C++ nonlohmann json读取子对象

Question

我实际上正在开发一个小程序，我需要读取一个 json 文件。我正在使用 C++ 和 nlohmann json 库。

我当前的代码

int main(int argc, const char** argv){
    ifstream ifs("Myjson.json");
    json j = json::parse(ifs);
    cout << "Front image path : "<< j["front"]["imagePath"]  << "\n";
    cout << "Back image path : " << j["back"]["imagePath"] << "\n";

    system("PAUSE");
    return 0;
}

MyJson.json

{
    "Side": [
        {
            "camera": "22344506",
            "width": 19860,
            "nbParts": 662,
            "wParts": 30,
            "height": 1600,
            "imagePath": "./Tchek_buffer/22344506.png"
        },
        {
            "camera": "22344509",
            "width": 5296,
            "nbParts": 662,
            "wParts": 8,
            "height": 1600,
            "imagePath": "./Tchek_buffer/22344509.png"
        },
    ],
    "front": {
        "camera": "22344513",
        "image": null,
        "width": 1200,
        "height": 1600,
        "imagePath": "./Tchek_buffer/22344513.png"
    },
    "back": {
        "camera": "22344507",
        "image": null,
        "width": 1600,
        "height": 1200,
        "imagePath": "./Tchek_buffer/22344507.png"
    },
}

我可以轻松读取和显示“背面”和“正面”对象，但无法读取扫描仪对象。我想获取所有“扫描仪”对象的“imagePath”

我试过类似的东西

cout << "scanner image path : " << j["scanner"]["imagePath"] << "\n";
cout << "scanner image path : " << j["scanner[1]"]["imagePath"] << "\n";
cout << "scanner image path : " << j["scanner"[1]]["imagePath"] << "\n";

我只得到“空”结果

如果有人可以帮助我并向我解释如何让它发挥作用。

Answer 1

假设 scanner 实际上是 json 中的 Side。

您的试验执行了以下操作:

• 访问列表的“imagePath”属性
• 访问列表的“scanner[1]”属性
• 访问列表的“c”(第二个字符)属性。

所以肮脏的方法是:

cout << "scanner image path : " << j["Side"][0]["imagePath"] << "\n";
cout << "scanner image path : " << j["Side"][1]["imagePath"] << "\n";

正确的应该是:

for (auto& element : j["Side"])
  cout << "scanner image path : " << element["imagePath"] << "\n";