c++ - rapidjson - 将键更改为另一个值

Question

这里是rapidjson 的hello world。如何更改密钥"hello"并"goodbye"从 json 获取字符串？我的意思是我想解析 json，更改一些键并获取 json 字符串，如{"goodbye" : "world"}.

const char json[] = "{ \"hello\" : \"world\" }";

rapidjson::Document d;
d.Parse<0>(json);

Answer 1

  const char *json = R"({"hello": "world"})";
  rapidjson::Document d;
  d.Parse<0> (json);

  rapidjson::Value::Member* hello = d.FindMember ("hello"); if (hello) {
    d.AddMember ("goodbye", hello->value, d.GetAllocator());
    d.RemoveMember ("hello");
  }

  typedef rapidjson::GenericStringBuffer<rapidjson::UTF8<>, rapidjson::MemoryPoolAllocator<>> StringBuffer;
  StringBuffer buf (&d.GetAllocator());
  rapidjson::Writer<StringBuffer> writer (buf, &d.GetAllocator());
  d.Accept (writer);
  json = buf.GetString();

PS您可能应该在json之后复制它，因为它的内存将与d.

PPS 您还可以就地替换字段名称，而不将其删除：

rapidjson::Value::Member* hello = d.FindMember ("hello");
if (hello) hello->name.SetString ("goodbye", d.GetAllocator());

或在迭代期间：

for (auto it = d.MemberBegin(); it != d.MemberEnd(); ++it)
  if (strcmp (it->name.GetString(), "hello") == 0) it->name.SetString ("goodbye", d.GetAllocator());

Answer 2

在我的例子中，有一个名为的键字典keyDict，它存储对象键应该被替换的值。

std::string line;
std::map<std::string, int>  keyDict;

.....................
.........................

rapidjson::Document         doc;
doc.Parse<0>(line.c_str());
rapidjson::Value::MemberIterator itr;


for (itr = doc.MemberonBegin(); itr != doc.MemberonEnd(); ++itr)
{
    std::string keyCode = std::to_string(keyDict[itr->name.GetString()]);
    itr->name.SetString(keyCode.c_str(), keyCode.size(), doc.GetAllocator());
}