我有以下名为header的字符串:SELECT s.ymd, s.symbol, s.price_close FROM stocks s
DISTRIBUTE BY s.symbol
SORT BY s.symbol ASC, s.ymd ASC;
我需要从这个字符串中获取文件名和文件类型,但我的解决方案似乎非常混乱(伪代码):
"bla bla hello, just more characters filename="myfile.1.2.doc" more characters"
如何从cpp的结尾看?
答案 0 :(得分:2)
我认为如果你使用正则表达式会更好
例如:我们有更复杂的字符串,文件名和文件名之外的(“)等字符混淆。
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();
}
第一次迭代为您提供:
name = myfile.1.2
ext = doc
第二:
name = newfile
ext = exe
答案 1 :(得分:0)
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
}