I receive a JSON object from an AJAX call to a REST server. This object has property names that match my TypeScript class (this is a follow-on to this question).
Wha
This seems to be the most maintainable method: add a constructor that takes as parameter the json structure, and extend the json object. That way you can parse a json structure into the whole application model.
There is no need to create interfaces, or listing properties in constructor.
export class Company
{
Employees : Employee[];
constructor( jsonData: any )
{
jQuery.extend( this, jsonData);
// apply the same principle to linked objects:
if ( jsonData.Employees )
this.Employees = jQuery.map( jsonData.Employees , (emp) => {
return new Employee ( emp ); });
}
calculateSalaries() : void { .... }
}
export class Employee
{
name: string;
salary: number;
city: string;
constructor( jsonData: any )
{
jQuery.extend( this, jsonData);
// case where your object's property does not match the json's:
this.city = jsonData.town;
}
}
In your ajax callback where you receive a company to calculate salaries:
onReceiveCompany( jsonCompany : any )
{
let newCompany = new Company( jsonCompany );
// call the methods on your newCompany object ...
newCompany.calculateSalaries()
}