如何从对象中删除未定义的属性？

Question

我有这个对象：

  this.prepaidObject = {
            'customerType' : this.prepaidDetailForm.prepaidDetails.customerType,
            'firstName' : this.prepaidDetailForm.prepaidDetails.firstName,
            'lastName' : this.prepaidDetailForm.prepaidDetails.lastName,
            'note' : this.prepaidDetailForm.prepaidDetails.note,
            'created': this.prepaidDetailForm.prepaidDetails.created
        };

现在有时某些属性是未定义的。 我想要的是如果this.prepaidDetailForm.prepaidDetails属性之一未定义不显示它们。 因此，例如，如果this.prepaidDetailForm.prepaidDetails.firsName未定义，我不需要创建'firstName'属性 isside 对象。 任何建议我该怎么做？

Answer 1

检查您的对象：

    for( var m in this.prepaidObject ) {
        if ( this.prepaidObject[m] == undefined ) {
            delete this.prepaidObject[m];
        }
    } 

Answer 2

执行此操作的最短方法是在stringify之后parse它。

确切的语法：

let prepaidObjectNew = JSON.parse(JSON.stringify(prepaidObject))

示例

之前：

var prepaidObject = {
    'customerType' : 'TestCustType',
    'firstName' : "sample FirstName",
    'lastName' : "sample LastName",
    'note' : undefined
 };

后：

{
    'customerType' : 'TestCustType',
    'firstName' : "sample FirstName",
    'lastName' : "sample LastName"
 };

Answer 3

一种方法是遍历键并只附加定义的键：

this.prepaidObject = { };
Object.keys(this.prepaidDetailForm.prepaidDetails)
     .forEach(function(key) {
           var val = this.prepaidDetailForm.prepaidDetails[key];
           if (val !== undefined) {
                this.prepaidObject[key] = val;
           }
     });

这是假设在键prepaidObject是一样prepaidDetails和所有的人都跟着你提到的同样的规则。

特别是如果您使用的是 ES6，您可以使用map和reduce使其更加优雅，如下所示：

this.prepaidObject = Object.keys(this.prepaidDetailForm.prepaidDetails)
     .map(key => ({key, val: this.prepaidDetailForm.prepaidDetails[key]}))
     .reduce((obj, {key, val}) => {
          if (val !== undefined) {
               obj[key] = val;
          }
          return obj;
     }, {});

还有一种使用更多 ES6 特性的更简洁的方法：

this.prepaidObject = Object.keys(this.prepaidDetailForm.prepaidDetails)
     .map(key => ({key, val: this.prepaidDetailForm.prepaidDetails[key]}))
     .filter(({_, val}) => val !== undefined)
     .reduce((obj, {key, val}) => Object.assign(obj, { [key]: val }), {});

Answer 4

使用Lodash使它在一行中变得太容易了：

_.pickBy(this.prepaidObject, _.identity);

这将删除所有虚假值

Answer 5

您可能需要检查类型。

(typeof(this.prepaidDetailForm.prepaidDetails.firsName) != "undefined") ? this.prepaidDetailForm.prepaidDetails.firsName : ''

它基本上是：

var firstName = '';
if(typeof(this.prepaidDetailForm.prepaidDetails.firsName) != "undefined")
     firstName = this.prepaidDetailForm.prepaidDetails.firsName;

Answer 6

请尝试以下操作：

 for (var propName in this.prepaidObject) { 
    if (this.prepaidObject[propName] === undefined) {
      delete this.prepaidObject[propName];
    }
  }

并查看以下答案： Remove blank attributes from an Object in Javascript

Answer 7

this.prepaidObject = {};
for (var i in this.prepaidDetailForm.prepaidDetails) {
  if (typeof(this.prepaidDetailForm.prepaidDetails[i]) !== 'undefined') {
    this.prepaidObject[i] = this.prepaidDetailForm.prepaidDetails[i];
  } 
} 

如果您还想排除空值和未定义，请将条件更改为：

if (this.prepaidDetailForm.prepaidDetails[i] != null)

不是双 ==。 因为null == undefined但null !== undefined

Answer 8

假设你有一个像下面这样的对象，

var prepaidObject = {
    'customerType' : 'TestCustType',
    'firstName' : "sample FirstName",
    'lastName' : "sample LastName",
    'note' : undefined
 };

您可以使用 JSON.parse 和 JSON.stringify 方法删除 undefined，

JSON.parse(JSON.stringify(prepaidObject));
console.log(prepaidObject)

输出将是，

{
    'customerType' : 'TestCustType',
    'firstName' : "sample FirstName",
    'lastName' : "sample LastName"
 }