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我想从一个字符串创建一个服务树，这些字符串是人的属性。我设法做到了，但是代码很丑陋，这对我来说是一种实践之王，因为我想使用“Object.create()”来使用“类”对象来制作它，而类会看起来像这样：

let service = {
        serviceFather: "",
        serviceChildren: [],
        people: [],
        }
    };

输入样本是：

[
  {
    "name": "John Doe",
    "service": "EE",
  },
  {
    "name": "Jane Doe",
    "service": "EE.EA",
  },
  {
    "name": "Jack Smith",
    "service": "EE.EA.EB",
  },
  {
    "name": "Jill Smith",
    "service": "EE.EA.EC"
  },
  {
    "name": "Jake Smith",
    "serviceLevel": "EE.EA.EC"
  }
]

预期的输出将是：

[
{
    "name": "EE",
    "serviceFather": "root",
    "people": [
    {
        "name": "John Doe"
    }],
    "serviceChildren": [
    {
        "name": "EA",
        "serviceFather": "EE",
        "people": [
        {
            "name": "Jane Doe"
        }],
        "serviceChildren": [
        {
            "name": "EB",
            "serviceFather": "EA",
            "people": [
            {
                "name": "Jack Smith"
            }],
            "sousService": ""
        },
        {
            "name": "EC",
            "serviceFather": "EA",
            "people": [
            {
                "name": "Jill Smith"
            },
            {
                "name": "Jake Smith"
            }],
            "sousService": ""
        }]
    }]
}]
4

2 回答 2

1

您可以使用spliton 服务属性，然后forEach循环并reduce迭代嵌套树并添加到数组。

const data = [{ "name": "John Doe", "service": "EE" }, { "name": "Jane Doe", "service": "EE.EA" }, { "name": "Jack Smith", "service": "EE.EA.EB" },
  { "name": "Jill Smith", "service": "EE.EA.EC" }, { "name": "Jake Smith", "service": "EE.EA.EC" }
]

let service = {
  serviceFather: "",
  serviceChildren: [],
  people: [],
};

function create(data) {
  const res = []
  data.forEach(obj => {
    obj.service.split('.').reduce((r, e, i, a) => {
      const match = r.find(({ name }) => name == e);
      if(!match) {
        const o = Object.create(service);
        o.name = e;
        o.serviceFather = (i == 0 ? 'root' : a[i - 1])
        o.people = [{ name: obj.name }]
        o.serviceChildren = [];
        r.push(o)
        return r;
      } else {
        if(!a[i + 1]) match.people.push({ name: obj.name })
        return match.serviceChildren
      }
    }, res)
  })
  return res;
}

const result = create(data)
console.log(result)

于 2019-03-21T11:13:49.723 回答
1

您可以获取一个虚拟对象并迭代服务级别。

var data = [{ name: "John Doe", service: "EE" }, { name: "Jane Doe", service: "EE.EA" }, { name: "Jack Smith", service: "EE.EA.EB" }, { name: "Jill Smith", service: "EE.EA.EC" }, { name: "Jake Smith", service: "EE.EA.EC" }],
    result = [];

data.forEach(({ name, service }) => {
    service
        .split('.')
        .reduce((o, k, i, a) => {
            var temp = (o.serviceChildren = o.serviceChildren || []).find(({ name }) => name === k);
            if (!temp) {
                o.serviceChildren.push(temp = {
                    name: k,
                    serviceFather: a[i - 1] || 'root',
                    people: []
                });
            }
            return temp;
        }, { serviceChildren: result })
        .people.push({ name });
});

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

于 2019-03-21T11:14:52.040 回答