CODEWITHSUNDEEP
javalistjava-8
Jeeppp
Jeeppp

Reputation: 1573

Stream on List to get element and also from inner elements

I have a List of Employee object populated from a Json.

{
  "emp": [
    {
      "name": "John",
      "id": "123",
      "reportingEmp": [
        {
          "name": "Mark",
          "id": "342"
        },
        {
          "name": "Mike",
          "id": "342"
        },
        {
          "name": "Lindsey",
          "id": "342"
        }
      ]
    },
    {
      "name": "Steve",
      "id": "123"
    },
    {
      "name": "Andrew",
      "id": "123"
    }
  ]
}

class Employee {
private String name;
private String id;
private List<Employee> reportingEmp;
}

Now I wanted to get the names of all employess including the reporting employees in a stream.

I can use the below to get all the employees but it would miss the reporting employees

List<String> names = emp.stream().map(Employee::getName).collect(toList());

But the above would give only John,Steve and Andrew but I want the below list

John
Mark
Mike
Lindsey
Steve
Andrew

Upvotes: 0

Views: 460

Answers (1)

Sweeper
Sweeper

Reputation: 271635

flatMap is the thing to use, when you want to map one thing (an Employee) to multiple things (the employee's name + all the reporting employees' names).

In the flatMap lambda, you just need to create a stream that contains the things you want to map that employee to, so:

List<String> names = employees.stream().flatMap(emp ->
    Stream.concat(
        // the employee's name,
        Stream.of(emp.getName()), // and
        // the employee's reporting employees' names
        emp.getReportingEmp().stream().map(Employee::getName)
    )
).collect(Collectors.toList());

If getReportingEmp can return null, then you can do:

List<String> names = employees.stream().flatMap(emp ->
    Stream.concat(
        Stream.of(emp.getName()),
        Optional.ofNullable(emp.getReportingEmp()).map(
            x -> x.stream().map(Employee::getName)
        ).orElse(Stream.of())
    )
).collect(Collectors.toList());

Upvotes: 1

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