将 json 扁平化为 csv 格式

我正在尝试根据用户选择的字段将 json 值转换为平面 csv。我的 json 看起来像

var data = {
"_index": "test",
"_type": "news",
"_source": {
    "partnerName": "propertyFile 9",
    "relatedSources": "null",
    "entityCount": "50",
    "Categories": {
        "Types": {
            "Events": [{
                "count": 1,
                "term": "Time",
                "Time": [{
                    "term": "Dec 9",
                    "Dec_9": [{
                        "count": 1,
                        "term": "2012"
                    }]
                    }]
                }, {
                "count": 4,
                "term": "News",
                "News": [{
                    "term": "Germany",
                    "Germany": [{
                        "count": 1,
                        "term": "Election"
                    }],
                    "currency": "Euro (EUR)"
                }, {
                    "term": "Egypt",
                    "Egypt": [{
                        "count": 1,
                        "term": "Revolution"
                    }]
                    }]
                }]
            }
    }
}};

我已经能够收集所有出现的值并将其存储为 csv，但我想保存根本身的详细信息。

如果我选择时间，则 csv 输出应如下所示：

"test", "news", "propertyFile 9","null", "50", "Events": "Time", "Dec 9", "2012"

是否可以压平 json..我将添加 json 小提琴链接来显示我已经到达了这个东西..http://jsfiddle.net/JHCwM/

这是将对象扁平化为键/值对的另一种方法，其中键是属性的完整路径。

let data = {
  pc: "Future Crew",
  retro: {
    c64: "Censor Design",
    amiga: "Kefrens"
  }
};

let flatten = (obj, path = []) => {
  return Object.keys(obj).reduce((result, prop) => {
    if (typeof obj[prop] !== "object") {
      result[path.concat(prop).join(".")] = obj[prop];
      return result;
    }
    return Object.assign(result, flatten(obj[prop], path.concat(prop), result));
  }, {});
}

console.log(
  flatten(data)
);