我需要逐行拆分。 我以前用以下方式做的事情:
int doSegment(char *sentence, int segNum)
{
assert(pSegmenter != NULL);
Logger &log = Logger::getLogger();
char delims[] = "\n";
char *line = NULL;
if (sentence != NULL)
{
line = strtok(sentence, delims);
while(line != NULL)
{
cout << line << endl;
line = strtok(NULL, delims);
}
}
else
{
log.error("....");
}
return 0;
}
我输入“我们是一个。\ n \ n我们是。”并调用doSegment方法。但是当我调试时,我发现句子参数是“我们是一个。\\ nyes we are”,并且拆分失败了。有人能告诉我为什么会这样,我该怎么办。有没有其他我可以用来在C ++中拆分字符串。谢谢!
答案 0 :(得分:47)
我想使用std :: getline或std :: string :: find来查看字符串。 下面的代码演示了getline函数
int doSegment(char *sentence)
{
std::stringstream ss(sentence);
std::string to;
if (sentence != NULL)
{
while(std::getline(ss,to,'\n')){
cout << to <<endl;
}
}
return 0;
}
答案 1 :(得分:13)
您可以在循环中调用std::string::find
并使用std::string::substr
。
std::vector<std::string> split_string(const std::string& str,
const std::string& delimiter)
{
std::vector<std::string> strings;
std::string::size_type pos = 0;
std::string::size_type prev = 0;
while ((pos = str.find(delimiter, prev)) != std::string::npos)
{
strings.push_back(str.substr(prev, pos - prev));
prev = pos + 1;
}
// To get the last substring (or only, if delimiter is not found)
strings.push_back(str.substr(prev));
return strings;
}
参见示例here。
答案 2 :(得分:1)
#include <iostream>
#include <string>
#include <regex>
#include <algorithm>
#include <iterator>
using namespace std;
vector<string> splitter(string in_pattern, string& content){
vector<string> split_content;
regex pattern(in_pattern);
copy( sregex_token_iterator(content.begin(), content.end(), pattern, -1),
sregex_token_iterator(),back_inserter(split_content));
return split_content;
}
int main()
{
string sentence = "This is the first line\n";
sentence += "This is the second line\n";
sentence += "This is the third line\n";
vector<string> lines = splitter(R"(\n)", sentence);
for (string line: lines){cout << line << endl;}
}
// 1) We have a string with multiple lines
// 2) we split those into an array (vector)
// 3) We print out those elements in a for loop
// My Background. . .
// github.com/Radicalware
// Radicalware.net
// https://www.youtube.com/channel/UCivwmYxoOdDT3GmDnD0CfQA/playlists
答案 3 :(得分:0)
这种相当低效的方法只是循环遍历字符串,直到遇到\ n换行符。然后创建一个子字符串并将其添加到向量中。
std::vector<std::string> Loader::StringToLines(std::string string)
{
std::vector<std::string> result;
std::string temp;
int markbegin = 0;
int markend = 0;
for (int i = 0; i < string.length(); ++i) {
if (string[i] == '\n') {
markend = i;
result.push_back(string.substr(markbegin, markend - markbegin));
markbegin = (i + 1);
}
}
return result;
}
答案 4 :(得分:0)
std::vector<std::string> split_string_by_newline(const std::string& str)
{
auto result = std::vector<std::string>{};
auto ss = std::stringstream{str};
for (std::string line; std::getline(ss, line, '\n');)
result.push_back(line);
return result;
}