Ignore Case in Typescript when mapping data from API-CodePudding

I have the below json example from my API:

{"Name":"John","Address":"New Town"...} - so the properties like Name and address starts with an upper-case.

I tried to set the return value into the below dto and it does not work. However if I change the name to Name and address to Address, it works. As per my coding convention, I have to use camel case. Is there a way to ignore the case in the mapping?

export interface Employees{
    name: string;
    address: string;
}

Here is how I do the mapping:

employeeResults: Employees[] = [];

this.employeeService.getEmployeeResults().subscribe((employee: Employees[]) => {
  this.employeeResults= hotels;
})

CodePudding user response：

you can use map like this:


this.employeeService.getEmployeeResults().subscribe((employees: any[]) => {
 let mappedEmpeloyees = employees.map( (e) => {return {name: e.Name,address:e.Address};})
  this.employeeResults=  mappedEmpeloyees;
})

CodePudding user response：

As JS is case-sensitive I would suggest using some helper function which can get an object's property by the key in an insensitive way.

const getPopertyByKey = (obj, key) => obj[Object.keys(obj).find(k => k.toLowerCase() === key.toLowerCase())];

And then you can map service's result to your interface:

    employeeResults: Employees[] = [];
    
    this.employeeService.getEmployeeResults()
        .subscribe((employees: Employees[]) => {
            this.employeeResults = employees
                .map((e: Employees) => (
                    {
                        name: getPopertyByKey(e, 'nAmE'),
                        address: getPopertyByKey(e, 'AdDrEsS')
                    } as Employees
                )
            );
    });