Question

我有以下保持相同结构的对象数组：

var fieldObjects = [{
  AllowGridEditing: 'FALSE',
  DisplayName: 'Submit',
  RealFieldName: 'SubmitField',
  Name: 'SubmitField',
  FieldType: 'Text',
  Type: 'Text',
  Filterable: 'FALSE',
  Sortable: 'FALSE',
  ReadOnly: 'TRUE',
}, {
  AllowGridEditing: 'FALSE',
  DisplayName: 'Hours Up',
  RealFieldName: 'HoursUpField',
  Name: 'HoursUpField',
  FieldType: 'Text',
  Type: 'Text',
  Filterable: 'FALSE',
  Sortable: 'FALSE',
  ReadOnly: 'TRUE',
}, {
  AllowGridEditing: 'FALSE',
  DisplayName: 'Personal Hours',
  RealFieldName: 'PersonalHoursField',
  Name: 'PersonalHoursField',
  FieldType: 'Text',
  Type: 'Text',
  Filterable: 'FALSE',
  Sortable: 'FALSE',
  ReadOnly: 'TRUE',
}, {
  AllowGridEditing: 'FALSE',
  DisplayName: 'Hours Down',
  RealFieldName: 'HoursDownField',
  Name: 'HoursDownField',
  FieldType: 'Text',
  Type: 'Text',
  Filterable: 'FALSE',
  Sortable: 'FALSE',
  ReadOnly: 'TRUE',
}]

我想以一种更具可读性的方式表示这一点。除DisplayName，RealFieldName和Name外，所有属性都应保持不变。如何在不声明每个对象的整个结构的情况下重新创建此数组？

Answer 1

您可以使用spread运算符：

let result = fieldObjects.map(({DisplayName, RealFieldName, Name, ...rest}) => rest);

var fieldObjects = [{
  AllowGridEditing: 'FALSE',
  DisplayName: 'Submit',
  RealFieldName: 'SubmitField',
  Name: 'SubmitField',
  FieldType: 'Text',
  Type: 'Text',
  Filterable: 'FALSE',
  Sortable: 'FALSE',
  ReadOnly: 'TRUE',
}, {
  AllowGridEditing: 'FALSE',
  DisplayName: 'Hours Up',
  RealFieldName: 'HoursUpField',
  Name: 'HoursUpField',
  FieldType: 'Text',
  Type: 'Text',
  Filterable: 'FALSE',
  Sortable: 'FALSE',
  ReadOnly: 'TRUE',
}, {
  AllowGridEditing: 'FALSE',
  DisplayName: 'Personal Hours',
  RealFieldName: 'PersonalHoursField',
  Name: 'PersonalHoursField',
  FieldType: 'Text',
  Type: 'Text',
  Filterable: 'FALSE',
  Sortable: 'FALSE',
  ReadOnly: 'TRUE',
}, {
  AllowGridEditing: 'FALSE',
  DisplayName: 'Hours Down',
  RealFieldName: 'HoursDownField',
  Name: 'HoursDownField',
  FieldType: 'Text',
  Type: 'Text',
  Filterable: 'FALSE',
  Sortable: 'FALSE',
  ReadOnly: 'TRUE',
}]

let result = fieldObjects.map(({DisplayName, RealFieldName, Name, ...rest}) => rest);
console.log(result);

从相同的结构创建对象数组

1 个答案: