Question

我有这样的对象数组

<?php

namespace App\Normalizer;

use Symfony\Component\Serializer\Normalizer\NormalizerInterface;
use Symfony\Component\Serializer\Normalizer\ObjectNormalizer;

use App\Entity\Event;


class EventNormalizer implements NormalizerInterface {

    private $normalizer;

    public function __construct(ObjectNormalizer $normalizer){
        $this->normalizer = $normalizer;
    }

    public function normalize($event, $format = null, array $context = [])
    {
        $data = $this->normalizer->normalize($event, $format, $context);

        return $data;
    }

    public function supportsNormalization($data, $format = null, array $context = [])
    {
        return $data instanceof Event;
    }
}

我要做的就是用newArray = [3,4]的多数量参数来获得这样的结果

[ { id: 2,
    name: 'woda',
    price: 45.98,
    quantity: 1,
    createdAt: 2019-07-23T17:50:43.033Z,
    updatedAt: 2019-07-23T17:50:43.033Z,
    shopId: 2 },
  { id: 3,
    name: 'sham',
    price: 69.98,
    quantity: 1,
    createdAt: 2019-07-23T17:50:43.033Z,
    updatedAt: 2019-07-23T17:50:43.033Z,
    shopId: 2 } ]

Answer 1

如果要通过数组的相同索引处的数字更新数组的每个索引，则简单的for循环非常有用。但是如果mult比数组短

，您就会遇到问题

var arr = [ { id: 2,
    name: 'woda',
    price: 45.98,
    quantity: 1,
    createdAt: '2019-07-23T17:50:43.033Z',
    updatedAt: '2019-07-23T17:50:43.033Z',
    shopId: 2 },
  { id: 3,
    name: 'sham',
    price: 69.98,
    quantity: 1,
    createdAt: '2019-07-23T17:50:43.033Z',
    updatedAt: '2019-07-23T17:50:43.033Z',
    shopId: 2 } ];

var mult = [3,4];

for (var i = 0; i<arr.length; i++) {
  arr[i]['quantity'] = mult[i] * arr[i]['quantity'];
}

console.log(arr);

Answer 2

var arr=[ { id: 2,
    name: 'woda',
    price: 45.98,
    quantity: 1,
    createdAt: "2019-07-23T17:50:43.033Z",
    updatedAt: "2019-07-23T17:50:43.033Z",
    shopId: 2 },
  { id: 3,
    name: 'sham',
    price: 69.98,
    quantity: 1,
    createdAt: "2019-07-23T17:50:43.033Z",
    updatedAt: "2019-07-23T17:50:43.033Z",
    shopId: 2 } ]
    
    function sometingCool(arr,multipleByThisArr){
	var i=0;
	return arr.map(x=>{
		x.quantity=x.quantity*multipleByThisArr[i];
		i++;
		return x; 
	})

}
console.log(sometingCool(arr,[3,4]))

Answer 3

执行此操作有两种选择。首选是使用map运算符：

let fooArray = [ { id: 2,
        name: 'woda',
        price: 45.98,
        quantity: 1,
        createdAt: '2019-07-23T17:50:43.033Z',
        updatedAt: '2019-07-23T17:50:43.033Z',
        shopId: 2 },
      { id: 3,
        name: 'sham',
        price: 69.98,
        quantity: 1,
        createdAt: '2019-07-23T17:50:43.033Z',
        updatedAt: '2019-07-23T17:50:43.033Z',
        shopId: 2 } ];
let newArray = [3,4];

使用destructuring assignment和...rest运算符（当您要使用嵌套对象时，这种方式会很有用）：

fooArray = fooArray.map(({
    quantity,
    ...rest
}, i) => ({ ...rest, quantity : quantity * newArray[i]}));
console.log(fooArray);

或更简单，更易读的map()函数，没有...rest运算符和破坏性分配：

fooArray = fooArray.map((f,i) => {
    f.quantity = f.quantity * newArray[i]
    return f;
});
console.log(fooArray);

，第二种方法是使用foreach方法。正如Cerbrus所建议的那样，因为它在语义上更好，代码更短，只想编辑一个值时更容易理解：

fooArray.forEach((e, i) => {
    e.quantity = e.quantity * newArray[i];
})
console.log(fooArray);

使用什么取决于您。我们只是展示了您可以实现所需的方法。

在对象数组中的其他对象中更改某些对象的参数

3 个答案: