{"account":{"name":"John","age":30},"hobbies":["reading","swimming"]}如何通过RapidJSON读取上述json字符串,并且将其中的account字段变成一个新的json字符串
时间: 2023-12-04 08:02:21 浏览: 62
把对象转换为JSON字符串
你可以使用RapidJSON库来读取JSON字符串并将其解析为C++对象。以下是一个示例代码,它将读取给定的JSON字符串并将account字段转换为新的JSON字符串:
```cpp
#include "rapidjson/document.h"
#include "rapidjson/writer.h"
#include "rapidjson/stringbuffer.h"
#include <iostream>
#include <string>
using namespace rapidjson;
using namespace std;
int main() {
// 定义JSON字符串
string jsonStr = "{\"account\":{\"name\":\"John\",\"age\":30},\"hobbies\":[\"reading\",\"swimming\"]}";
// 解析JSON字符串
Document doc;
doc.Parse(jsonStr.c_str());
// 获取account字段
Value& account = doc["account"];
// 将account字段转换为新的JSON字符串
StringBuffer buffer;
Writer<StringBuffer> writer(buffer);
account.Accept(writer);
string accountStr = buffer.GetString();
// 输出新的JSON字符串
cout << accountStr << endl;
return 0;
}
```
输出结果为:
```
{"name":"John","age":30}
```
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