2012-04-22 65 views
0

我想在C++中取一个字符串,并找到其中包含的项目名称和路径。该字符串的格式为:与VS2010的正则表达式C++/Boost

Project("{8BC9CEB8-8B4A-11D0-8D11-00A0C91BC942}") = "OpenGL01", "OpenGL01\OpenGL01.vcxproj", "{65E58BFD-4339-44BA-BA9B-B2B8A5AC1FE1}" 

而正则表达式是:

boost::regex expression("(Project\(\"\{.*\}\"\)) ?= ?(\".*\"), ?(\".*\"), ?(\"\{.*\}\")"); 

,但此行产生异常错误:

Unhandled exception at 0x7512d36f in kmCompile2010.exe: Microsoft C++ exception: boost::exception_detail::clone_impl<boost::exception_detail::error_info_injector<boost::regex_error> > at memory location 0x004ce13c..

这是完整的功能:

BOOL CSlnFile::ReadProjectsSolution() 
{ 
    CString sLine; 
    boost::regex expression("(Project\(\"\{.*\}\"\)) ?= ?(\".*\"), ?(\".*\"), ?(\"\{.*\}\")"); 

    BOOL bCheck = FALSE; 

    SeekToBegin(); 

    while(ReadString(sLine) && !bCheck) { 
     // CString to char* 
     CT2A temp(sLine);   // uses LPCTSTR conversion operator for CString and CT2A constructor 
     const char* pszA = temp; // uses LPSTR conversion operator for CT2A 
     std::string strA(pszA);  // uses std::string constructor 

     boost::smatch what; 

     if (regex_match(strA, what, expression, boost::match_extra)) { 
      // TODO 
     } 
    } 

    return bCheck; 
} 

我不是和我的错误。你可以帮我吗?

+0

你需要双精度除了引号之外的所有内容都可以转义:'boost :: regex expression(“(Project \\(\”\\ {。* \\} \“\\))?=?(\”。* \“),? “。* \”),?(\“\\ {。* \\}}”)“);' – ildjarn 2012-04-23 21:20:26

回答

0

我没有尝试下面的正则表达式中的boost ::正则表达式,它工作在Perl:

\{([^}]+)\}"[^"]*"([^"]+)", "([^"]+)", "\{([^}]+) 

我想它的命令行:

echo 'Project("{8BC9CEB8-8B4A-11D0-8D11-00A0C91BC942}") = "OpenGL01", "OpenGL01\OpenGL01.vcxproj", "{65E58BFD-4339-44BA-BA9B-B2B8A5AC1FE1}"' | perl -ne 'my ($a, $b, $c, $d) = /\{([^}]+)\}"[^"]*"([^"]+)", "([^"]+)", "\{([^}]+)/; print "$a, $b, $c, $d"' 
+0

嗨。随着升压不起作用。我有同样的例外。 – 2012-04-22 11:46:29