从数组中的所有对象中删除键 [英] Remove key from all objects in array

问题描述

我有以下对象数组：

[{id:1, value:"100", name:"dog" ...},
{id:2, value:"200", name:"cat" ...},
{id:3, value:"300", name:"fish"....},
{id:4, value:"400", name:"mouse" ...},
{id:5, value:"500", name:"snake"...}]

我想过滤对象数组并只保留两个键， id 和 value 得到这样的结果：

I want to filter the object array and keep only two keys, id and value to get something like this:

[{id:1, value:"100"},
{id:2, value:"200"},
{id:3, value:"300"},
{id:4, value:"400"},
{id:5, value:"500"}]

目前，我正在使用for循环遍历对象数组并执行 push（）带有新变量的空数组。有没有更简单的方法呢？

Currently, I'm traversing through the object array with a for loop and doing a push() to an empty array with the new variables. Is there an easier way to do this?

我想使用Lodash，如 _。pluck（PetList，'id'，'value'） ; 但是lodash仅提取值，而不是密钥。

I wanted to use Lodash like _.pluck(PetList, 'id', 'value'); but lodash extracts the value only, not the key.

推荐答案

数组#filter 过滤出数组中的各个项目，而不是来自数组中对象的某些键。您可以使用 方法Array＃map 用于转换对象并仅保留所需的键。 map 旨在将数组的每个元素转换为新的值，将旧值映射为新值：

Array#filter filters individual items out of the array, not certain keys from an object in the array. You could use the method Array#map to transform the objects and only keeping the keys you want. map is designed to transform each element of an array to something new, "mapping" the old value to a new value:

let newPetList = PetList.map(pet => ({ 
    id: pet.id,
    value: pet.value
}));

上面遍历数组，并将当前对象存储在 pet 。然后它从箭头函数这将是 newPetList 中相应的转换对象，只有 id 和值来自 pet 。结果是旧数组中的所有对象都映射到一个没有 name 键的新对象。

The above traverses the array, and stores the current object in pet. It then returns a new object from an arrow function which will be the corresponding transformed object in newPetList with only keys id and value from pet. The result is that all objects in the old array are mapped to a new object with no name key.

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