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);
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());
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());
}