Create object array from same structure

Question

I have the following object array that maintains the same structure:

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',
}]

I would like to represent this in a more readable manner. All properties should remain the same except DisplayName, RealFieldName, and Name. How can I recreate this array without declaring the entire structure for each object?

@TKoL I would like to recreate this variable without showing properties that have the same values for each object. — cfoster5
– cfoster5, Commented Mar 19, 2020 at 17:29
So you want each object, but with only DisplayName, RealFieldName, and Name properties? — TKoL
– TKoL, Commented Mar 19, 2020 at 17:30

mickl · Accepted Answer · 2020-03-19 17:27:47Z

1

You can use spread operator:

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);

answered Mar 19, 2020 at 17:27

mickl

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2 Comments

TKoL Over a year ago

based on what he's said in the comments, I feel your answer would be what he's looking for if you just changed this bit: let result = fieldObjects.map(({DisplayName, RealFieldName, Name, ...rest}) => {DisplayName,RealFieldName,Name});. Those are the fields he seems to want to keep

mickl Over a year ago

@TKoL not sure either "without showing properties that have the same values for each object. " means those should be skipped but either way spread would be the way to go

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