将数组中的变量，拿出来生成一个对象？

Question

问题描述

**js 替换，将

数组他替换成对象
 [
   {
     name: "aaa"
     sqlVal: 134
   },
   {
     name: "vvv"
     sqlVal: 567
   }
 ]
 {
   aaa: 134,
   vvv:567
 }

如何替换？

**

问题出现的环境背景及自己尝试过哪些方法

相关代码

你期待的结果是什么？实际看到的错误信息又是什么？

Answer 1

const toObject = (items, keyField, valueField) => 
    Object.fromEntries(items.map(item => [item[keyField], item[valueField]]));

// toObject([{name: 'aaa', sqlVal: 134}], 'name', 'sqlVal');

Answer 2

let array = [
  {
    name: "aaa",
     sqlVal: 134
  },
  {
    name: "vvv",
     sqlVal: 567
  }
]

let map = new Map()
array.forEach(item => {
  map.set(item.name, item.sqlVal)
})

let obj = Object.fromEntries(map)

console.log(obj)

Answer 3

var arr = [
   {
     name: "aaa",
     sqlVal: 134
   },
   {
     name: "vvv",
     sqlVal: 567
   }
 ]
function getObj(arr) {
    var obj = {}
    arr.forEach(item => {
        obj[item.name] = item.sqlVal
    })
    return obj
}
console.log(getObj(arr))

Answer 4

const arr = [
  { name: "aaa", sqlVal: 134 },
  { name: "vvv", sqlVal: 567 }
];

const obj = arr.reduce((acc, curr) => {
  acc[curr.name] = curr.sqlVal;
  return acc;
}, {});

console.log(obj); // { aaa: 134, vvv: 567 }

首先，我们定义一个空对象作为reduce()方法的初始值。然后，使用reduce()方法遍历数组中的每个元素，并将每个元素的name属性作为对象的键，将sqlVal属性作为对象的值，最终返回一个新的对象。

Answer 5

const arr =  [
  { name: "aaa", sqlVal: 134 },
  { name: "vvv", sqlVal: 567 }
]

const result = arr.reduce((prev, curr) => {
  prev[curr.name] = curr.sqlVal
  return prev
}, {})
console.log(result);

Answer 6

const arr = [
    { name: "aaa", sqlVal: 134 },
    { name: "vvv", sqlVal: 567 }
];

const obj = Object.fromEntries(arr.map(({ name, sqlVal }) => [name, sqlVal]));
console.log(obj);