Question

我有下面提到的一系列对象

Flag =[{type: "folder", name: "Real Estate", order: "1", delete: "false", canModify: "true"}
 {type: "folder", name: "Legal", order: "2", delete: "false", canModify: "false"}
 {type: "folder", name: "Construction", order: "3", delete: "false", canModify: "true"}
 {type: "folder", name: "Design", order: "4", delete: "false", canModify: "true"}];

我要从输出中删除名称，顺序的数组

Flag =[{type: "folder",  delete: "false", canModify: "true"}
 {type: "folder",   delete: "false", canModify: "false"}
 {type: "folder",  delete: "false", canModify: "true"}
 {type: "folder",  delete: "false", canModify: "true"}];

array.splice（）或 array.pop（）

Answer 1

我不确定如何使用splice和pop，我会使用rest运算符删除属性：

const Flag = [{
  type: "folder",
  name: "Real Estate",
  order: "1",
  delete: "false",
  canModify: "true"
}, {
  type: "folder",
  name: "Legal",
  order: "2",
  delete: "false",
  canModify: "false"
}, {
  type: "folder",
  name: "Construction",
  order: "3",
  delete: "false",
  canModify: "true"
}, {
  type: "folder",
  name: "Design",
  order: "4",
  delete: "false",
  canModify: "true"
}, ];

const result = Flag.map(x => {
  const {
    name,
    order,
    ...rest,
  } = x;

  return rest;
});

console.dir(result);

Answer 2

更好的方法是使用delete运算符将属性删除到对象中

Flag =[{type: "folder", name: "Real Estate", order: "1", delete: "false", canModify: "true"},
 {type: "folder", name: "Legal", order: "2", delete: "false", canModify: "false"},
 {type: "folder", name: "Construction", order: "3", delete: "false", canModify: "true"},
 {type: "folder", name: "Design", order: "4", delete: "false", canModify: "true"}];
 
 Flag.forEach( obj => { 
   delete obj.name; 
   delete obj.order; 
 });
 
 console.log(Flag)

array.splice()或array.pop()会将对象删除到您的数组中，而不是属性中

Answer 3

您可以使用delete关键字删除对象的属性

let Flag =[{type: "folder", name: "Real Estate", order: "1", delete: "false", canModify: "true"},
 {type: "folder", name: "Legal", order: "2", delete: "false", canModify: "false"},
 {type: "folder", name: "Construction", order: "3", delete: "false", canModify: "true"},
 {type: "folder", name: "Design", order: "4", delete: "false", canModify: "true"}];
 
Flag.forEach(obj => {
  delete obj.name;
  delete obj.order;
})
console.log(Flag);

Answer 4

使用删除

Flag = [{
    type: "folder",
    name: "Real Estate",
    order: "1",
    delete: "false",
    canModify: "true"
  },
  {
    type: "folder",
    name: "Legal",
    order: "2",
    delete: "false",
    canModify: "false"
  },
  {
    type: "folder",
    name: "Construction",
    order: "3",
    delete: "false",
    canModify: "true"
  },
  {
    type: "folder",
    name: "Design",
    order: "4",
    delete: "false",
    canModify: "true"
  }
];

function del() {
  for (i = 0; i < Flag.length; i++) {
    if (Flag[i].name) delete Flag[i].name;
    if (Flag[i].order) delete Flag[i].order;
  }
  return Flag;
}
console.log(del());

Answer 5

方法1

Flag = [{
    type: "folder",
    name: "Real Estate",
    order: "1",
    delete: "false",
    canModify: "true"
  },
  {
    type: "folder",
    name: "Legal",
    order: "2",
    delete: "false",
    canModify: "false"
  },
  {
    type: "folder",
    name: "Construction",
    order: "3",
    delete: "false",
    canModify: "true"
  },
  {
    type: "folder",
    name: "Design",
    order: "4",
    delete: "false",
    canModify: "true"
  }
];
var removeKeys = ['name', 'order'];
myArray = Flag.filter(function(obj) {
  removeKeys.forEach(e => delete obj[e]);
  return obj;
});
console.log(myArray);

方法2

Flag = [{
    type: "folder",
    name: "Real Estate",
    order: "1",
    delete: "false",
    canModify: "true"
  },
  {
    type: "folder",
    name: "Legal",
    order: "2",
    delete: "false",
    canModify: "false"
  },
  {
    type: "folder",
    name: "Construction",
    order: "3",
    delete: "false",
    canModify: "true"
  },
  {
    type: "folder",
    name: "Design",
    order: "4",
    delete: "false",
    canModify: "true"
  }
];
Flag = Flag.filter(function(obj) {
  delete obj.name;
  delete obj.order
  return obj;
});
console.log(Flag);

从对象数组中删除属性

5 个答案: