Question

我是ES6的新手，我有一系列对象，如下所示：

locations: [
  {
    "is_moving": true,
    "uuid": "82fa9dda-e57b-4b3f-99a0-a1db98ae4a19",
    "timestamp": "2017-08-05T04:48:25.526Z",
    "odometer": 0,
    "sample": true,
    "coords": {
      "latitude": 32.7323862,
      "longitude": -117.1939315,
      "accuracy": 1000,
      "speed": -1,
      "heading": -1,
      "altitude": -1
    },
    "activity": {
      "type": "in_vehicle",
      "confidence": 54
    },
    "battery": {
      "is_charging": false,
      "level": 0.5
    },
    "extras": {}
  },
  {
    "event": "motionchange",
    "is_moving": true,
    "uuid": "57a0146a-28b9-4baf-86de-e037843c2d32",
    "timestamp": "2017-08-05T04:48:25.526Z",
    "odometer": 0,
    "coords": {
      "latitude": 32.7323862,
      "longitude": -117.1939315,
      "accuracy": 1000,
      "speed": -1,
      "heading": -1,
      "altitude": -1
    },
    "activity": {
      "type": "in_vehicle",
      "confidence": 54
    },
    "battery": {
      "is_charging": false,
      "level": 0.5
    },
    "extras": {}
  }
]

我最终想要的是：

locations: [
  {
    "timestamp": "2017-08-05T04:48:25.526Z",
    "odometer": 0,
    "latitude": 32.7323862,
    "longitude": -117.1939315,
    "accuracy": 1000,
    "speed": -1,
    "heading": -1,
    "altitude": -1
  },
  {
    "timestamp": "2017-08-05T04:48:25.526Z",
    "odometer": 0,
    "latitude": 32.7323862,
    "longitude": -117.1939315,
    "accuracy": 1000,
    "speed": -1,
    "heading": -1,
    "altitude": -1
  }
]

所以我知道我可以过滤掉我不想做的键/值对（感谢https://stackoverflow.com/a/39333664/994275）：

locations.map(({ timestamp, odometer, coords }) => ({ timestamp, odometer, coords }))

我知道我可以通过这样做来展平对象（感谢https://stackoverflow.com/a/33037683/994275）：

Object.assign(
  {}, 
  ...function _flatten(location) { 
    return [].concat(...Object.keys(location)
      .map(k => 
        typeof location[k] === 'object' ?
          _flatten(location[k]) : 
          ({[k]: location[k]})
      )
    );
  }(location)
)

但是我试图将两者结合起来并且悲惨地失败。我在地图中添加了flatten，但这只是返回一个undefined数组。

我确定这是一个简单的解决方法，但此时它已经躲过了我。任何帮助将不胜感激！

这里有效（感谢似乎已删除评论的用户）：

let newLocations = locations.map(({ is_moving, uuid, timestamp, odometer, coords }) => ({ is_moving, uuid, timestamp, odometer, coords }));
let test = newLocations.map((location) => {
  return Object.assign(
    {}, 
    ...function _flatten(location) { 
      return [].concat(...Object.keys(location)
        .map(k => 
          typeof location[k] === 'object' ?
            _flatten(location[k]) : 
            ({[k]: location[k]})
        )
      );
    }(location)
  )
});

有没有办法压缩过滤和展平？

Answer 1

function body of an arrow function有两种语法：一种带有“块体”，另一种带有“简洁体”。当您使用“块体”时，您必须使用花括号{ }将所包含的语句包围在块中，如果不使用，则需要显式的return语句想要返回undefined。

您可以将两个操作合并为一个map并使用“简洁的正文”语法：

let newLocations = locations.map(({ is_moving, uuid, timestamp, odometer, coords }) =>
  Object.assign(
    {}, 
    ...function _flatten(location) { 
      return [].concat(...Object.keys(location)
        .map(k => 
          typeof location[k] === 'object' ?
            _flatten(location[k]) : 
            ({[k]: location[k]})
        )
      );
    }({ is_moving, uuid, timestamp, odometer, coords })
  )
);

尝试使用ES6过滤和展平嵌套对象数组

1 个答案: