如何從對象中刪除未定義的屬性？

Question

我有這樣的對象：

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現在有時會出現一些此酒店是不確定的。我想要的是如果其中一個this.prepaidDetailForm.prepaidDetails屬性未定義不顯示它們。因此，例如，如果this.prepaidDetailForm.prepaidDetails.firsName未定義，我不需要創建'firstName'屬性isnide對象。任何建議我該怎麼做？

Answer 1

檢查對象：

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

Answer 2

您可能需要檢查類型。

(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 3

的方法之一是要經過鍵，只追加定義的：

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

這是假設中的密鑰與prepaidDetails相同，它們都遵循您提到的相同規則。使用更多的功能ES6

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; 
    }, {}); 

而且一個更簡潔的方法：

特別是如果你使用ES6你可能會使用map和reduce像這樣使這更優雅

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

請請嘗試以下方法：

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

And lo確定在以下答案： Remove blank attributes from an Object in Javascript

Answer 5

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

如果您還想排除空值和未定義的條件更改爲：

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

不是雙==。因爲null == undefined但null !== undefined

Answer 6

假設你有一個像下面的一個對象，

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

可以使用JSON.parse和JSON除去不確定的。字符串化方法，

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

輸出將是，

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

Answer 7

使用Lodash使它很容易在同一行：

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

這將刪除所有falsey值