使用下划线或 lodash 将一种 JSON 结构转换为另一种

Question

我在尝试将当前 JSON 结构转换为另一个时遇到问题

var data = [
  {
    "url": "asset/01.flv",
    "pic": "asset/01.jpg"
  },
  {
    "url": "asset/02.flv",
    "pic": "asset/02.jpg"
  },
  {
    "url": "asset/03.flv",
    "pic": "asset/03.jpg"
  },
  {
    "url": "asset/04.flv|asset/05.flv|asset/06.flv|asset/07.flv|asset/08.flv",
    "pic": "asset/04.jpg|asset/05.jpg|asset/06.jpg|asset/07.jpg|asset/08.jpg"
  },
  {
    "url": "asset/09.flv|asset/10.flv",
    "pic": "asset/09.jpg|asset/10.jpg"
  }
]

我想将数据转换成这样的输出结构。
这是我想要达到的结果的目标。

var data = [
  {
    "url": "asset/01.flv",
    "pic": "asset/01.jpg"
  },
  {
    "url": "asset/02.flv",
    "pic": "asset/02.jpg"
  },
  {
    "url": "asset/03.flv",
    "pic": "asset/03.jpg"
  },
  {
    "url": "asset/04.flv",
    "pic": "asset/04.jpg"
  },
  {
    "url": "asset/05.flv",
    "pic": "asset/05.jpg"
  },
  {
    "url": "asset/06.flv",
    "pic": "asset/06.jpg"
  },
  {
    "url": "asset/07.flv",
    "pic": "asset/07.jpg"
  },
  {
    "url": "asset/08.flv",
    "pic": "asset/08.jpg"
  },
  {
    "url": "asset/09.flv",
    "pic": "asset/09.jpg"
  },
  {
    "url": "asset/10.flv",
    "pic": "asset/10.jpg"
  }
]

这对我来说很难，有人可以在这件事上帮助我吗？我已经尝试了几个小时。使用 lodash 或下划线或纯 JS 是可以的。提前致谢。

Answer 1

您可以使用Array#reduce 来完成此操作。想法是遍历数组并检查 url 和 pic 的值是否包含 |。

如果是这样，拆分值并将所有项目推送到结果。如果没有，请按原样推送url 和pic。

var data = [{"url":"asset/01.flv","pic":"asset/01.jpg"},{"url":"asset/02.flv","pic":"asset/02.jpg"},{"url":"asset/03.flv","pic":"asset/03.jpg"},{"url":"asset/04.flv|asset/05.flv|asset/06.flv|asset/07.flv|asset/08.flv","pic":"asset/04.jpg|asset/05.jpg|asset/06.jpg|asset/07.jpg|asset/08.jpg"},{"url":"asset/09.flv|asset/10.flv","pic":"asset/09.jpg|asset/10.jpg"}];

var result = data.reduce((r, {url, pic}) => {
  if (url.includes('|') && pic.includes('|')) {
    var urls = url.split('|');
    var pics = pic.split('|');
    
    urls.forEach((u, i) => {
      r.push({url: u, pic: pics[i]});
    });
  } else {
    r.push({url, pic})
  }
  
  return r;
}, []);

console.log(result);

Answer 2

您可以使用Array.reduce 自己进行操作。

const arr = [{
    "url": "asset/01.flv",
    "pic": "asset/01.jpg"
  },
  {
    "url": "asset/02.flv",
    "pic": "asset/02.jpg"
  },
  {
    "url": "asset/03.flv",
    "pic": "asset/03.jpg"
  },
  {
    "url": "asset/04.flv|asset/05.flv|asset/06.flv|asset/07.flv|asset/08.flv",
    "pic": "asset/04.jpg|asset/05.jpg|asset/06.jpg|asset/07.jpg|asset/08.jpg"
  },
  {
    "url": "asset/09.flv|asset/10.flv",
    "pic": "asset/09.jpg|asset/10.jpg"
  }
];

console.log(arr.reduce((tmp, {
  pic,
  url,
}) => {
  const urlSplit = url.split('|');

  return [
    ...tmp,

    ...pic.split('|').map((x, xi) => ({
      pic: x,
      url: urlSplit[xi],
    })),
  ];
}, []));

Answer 3

恕我直言，您可以使用 Array#reduce() 执行类似操作：

var data = [{"url":"asset/01.flv","pic":"asset/01.jpg"},{"url":"asset/02.flv","pic":"asset/02.jpg"},{"url":"asset/03.flv","pic":"asset/03.jpg"},{"url":"asset/04.flv|asset/05.flv|asset/06.flv|asset/07.flv|asset/08.flv","pic":"asset/04.jpg|asset/05.jpg|asset/06.jpg|asset/07.jpg|asset/08.jpg"},{"url":"asset/09.flv|asset/10.flv","pic":"asset/09.jpg|asset/10.jpg"}];

var parsedData = data.reduce((accumulator, e) => {
  e['url'].split('|').forEach((ele, i) => {
    accumulator.push({"url":ele, "pic": e['pic'].split('|')[i]})
  })
  return accumulator;
}, [])

console.log(parsedData);

使用Underscore.js 的_.zip() 函数的相同解决方案。

var data = [{"url":"asset/01.flv","pic":"asset/01.jpg"},{"url":"asset/02.flv","pic":"asset/02.jpg"},{"url":"asset/03.flv","pic":"asset/03.jpg"},{"url":"asset/04.flv|asset/05.flv|asset/06.flv|asset/07.flv|asset/08.flv","pic":"asset/04.jpg|asset/05.jpg|asset/06.jpg|asset/07.jpg|asset/08.jpg"},{"url":"asset/09.flv|asset/10.flv","pic":"asset/09.jpg|asset/10.jpg"}];

var parsedData = data.reduce((accumulator, e) => {
  return accumulator.concat(...(_.zip(e['url'].split('|'), e['pic']
             .split('|'))
             .map(([url, pic]) => ({url, pic}))))
}, []);

console.log(parsedData)

<script src="https://underscorejs.org/underscore-min.js"></script>

Answer 4

将flatmap 与zipWith 一起使用怎么样？？

const sortedData = _.flatMap(data, obj =>
    _.zipWith(
        obj.url.split("|"),
        obj.pic.split("|"),
        (url, pic) => ({url, pic})
    )
);

var data = [
  {
    "url": "asset/01.flv",
    "pic": "asset/01.jpg"
  },
  {
    "url": "asset/02.flv",
    "pic": "asset/02.jpg"
  },
  {
    "url": "asset/03.flv",
    "pic": "asset/03.jpg"
  },
  {
    "url": "asset/04.flv|asset/05.flv|asset/06.flv|asset/07.flv|asset/08.flv",
    "pic": "asset/04.jpg|asset/05.jpg|asset/06.jpg|asset/07.jpg|asset/08.jpg"
  },
  {
    "url": "asset/09.flv|asset/10.flv",
    "pic": "asset/09.jpg|asset/10.jpg"
  }
]

var res = _.flatMap(data, o=>_.zipWith(o.url.split("|"), o.pic.split("|"), (url, pic) => ({url, pic})));

console.log(res);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>