undefined not assignable issue in Typescript

Question

I have an interface Employee:

export interface Employee {
  firstName: string;
  lastName: string;
}

I am trying to push elements of Type Employee into an array as below:

  let allEmployees: Employee[] = [];
  allEmployees = emp.data?.map((employee) => {
    return { firstName: employee.firstName, lastName: employee.lastName }
  });

I am getting the following typescript compile time error:

type 'undefined' is not assignable to type 'Employee[]'

I have used .map function to return final value of the array. How do I resolve this issue ?

Thanks

Please consider editing the code in the question to constitute a minimal reproducible example that can be dropped into a standalone IDE like The TS Playground so others can see the issue for themselves. The type of emp is important here. I presume its data property might itself be undefined, and therefore the optional chaining operator emp.data?.map() will possibly set allEmployees to undefined without ever running map(). Also, I don't see you trying to push any elements; instead you're trying to reassign allEmployees. Do you want to try to push() instead? — jcalz
– jcalz, Commented Jul 29, 2020 at 15:53
The push method did work. But I wanted to reuse the .map function. — Ashy Ashcsi
– Ashy Ashcsi, Commented Jul 30, 2020 at 8:21

kaveh · Accepted Answer · 2020-07-29 15:55:43Z

1

? in emp.data? means data is optional and might not exist in emp, which tells typescript not to go through the chained function, i.e map. So your assignment can return undefined or Employee[], which cannot be assigned to allEmployees.

edited Jul 29, 2020 at 15:55

answered Jul 29, 2020 at 15:54

kaveh

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1 Comment

T.J. Crowder Over a year ago

One of many possible fixes: let allEmployes = (emp.data ?? []).map(/*...*/);

tmhao2005 · Accepted Answer · 2020-07-29 15:55:22Z

0

It's because emp.data? possibly returns void 0 as well. That's why you don't have to use proposal-optional-chaining, just check like below:

allEmployees = emp.data ? emp.data.map((employee) => {
  return { firstName: employee.firstName, lastName: employee.lastName }
}) : [];

answered Jul 29, 2020 at 15:55

tmhao2005

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