();
}
其他回答
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;
}
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;
}
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;
}
你可以使用next函数拆分字符串:
vector<string> split(const string& str, const string& delim)
{
vector<string> tokens;
size_t prev = 0, pos = 0;
do
{
pos = str.find(delim, prev);
if (pos == string::npos) pos = str.length();
string token = str.substr(prev, pos-prev);
if (!token.empty()) tokens.push_back(token);
prev = pos + delim.length();
}
while (pos < str.length() && prev < str.length());
return tokens;
}
#include<iostream>
#include<algorithm>
using namespace std;
int split_count(string str,char delimit){
return count(str.begin(),str.end(),delimit);
}
void split(string str,char delimit,string res[]){
int a=0,i=0;
while(a<str.size()){
res[i]=str.substr(a,str.find(delimit));
a+=res[i].size()+1;
i++;
}
}
int main(){
string a="abc.xyz.mno.def";
int x=split_count(a,'.')+1;
string res[x];
split(a,'.',res);
for(int i=0;i<x;i++)
cout<<res[i]<<endl;
return 0;
}
注:仅当分割后的字符串长度相等时才有效
