Question

如何使用rapidjson将对象序列化为std :: string？我已经实施了

class Person{
public:
    std::string name;
    uint64 id; // uint64 is typedef

    template <typename Writer>
    void Serialize(Writer& writer) const {
        writer.StartObject();
        writer.String("name");
        writer.String(name);
        writer.String(("id"));
        writer.Uint64(id);
        writer.EndObject();
    }
     std::string serialize(){
        FileStream s(stdout);
        PrettyWriter<FileStream> writer(s);   
        Serialize(writer);
        return  ? /// There is a problem

    }
}

问题是序列化函数要返回什么？

Answer 1

无。您已将其发送至stdout。

如果您不想将输出流式传输到文件，请不要使用FileStream;使用PrettyWriter的其他模板参数，存储并允许您提取字符串。

通过快速浏览文档，StringBuffer看起来很有希望。它是GenericStringBuffer<UTF8<> >的类型别名。

Answer 2

问题很老，如果你还在寻找答案，那么这里是@Lightness Races in Orbit建议的那个

class Person{
public:
    std::string name;
    uint64 id; // uint64 is typedef

    template <typename Writer>
    void Serialize(Writer& writer) const {
        writer.StartObject();
        writer.String("name");
        writer.String(name);
        writer.String(("id"));
        writer.Uint64(id);
        writer.EndObject();
    }
     std::string serialize(){
        StringBuffer s;
        Writer<StringBuffer> writer(s);
        Serialize(writer);
        return  s.GetString();
    }
}

您还可以在此处查看示例代码：simplewriter.cpp

Answer 3

试试这个：

std::string serialize() {
  GenericStringBuffer<UTF8<> > buffer;
  Writer<GenericStringBuffer<UTF8<> > > writer(buffer);

  Serialize(writer);

  return buffer.GetString();
}

如何使用rapidjson（实现序列化）将对象序列化为std :: string？

3 个答案: