What would be the right way to present business objects in the front end?
There is a backend that returns list of persons.
It gets automatically mapped into collection of TypeScript busness object class (Person) objects using Angular/rxjs.
export class Person {
Id: string;
Name: string;
Age: number;
}
There is a need to display these entities in a list with an additional column, stating whether records are selected or not - required for further processing.
The question is - what would be the right way to do this?
Should I create a PersonModel
class that inherits from Person
and add additional field in there?
export class PersonModel extends Person {
Selected: boolean;
}
Or, should I ignore the fact that back-end has nothing to do with the Selected
property and simply add the property to the Person
class?
export class Person {
Id: string;
Name: string;
Age: number;
Selected: boolean;
}
Are there other, more proper ways of dealing with such situation?
angular typescript architecture rxjs angular6
add a comment |
There is a backend that returns list of persons.
It gets automatically mapped into collection of TypeScript busness object class (Person) objects using Angular/rxjs.
export class Person {
Id: string;
Name: string;
Age: number;
}
There is a need to display these entities in a list with an additional column, stating whether records are selected or not - required for further processing.
The question is - what would be the right way to do this?
Should I create a PersonModel
class that inherits from Person
and add additional field in there?
export class PersonModel extends Person {
Selected: boolean;
}
Or, should I ignore the fact that back-end has nothing to do with the Selected
property and simply add the property to the Person
class?
export class Person {
Id: string;
Name: string;
Age: number;
Selected: boolean;
}
Are there other, more proper ways of dealing with such situation?
angular typescript architecture rxjs angular6
1
Composition, perhaps, an object like{ person: Person, selected: boolean }
.
– jonrsharpe
Nov 22 '18 at 17:26
1
Create a PersonRow class/interface with a field person of type Person and a field selected of type boolean?
– JB Nizet
Nov 22 '18 at 17:26
add a comment |
There is a backend that returns list of persons.
It gets automatically mapped into collection of TypeScript busness object class (Person) objects using Angular/rxjs.
export class Person {
Id: string;
Name: string;
Age: number;
}
There is a need to display these entities in a list with an additional column, stating whether records are selected or not - required for further processing.
The question is - what would be the right way to do this?
Should I create a PersonModel
class that inherits from Person
and add additional field in there?
export class PersonModel extends Person {
Selected: boolean;
}
Or, should I ignore the fact that back-end has nothing to do with the Selected
property and simply add the property to the Person
class?
export class Person {
Id: string;
Name: string;
Age: number;
Selected: boolean;
}
Are there other, more proper ways of dealing with such situation?
angular typescript architecture rxjs angular6
There is a backend that returns list of persons.
It gets automatically mapped into collection of TypeScript busness object class (Person) objects using Angular/rxjs.
export class Person {
Id: string;
Name: string;
Age: number;
}
There is a need to display these entities in a list with an additional column, stating whether records are selected or not - required for further processing.
The question is - what would be the right way to do this?
Should I create a PersonModel
class that inherits from Person
and add additional field in there?
export class PersonModel extends Person {
Selected: boolean;
}
Or, should I ignore the fact that back-end has nothing to do with the Selected
property and simply add the property to the Person
class?
export class Person {
Id: string;
Name: string;
Age: number;
Selected: boolean;
}
Are there other, more proper ways of dealing with such situation?
angular typescript architecture rxjs angular6
angular typescript architecture rxjs angular6
asked Nov 22 '18 at 17:21
PatrykPatryk
1,11882967
1,11882967
1
Composition, perhaps, an object like{ person: Person, selected: boolean }
.
– jonrsharpe
Nov 22 '18 at 17:26
1
Create a PersonRow class/interface with a field person of type Person and a field selected of type boolean?
– JB Nizet
Nov 22 '18 at 17:26
add a comment |
1
Composition, perhaps, an object like{ person: Person, selected: boolean }
.
– jonrsharpe
Nov 22 '18 at 17:26
1
Create a PersonRow class/interface with a field person of type Person and a field selected of type boolean?
– JB Nizet
Nov 22 '18 at 17:26
1
1
Composition, perhaps, an object like
{ person: Person, selected: boolean }
.– jonrsharpe
Nov 22 '18 at 17:26
Composition, perhaps, an object like
{ person: Person, selected: boolean }
.– jonrsharpe
Nov 22 '18 at 17:26
1
1
Create a PersonRow class/interface with a field person of type Person and a field selected of type boolean?
– JB Nizet
Nov 22 '18 at 17:26
Create a PersonRow class/interface with a field person of type Person and a field selected of type boolean?
– JB Nizet
Nov 22 '18 at 17:26
add a comment |
1 Answer
1
active
oldest
votes
Instead of Class, You can create an interface which extends another interface.
export interface Person {
Id: string;
Name: string;
Age: number;
}
export interface PersonAttrib extends Person {
Selected: boolean;
}
add a comment |
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
Instead of Class, You can create an interface which extends another interface.
export interface Person {
Id: string;
Name: string;
Age: number;
}
export interface PersonAttrib extends Person {
Selected: boolean;
}
add a comment |
Instead of Class, You can create an interface which extends another interface.
export interface Person {
Id: string;
Name: string;
Age: number;
}
export interface PersonAttrib extends Person {
Selected: boolean;
}
add a comment |
Instead of Class, You can create an interface which extends another interface.
export interface Person {
Id: string;
Name: string;
Age: number;
}
export interface PersonAttrib extends Person {
Selected: boolean;
}
Instead of Class, You can create an interface which extends another interface.
export interface Person {
Id: string;
Name: string;
Age: number;
}
export interface PersonAttrib extends Person {
Selected: boolean;
}
answered Nov 22 '18 at 17:37
Suresh Kumar AriyaSuresh Kumar Ariya
4,4761215
4,4761215
add a comment |
add a comment |
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1
Composition, perhaps, an object like
{ person: Person, selected: boolean }
.– jonrsharpe
Nov 22 '18 at 17:26
1
Create a PersonRow class/interface with a field person of type Person and a field selected of type boolean?
– JB Nizet
Nov 22 '18 at 17:26