确定 Javascript 对象中的不完整部分

Question

我有一个具有多个属性的 JavaScript 对象：

{
  banking: {
    account_number: null,
    account_type: null,
    bank_name: null,
    debit_day: null
  },
  fitment: {
    date: null,
    terms: null
  },
  personal_info: {
    email: null,
    IDNumber: null,
    mobile: null,
    name: null,
    residential_address: null,
    surname: null,
    title: null,
    work_address: null,
    work_tel: null
  },
  vehicle: {
    brand: null,
    colour: null,
    model: null,
    registration: null,
    vin: null,
    year: null
  }
}

所有顶级属性和嵌套属性的默认值为

null

.

我正在尝试找出一种将属性分为三组的方法，即：

空，部分和完整。

“empty”是一个部分中的所有值都设置为

null

.

的情况

“partial”是部分中的某些值已设置（并非所有

null

）

“complete”是一个部分中没有任何值设置为

null

。都有价值。

我的第一次尝试是将 Underscore 库与

_.some()

一起使用，但是我似乎无法全神贯注于满足所有组场景。

一些帮助和指导将不胜感激。

谢谢！

Answer 1

您可以先检查

every

，然后再检查

some

，或者默认为

empty

。

const main = () => {
  const sectionStatus = Object.fromEntries(
    Object.entries(sections).map(([section, data]) => [
      section,
      Object.entries(data).every(([id, value]) => value !== null)
        ? 'complete'
        : Object.entries(data).some(([id, value]) => value !== null)
          ? 'partial'
          : 'empty'
    ]));
  
  console.log(sectionStatus);
};

const sections = {
  banking: {
    account_number: '555',
    account_type: null,
    bank_name: null,
    debit_day: null
  },
  fitment: {
    date: new Date(),
    terms: 'foobar'
  },
  personal_info: {
    email: null,
    IDNumber: null,
    mobile: null,
    name: null,
    residential_address: null,
    surname: null,
    title: null,
    work_address: null,
    work_tel: null
  },
  vehicle: {
    brand: null,
    colour: null,
    model: null,
    registration: null,
    vin: null,
    year: null
  }
};

main();

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

输出

{
  "banking": "partial",
  "fitment": "complete",
  "personal_info": "empty",
  "vehicle": "empty"
}

如果你想优化这个，你可以用

1

（不是

null

）或

0

（

null

）标记每个字段并减少值。

const main = () => {
  const sectionStatus = Object.fromEntries(
    Object.entries(sections).map(([section, data]) => { 
      const flags = Object.values(data).map((v) => v !== null ? 1 : 0);
      const state = flags.reduce((a, b) => a & b, 1) === 1 // or (a * b)
        ? 'complete'
        : flags.reduce((a, b) => a | b, 0) === 1 // or (a + b) > 0
          ? 'partial'
          : 'empty';
      return [section, state];
    }, []));
  
  console.log(sectionStatus);
};

const sections = {
  banking: {
    account_number: '555',
    account_type: null,
    bank_name: null,
    debit_day: null
  },
  fitment: {
    date: new Date(),
    terms: 'foobar'
  },
  personal_info: {
    email: null,
    IDNumber: null,
    mobile: null,
    name: null,
    residential_address: null,
    surname: null,
    title: null,
    work_address: null,
    work_tel: null
  },
  vehicle: {
    brand: null,
    colour: null,
    model: null,
    registration: null,
    vin: null,
    year: null
  }
};

main();

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

Answer 2

您还可以使用

_.groupBy

进一步分解这个问题。首先，一个函数只是为了确定单个部分的类别：

function category(section) {
    return _.every(section, _.isNull) ? 'empty' :
           _.some(section, _.isNull) ? 'partial' : 'complete';
}

然后，结合

_.groupBy

为每个类别创建一个原始对象数组：

_.groupBy(sections, category)

把它们放在一起：

var sections = {
  banking: {
    account_number: null,
    account_type: null,
    bank_name: null,
    debit_day: null
  },
  fitment: {
    date: null,
    terms: null
  },
  personal_info: {
    email: null,
    IDNumber: null,
    mobile: null,
    name: 'Jane Doe',
    residential_address: null,
    surname: null,
    title: null,
    work_address: null,
    work_tel: null
  },
  vehicle: {
    brand: null,
    colour: null,
    model: null,
    registration: null,
    vin: null,
    year: null
  }
};

function category(section) {
    return _.every(section, _.isNull) ? 'empty' :
           _.some(section, _.isNull) ? 'partial' : 'complete';
}

console.log(_.groupBy(sections, category));

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