2

I receive a JSON result message in the following format from an old database query that I do not have the ability to change at this time:

{
  "vsm1": "2429",
  "vsm2": "2488",
  "vsm3": "1968",
  "vsm4": "",
  "vsm5": "",
  "vsm6": "",
  "vsm7": "",
  "vsm8": "",
  "vsm9": "",
  "vsm10": "",
  "color1": "5",
  "color2": "4",
  "color3": "4",
  "color4": "0",
  "color5": "0",
  "color6": "0",
  "color7": "0",
  "color8": "0",
  "color9": "0", 
  "color10": "0",
  "p1mtime": "1549296004",
  "p2mtime": "1549296009",
  "p3mtime": "1549296014",
  "p4mtime": "",
  "p5mtime": "",
  "p6mtime": "",
  "p7mtime": "",
  "p8mtime": "",
  "p9mtime": "",
  "p10mtime": "",
  "inch1": "",
  "inch2": "",
  "inch3": "",
  "inch4": "",
  "inch5": "",
  "inch6": "",
  "inch7": "",
  "inch8": "",
  "inch9": "",
  "inch10": ""

}

I would like to re-format it to a more useable object, like so: 

{ id: 1, vsm: 2429, color: 5, pmtime: 1549296004, inch: 0  }
{ id: 2, vsm: 2488, color: 4, pmtime: 1549296009, inch: 0  }
{ id: 3, vsm: 1968, color: 4, pmtime: 1549296014, inch: 0  }

...and so on.

The incoming data is currently limited to ten of each 'section' (vsm1, vsm2, ...vsm10, color1, color2, ...color10, etc.), so a static loop of some sort over the ten elements in each section is how i started, but seemed rather ugly and certainly not flexible.

A smart snippet that would handle n-number of elements in each section would be even better just in case the data goes beyond ten elements, or drops to just three (due to absence of data or pruned data).

I'm thinking of something along the lines of using .forEach(), but admittedly my JSON / Object manipulation skills are rather poor, so I turn to the community in the hope that someone can point me in the right direction or knows of a cool, tight routine/function that achieves what I'm looking for. Thanks in advance for any insights.

4

4 回答 4

6

您可以获取所需键的数组，并使用占位符作为运行编号并构建新对象并将它们推送到结果集。

var data = { vsm1: "2429", vsm2: "2488", vsm3: "1968", vsm4: "", vsm5: "", vsm6: "", vsm7: "", vsm8: "", vsm9: "", vsm10: "", color1: "5", color2: "4", color3: "4", color4: "0", color5: "0", color6: "0", color7: "0", color8: "0", color9: "0", color10: "0", p1mtime: "1549296004", p2mtime: "1549296009", p3mtime: "1549296014", p4mtime: "", p5mtime: "", p6mtime: "", p7mtime: "", p8mtime: "", p9mtime: "", p10mtime: "", inch1: "", inch2: "", inch3: "", inch4: "", inch5: "", inch6: "", inch7: "", inch8: "", inch9: "", inch10: "" },
    keys = ['vsm*', 'color*', 'p*mtime', 'inch*'],
    result = [],
    id = 1;

while (keys[0].replace('*', id) in data) {
    result.push(Object.assign(
        { id },
        ...keys.map(k => ({ [k.replace('*', '')]: +data[k.replace('*', id)]  || 0 }))
    ));
    id++;
}

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

使用模板文字

var data = { vsm1: "2429", vsm2: "2488", vsm3: "1968", vsm4: "", vsm5: "", vsm6: "", vsm7: "", vsm8: "", vsm9: "", vsm10: "", color1: "5", color2: "4", color3: "4", color4: "0", color5: "0", color6: "0", color7: "0", color8: "0", color9: "0", color10: "0", p1mtime: "1549296004", p2mtime: "1549296009", p3mtime: "1549296014", p4mtime: "", p5mtime: "", p6mtime: "", p7mtime: "", p8mtime: "", p9mtime: "", p10mtime: "", inch1: "", inch2: "", inch3: "", inch4: "", inch5: "", inch6: "", inch7: "", inch8: "", inch9: "", inch10: "" },
    templates = [id => `vsm${id}`, id => `color${id}`, id => `p${id}mtime`, id => `inch${id}`],
    result = [],
    id = 1;

while (templates[0](id) in data) {
    result.push(Object.assign(
        { id },
        ...templates.map(t => ({ [t('')]: +data[t(id)]  || 0 }))
    ));
    id++;
}

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

于 2019-02-04T15:31:08.763 回答
2

试试这个，用 oldObject 你想清理的对象：

var cleanedObject = {};
for (let [key, value] of Object.entries(oldObject)) {
  let index = key.match('[0-9]+');
  cleanedObject[index] = cleanedObject[index] || {};
  cleanedObject[index][key.replace(index, '')] = value;
}

结果将是一个对象 where cleanedObject['1'] = { vsm: 2429, color: 5, pmtime: 1549296004, inch: '' }，依此类推。

于 2019-02-04T15:39:38.330 回答
1

我现在看到我的答案与 Ninas 基本相同，以前没有见过模板，所以这很酷，但是看到这是我第一次尝试在这里回答问题，无论如何我都会分享它。

作为 Ninas，这可以处理任何长度的数据。

const data = {"vsm1": "2429",
  "vsm2": "2488",
  "vsm3": "1968",
  "vsm4": "",
  "color1": "5",
  "color2": "4",
  "color3": "4",
  "color4": "0",
  "p1mtime": "1549296004",
  "p2mtime": "1549296009",
  "p3mtime": "1549296014",
  "p4mtime": "",
  "inch1": "",
  "inch2": "",
  "inch3": "",
  "inch4": "",
  };


const vsmRegex = new RegExp("(vsm\\d)");
const keys = Object.keys(data);

const result = [];
let id= 1;

for(let i = 0; i < keys.length; i++) {
  if(keys[i].match(vsmRegex)) {
    let object = {
      id: id,
      vsm: Number(data[`vsm${id}`]) || 0,
      color: Number(data[`color${id}`]) || 0,
      pmtime: Number(data[`p${id}mtime`]) || 0,
      inch: Number(data[`inch${id}`]) || 0
    };
    result.push(object);
    id++;
  } else {
    break;
  }
}

console.log(result);

于 2019-02-05T07:38:21.973 回答
1

该解决方案与 Nina Sholz 的解决方案具有不同的灵活性。Nina's 允许您匹配任何样式的包含数字的键。但它也需要您添加一个模板才能这样做。我的将处理仅包含单个数字但没有更复杂的任何键。但它不需要你做任何事情来处理这样的模板。

const reformat = data => Object.values(Object.keys(data)
  .reduce(
    (a, k, i, _, d = k.match(/\d+/)[0])  => ({
      ...a, 
      [d]: {...(a[d] || {id: Number(d)}), [k.replace(/\d+/, '')]: data[k]}
    }), {})).sort((a, b) => a.id - b.id)

const data = {"vsm1":"2429","vsm2":"2488","vsm3":"1968","vsm4":"","vsm5":"","vsm6":"","vsm7":"","vsm8":"","vsm9":"","vsm10":"","color1":"5","color2":"4","color3":"4","color4":"0","color5":"0","color6":"0","color7":"0","color8":"0","color9":"0","color10":"0","p1mtime":"1549296004","p2mtime":"1549296009","p3mtime":"1549296014","p4mtime":"","p5mtime":"","p6mtime":"","p7mtime":"","p8mtime":"","p9mtime":"","p10mtime":"","inch1":"","inch2":"","inch3":"","inch4":"","inch5":"","inch6":"","inch7":"","inch8":"","inch9":"","inch10":""}

console.log(reformat(data))

我不知道你是否需要任何一种灵活性，但这些都是有趣的替代方案。

于 2019-02-04T17:02:00.883 回答