如何在不合并对象属性的情况下推入对象数组以创建新数组？

Question

我正在尝试将循环输出保存到一个数组中。如果我console.log（arr2）我得到了想要的结果，但是得到了多个数组。我想要所有这些放在一个阵列中。当我尝试使用.push时，所有数组的值都相同。

也许我可以用其他方式做到这一点，但是我不知道怎么做。

    let b = {};
    let arr3 = [];
    let arr1 = [{value1: 123, value2: 34, value3: 90, value4: 23, value5: 32},{value1: 10, value2: 09, value3: 54, value4: 32, value5: 311}];
    for (let a = 0; a <=4; a++){
       b = {"value1": a*a, "value3": a+12, "value4": a/2};
       arr2 = arr1.map(function(record){
       record.value1 = b.value1;
       record.value3 = b.value3;
       record.value4 = b.value4;
       return record;
       })
       // console.log(arr2);
       arr3.push(arr2);

}
console.log(arr3);

这是我用.push完成的，这不是我想要的。

  [
  { value1: 0, value2: 34, value3: 12, value4: 0, value5: 32 },
  { value1: 0, value2: 9, value3: 12, value4: 0, value5: 311 }
]
[
  { value1: 1, value2: 34, value3: 13, value4: 0.5, value5: 32 },
  { value1: 1, value2: 9, value3: 13, value4: 0.5, value5: 311 }
]
[
  { value1: 4, value2: 34, value3: 14, value4: 1, value5: 32 },
  { value1: 4, value2: 9, value3: 14, value4: 1, value5: 311 }
]
[
  { value1: 9, value2: 34, value3: 15, value4: 1.5, value5: 32 },
  { value1: 9, value2: 9, value3: 15, value4: 1.5, value5: 311 }
]
[
  { value1: 16, value2: 34, value3: 16, value4: 2, value5: 32 },
  { value1: 16, value2: 9, value3: 16, value4: 2, value5: 311 }
]

当我在for循环内console.log（arr2）时，我几乎得到了想要的结果，想要的结果却不在单个数组中。

   [
  { value1: 0, value2: 34, value3: 12, value4: 0, value5: 32 },
  { value1: 0, value2: 9, value3: 12, value4: 0, value5: 311 }
]
[
  { value1: 1, value2: 34, value3: 13, value4: 0.5, value5: 32 },
  { value1: 1, value2: 9, value3: 13, value4: 0.5, value5: 311 }
]
[
  { value1: 4, value2: 34, value3: 14, value4: 1, value5: 32 },
  { value1: 4, value2: 9, value3: 14, value4: 1, value5: 311 }
]
[
  { value1: 9, value2: 34, value3: 15, value4: 1.5, value5: 32 },
  { value1: 9, value2: 9, value3: 15, value4: 1.5, value5: 311 }
]
[
  { value1: 16, value2: 34, value3: 16, value4: 2, value5: 32 },
  { value1: 16, value2: 9, value3: 16, value4: 2, value5: 311 }
]

预期结果：

[
  { value1: 0, value2: 34, value3: 12, value4: 0, value5: 32 },
  { value1: 0, value2: 9, value3: 12, value4: 0, value5: 311 },
  { value1: 1, value2: 34, value3: 13, value4: 0.5, value5: 32 },
  { value1: 1, value2: 9, value3: 13, value4: 0.5, value5: 311 },
  { value1: 4, value2: 34, value3: 14, value4: 1, value5: 32 },
  { value1: 4, value2: 9, value3: 14, value4: 1, value5: 311 },
  { value1: 9, value2: 34, value3: 15, value4: 1.5, value5: 32 },
  { value1: 9, value2: 9, value3: 15, value4: 1.5, value5: 311 },
  { value1: 16, value2: 34, value3: 16, value4: 2, value5: 32 },
  { value1: 16, value2: 9, value3: 16, value4: 2, value5: 311 }
]

Answer 1

这是因为对象通过引用来工作。这应该工作： 替换为：

 arr2 = arr1.map(function(record){
   record.value1 = b.value1;
   record.value3 = b.value3;
   record.value4 = b.value4;
   return record;
   })

通过此代码：

 arr2 = arr1.map(function(record) {
const tempRecord = {...record}
tempRecord.value1 = b.value1;
tempRecord.value3 = b.value3;
tempRecord.value4 = b.value4;
return tempRecord;
})

您可以通过执行以下操作来获得最终结果： arr3.flat（）;

Answer 2

这是您想要的吗？

let b = {};
let arr3 = [];
let arr1 = [
  { value1: 123, value2: 34, value3: 90, value4: 23, value5: 32 },
  { value1: 10, value2: 09, value3: 54, value4: 32, value5: 311 }
];

let result = [];

for (let a = 0; a <= 4; a++) {
  b = {
    "value1": a^2,
    "value3": a+12,
    "value4": a/2
  };

  arr3.push(b);
}

console.log(arr3);

Answer 3

我认为这是您想要实现的。

  var b = {};
var arr3 = [];
var arr1 = [{value1: 123, value2: 34, value3: 90, value4: 23, value5: 32},{value1: 10, value2: 09, value3: 54, value4: 32, value5: 311}];
for (let a = 0; a <=4; a++){
   b = {"value1": a*a, "value3": a+12, "value4": a/2};
  arr3.push(b);
	}
console.log(arr3);