如何从cpp中给定的字符串中提取特定的字符串?
我有以下字符串名为标题:"bla bla hello, just more characters filename="myfile.1.2.doc" more characters"
如何从cpp中给定的字符串中提取特定的字符串?
我需要的文件名,并从该字符串的文件类型,但我的解决办法似乎很凌乱(伪代码):
unsigned int end = header.find("filename=");
unsigned int end2 = header.find(" " ", end + sizeof("filename=") + 1) // how to search for ' " ' ?!
std::string fullFileName = header.substr(end +sizeof("filename=") + 1 ,end2 -1);
//now look for the first "." from the end and split by that .
如何从cpp最后看?
我想如果你使用正则表达式会更好。
例如:我们有一个文件名之外的几个文件名和混乱的字符,如(“)更复杂的字符串
std::string str("bla bla hello, just more characters filename=\"myfile.1.2.doc\" more characters bla bla hello, just more characters filename=\"newFile.exe\" more char\"acters");
std::smatch match;
std::regex regExp("filename=\"(.*?)\\.([^.]*?)\"");
while (std::regex_search(str, match, regExp))
{
std::string name = match[1].str();
std::string ext = match[2].str();
str = match.suffix().str();
}
第一次迭代给你:
名= myfile.1.2
EXT = DOC
第二种:
名称= 的newfile
EXT = exe
我应该使用什么版本的Cpp?我正在尝试这样做,但这些都不是我的编译器所熟悉的。我看到的东西是std :: regexec() \t \t std :: regcomp() – user1386966
它需要C++ 11. 你使用什么编译器? – arturx64
size_t startpos = header.find("filename=");
if (startpos != header.npos)
{ // found filename
startpos += sizeof("filename=") - 1; // sizeof determined at compile time.
// -1 ignores the null termination on the c-string
if (startpos != header.length() && header[startpos] == '\"')
{ // next char, if there is one, should be "
startpos++;
size_t endpos = header.find('\"', startpos);
if (endpos != header.npos)
{ // found terminating ". get full file name
std::string fullfname = header.substr(startpos, endpos-startpos);
size_t dotpos = fullfname.find_last_of('.');
if (dotpos != fullfname.npos)
{ // found dot split string
std::string filename = fullfname.substr(0, dotpos);
//add extra brains here to remove path
std::string filetype = fullfname.substr(dotpos + 1, token.npos);
// dostuff
std::cout << fullfname << ": " << filename << " dot " << filetype << std::endl;
}
else
{
// handle error
}
}
else
{
// handle error
}
}
else
{
// handle error
}
}
else
{
// handle error
}
使用“as a”而不是解析令牌:\“ – user4581301
您需要*转义*字符串文本中的任何双引号字符。例如:'header.find(“\”“,// ...);' –