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
#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
其他回答
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;
}
这段代码从文本中分离行，并将每个行添加到一个向量中。
vector<string> split(char *phrase, string delimiter){
vector<string> list;
string s = string(phrase);
size_t pos = 0;
string token;
while ((pos = s.find(delimiter)) != string::npos) {
token = s.substr(0, pos);
list.push_back(token);
s.erase(0, pos + delimiter.length());
}
list.push_back(s);
return list;
}
调用:
vector<string> listFilesMax = split(buffer, "\n");
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。
下面是一个使用Boost string Algorithms库和Boost Range库将一个字符串与另一个字符串分割的示例。这个解决方案的灵感来自StringAlgo库文档，请参阅Split部分。
下面是split_with_string函数的完整程序，以及全面的测试-用godbolt试试:
#include <iostream>
#include <string>
#include <vector>
#include <boost/algorithm/string.hpp>
#include <boost/range/iterator_range.hpp>
std::vector<std::string> split_with_string(std::string_view s, std::string_view search)
{
if (search.empty()) return {std::string{s}};
std::vector<boost::iterator_range<std::string_view::iterator>> found;
boost::algorithm::ifind_all(found, s, search);
if (found.empty()) return {};
std::vector<std::string> parts;
parts.reserve(found.size() + 2); // a bit more
std::string_view::iterator part_begin = s.cbegin(), part_end;
for (auto& split_found : found)
{
// do not skip empty extracts
part_end = split_found.begin();
parts.emplace_back(part_begin, part_end);
part_begin = split_found.end();
}
if (part_end != s.end())
parts.emplace_back(part_begin, s.end());
return parts;
}
#define TEST(expr) std::cout << ((!(expr)) ? "FAIL" : "PASS") << ": " #expr "\t" << std::endl
int main()
{
auto s0 = split_with_string("adsf-+qwret-+nvfkbdsj", "");
TEST(s0.size() == 1);
TEST(s0.front() == "adsf-+qwret-+nvfkbdsj");
auto s1 = split_with_string("adsf-+qwret-+nvfkbdsj", "-+");
TEST(s1.size() == 3);
TEST(s1.front() == "adsf");
TEST(s1.back() == "nvfkbdsj");
auto s2 = split_with_string("-+adsf-+qwret-+nvfkbdsj-+", "-+");
TEST(s2.size() == 5);
TEST(s2.front() == "");
TEST(s2.back() == "");
auto s3 = split_with_string("-+adsf-+qwret-+nvfkbdsj", "-+");
TEST(s3.size() == 4);
TEST(s3.front() == "");
TEST(s3.back() == "nvfkbdsj");
auto s4 = split_with_string("adsf-+qwret-+nvfkbdsj-+", "-+");
TEST(s4.size() == 4);
TEST(s4.front() == "adsf");
TEST(s4.back() == "");
auto s5 = split_with_string("dbo.abc", "dbo.");
TEST(s5.size() == 2);
TEST(s5.front() == "");
TEST(s5.back() == "abc");
auto s6 = split_with_string("dbo.abc", ".");
TEST(s6.size() == 2);
TEST(s6.front() == "dbo");
TEST(s6.back() == "abc");
}
测试输出:
PASS: s0.size() == 1
PASS: s0.front() == "adsf-+qwret-+nvfkbdsj"
PASS: s1.size() == 3
PASS: s1.front() == "adsf"
PASS: s1.back() == "nvfkbdsj"
PASS: s2.size() == 5
PASS: s2.front() == ""
PASS: s2.back() == ""
PASS: s3.size() == 4
PASS: s3.front() == ""
PASS: s3.back() == "nvfkbdsj"
PASS: s4.size() == 4
PASS: s4.front() == "adsf"
PASS: s4.back() == ""
PASS: s5.size() == 2
PASS: s5.front() == ""
PASS: s5.back() == "abc"
PASS: s6.size() == 2
PASS: s6.front() == "dbo"
PASS: s6.back() == "abc"
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
