如何删除值和键而不影响原始数组[[副本]

"..." 深度复制

let downloadData = [...this.listData.filteredData];

let downloadDataNum = downloadData.length;

for( let i = 0; i < downloadDataNum; i++ ) {
  delete downloadData[i].id;
  delete downloadData[i].createdBy;
  delete downloadData[i].created_at;
  delete downloadData[i].updatedBy;
  delete downloadData[i].updated_at;
}

让我知道它是否有效。

 let downloadData = [...this.listData.filteredData];

for( let i = 0; i < downloadData.length; i++ ) {
  delete downloadData[i].id;
  delete downloadData[i].createdBy;
  delete downloadData[i].created_at;
  delete downloadData[i].updatedBy;
  delete downloadData[i].updated_at;
}

你可以试试这个。。。使用spread操作符,您可以将数据分散到downloadData变量,而无需改变listData

你必须复制原始数据。

您可以通过以下几种方式实现:

// Way 1
let downloadData = Array.from(this.listData.filteredData);

// Way 2
let downloadData = JSON.parse(JSON.stringify(this.listData.filteredData));

// Way 3 (using spread operator)
let downloadData = [...this.listData.filteredData];

// Way 4
let downloadData = this.listData.filteredData.slice(0);

downloadData 不会影响原始数据。

只需在编辑数组之前克隆它,因为数组是通过引用复制的:

let downloadData = this.listData.filteredData;

// Clone the array
let cloneDownloadData = downloadData.slice(0);

let downloadDataNum = cloneDownloadData.length;

for( let i = 0; i < downloadDataNum; i++ ) {
  delete cloneDownloadData[i].id;
  delete cloneDownloadData[i].createdBy;
  delete cloneDownloadData[i].created_at;
  delete cloneDownloadData[i].updatedBy;
  delete dcloneDownloadData[i].updated_at;
}

您应该使用 JSON.stringify() JSON.parse()

let downloadData = JSON.parse(JSON.stringify(this.listData.filteredData));

此外,还可以使用键数组来简化代码 forEach

let keys = ['id','createdBy','created_at','updatedBy','updated_at'];
downloadDataNum.forEach(x => {
     keys.forEach(key => delete x[key]);
})

使用 map() 和 reduce()

你可以用 地图() 并返回包含所需属性的新数组

let keys = ['id','createdBy','created_at','updatedBy','updated_at'];
let downloadData = this.listData.filteredData.map(x => {
     return keys.reduce((ac,a) => (ac[a] = x[a],ac),{})
});