Question

我必须读取以“;”分隔的值来自csv文件......例如

2039213;Hans;Meier;12.20.1943;2.4;
4039293;Jim;Raynor;31.12.2011;3.0;

int;char[];char[];char[],float

如何使用not string将一行拆分为单词，只是char []？然后我必须把那些拆分的值放到一个结构中，好吧，我认为这不是太难，但我如何分割值？我的代码：

struct Studentendaten {
    int matrnr;
    string name;
    string vorname;
    string datum;
    float note;
};
Studentendaten stud;
array<Studentendaten,100> studArray ;   


  if (pos != -1) 
  {

      sub1 = sub.substr(0,pos);               
      sub2  = sub.substr(pos+1,pos);
      sub3  = sub.substr(pos+1,pos);
      sub4  =sub.substr(pos+1,pos);
      sub5  =sub.substr(pos+1,pos);           
      stud.matrnr = std::to_string(sub1);
      stud.name = sub2;
      stud.vorname = sub3;
      stud.datum = sub4;
      stud.note = float(sub5);
  }

  if (ch == '\n') 
  {
      stud = {matrn,name,vorname,datum,note};
      studArray[i] = stud;
      i++;
  }

我还有问题从字符串转换为int，从字符串转换为float不起作用，无论我应用什么函数... 它经常说：

dateiLesen.cc:54:19: error: 'to_string' is not a member of 'std'

或

 dateiLesen.cc:58:27: error: invalid cast from type 'std::string {aka std::basic_string<char>}' to type 'float'

此外，我不知道我的结构有什么问题：

 dateiLesen.cc:13:9: note: main()::Studentendaten& main()::Studentendaten::operator=(const main()::Studentendaten&)
  struct Studentendaten {
 dateiLesen.cc:13:9: note:   no known conversion for argument 1 from '<brace-enclosed initializer  list>' to 'const main()::Studentendaten&'
dateiLesen.cc:13:9: note: main()::Studentendaten& main()::Studentendaten::operator=(main()::Studentendaten&&)
dateiLesen.cc:13:9: note:   no known conversion for argument 1 from '<brace-enclosed initializer list>' to 'main()::Studentendaten&&'

Answer 1

好的，让我们尝试解决那些编译错误

dateiLesen.cc:54:19: error: 'to_string' is not a member of 'std'

您没有包含定义std :: to_string的标头，因此编译器不知道它是什么。 [提示：试试谷歌！ ]

dateiLesen.cc:58:27: error: invalid cast from type 'std::string {aka std::basic_string<char>}' to type 'float'

您无法直接将字符串转换为浮点数。您需要使用boost::lexical_cast

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读取csv并以分号分隔

1 个答案: