如何在不合并对象属性的情况下推送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。如果我控制台.log（arr2），我得到了我想要的结果，但在多个数组中。我想把所有这些东西都放在一个数组中。当我尝试使用 .push 时，所有数组都会产生相同的值。也许我可以以其他方式做到这一点，但我不知道该怎么做。    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);这就是我用.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 }]当我控制台.log（arr2）在for循环中时，我得到了几乎我想要的东西，想要的结果，但不是在单个数组中。   [  { 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 }]

Answer 1

这是因为对象通过引用工作。这应该有效： 替换这个：

 arr2 = arr1.map(function(record){

   record.value1 = b.value1;

   record.value3 = b.value3;

   record.value4 = b.value4;

   return record;

   })

通过您的代码中的这个：

 arr2 = arr1.map(function(record) {

const tempRecord = {...record}

tempRecord.value1 = b.value1;

tempRecord.value3 = b.value3;

tempRecord.value4 = b.value4;

return tempRecord;

})

你可以通过这样做来完成最终结果：arr3.flat（）;