Question

我有这个对象

Object {
  "87291": "valid",
  "1873681927": "valid",
  "nwq89j8jw0qj9": "valid",
  "oVFYfWIUOsONE6JyMGYAbnsPMAr1": "valid",
}

我要做的是根据键值删除对象并以数组形式返回。

我使用了以下代码：

    let attay = {
      "87291": "valid",
      "1873681927": "valid",
      "nwq89j8jw0qj9": "valid",
      "oVFYfWIUOsONE6JyMGYAbnsPMAr1": "valid",
    }


    let aar = Object.entries(attay)
    attay = Object.keys(attay)

    for(var i = 0; i < attay.length; i++) {
      if(attay[i] == 'oVFYfWIUOsONE6JyMGYAbnsPMAr1'){
        console.log("found "+attay[i]+" at "+i)
        aar.splice(i, 1);
        console.log(aar)
      }else{
        console.log("NOT found at "+i)        
      }
    }

这段代码可以正常工作，但是remove元素之后的输出是这样的

Array [
  Array [
    "87291",
    "valid",
  ],
  Array [
    "1873681927",
    "valid",
  ],
  Array [
    "nwq89j8jw0qj9",
    "valid",
  ],
]

预期输出应类似于

Array [
      "87291": "valid",
      "1873681927": "valid",
      "nwq89j8jw0qj9": "valid",
    ]

Answer 1

也许我误会了，但似乎您对此过于复杂了。您要从对象中删除属性吗？

let arr = {
  "87291": "valid",
  "1873681927": "valid",
  "nwq89j8jw0qj9": "valid",
  "oVFYfWIUOsONE6JyMGYAbnsPMAr1": "valid",
}

let removeKey = "oVFYfWIUOsONE6JyMGYAbnsPMAr1"
delete arr[removeKey]; // Remove the property

console.log(arr);

Answer 2

如果相对较新的entries（和fromEntries）的运行时还可以，那么您可以轻松获得所需的内容。（猜测您实际上是针对对象输出，而不是基于问题语法的数组）。

let attay = {
  "87291": "valid",
  "1873681927": "valid",
  "nwq89j8jw0qj9": "valid",
  "oVFYfWIUOsONE6JyMGYAbnsPMAr1": "valid",
}

let removeKey = "oVFYfWIUOsONE6JyMGYAbnsPMAr1"

let pairs = Object.entries(attay).filter(e => e[0] != removeKey)
const obj = Object.fromEntries(pairs);
console.log(obj)

Answer 3

答案来自@ Pointy

delete attay.oVFYfWIUOsONE6JyMGYAbnsPMAr1

简单而且像魅力一样工作，还可以在React Native上工作

Answer 4

如果密钥与适当性匹配，则可以这样使用delete：

const obj = {
  "87291": "valid",
  "1873681927": "valid",
  "nwq89j8jw0qj9": "valid",
  "oVFYfWIUOsONE6JyMGYAbnsPMAr1": "valid",
}

const key = "87291";

delete obj[key]



console.log(obj)

Answer 5

您可以使用简单的fromEntries来避免受较少支持的reduce。

let attay = {
  "87291": "valid",
  "1873681927": "valid",
  "nwq89j8jw0qj9": "valid",
  "oVFYfWIUOsONE6JyMGYAbnsPMAr1": "valid",
};
const remove = "oVFYfWIUOsONE6JyMGYAbnsPMAr1";
const res = Object.entries(attay).reduce((acc, [k, v]) => k == remove ? acc : (acc[k] = v, acc), {});
console.log(res);

Javascript对象数组然后删除元素

5 个答案: