typescript - invalid field in function parameter with multiple types

Question

I have a code where I have these interfaces and type defined:

export interface ReportItem {
    title: string;
    earnings: number;
}

export interface CustomError {
    errorRaw: any;
}

export type SalesReport = ReportItem[];

Then I have a method that accepts a parameter, which may be of type SalesReport, but might also be a CustomError:

public getReport(data: SalesReport | CustomError) {
   if ('errorRaw' in data) {
       data.errorRaw // error
   }
}

The reference to errorRaw field in data throws an error. What change do I need to make so it would work properly?

Thanks.

vince · Accepted Answer · 2018-01-27 18:22:29Z

1

You're looking for a type guard to "narrow" your interface.

function isCustomErro(x: any): x is CustomError {
  return x.hasOwnProperty('errorRaw');
}

Then, you getReport method will look like this:

public getReport(data: SalesReport | CustomError) {
  if (isCustomError(data)) {
   throw new Error(data.errorRaw)
  }
}

Read more about type guards and narrowing interfaces here: https://medium.com/@OlegVaraksin/narrow-interfaces-in-typescript-5dadbce7b463

answered Jan 27, 2018 at 18:22

vince

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