check for a property in array of objects and return a string based on the value

Question

I have an array of objects like this ,

   let employee =  [
            {
                NodeType: "intern",
                NodeName: "Node1"
            }, {
                NodeType: "intern",
                NodeName: "Node2"
            }, {
                NodeType: "full-time",
                NodeName: "Node1"
            }, {
                NodeType: "contract",
                NodeName: "Node1"
            }
 ]

I need to be able to look through the array and see if all employees are full time then return "full time" or if the employee list is only "interns" return interns or if its mixed , return "mixed"

I tried

         var interntype = employee.find((obj) => {
                return obj.type == "intern" 
            });
         var fulltimetype = employee.find((obj) => {
                return obj.type == "full-time" 
            });
         var contracttype = employee.find((obj) => {
                return obj.type == "contract" 
            });

    if( internType) {
      return "intern";
    } else if (fulltimeType) {
          return "fullTime"
     } else return "mixed";

but is there a way where i dont do this multiple times and instead do it once

Look into the reduce function of an array.

Heretic Monkey
– Heretic Monkey

2017-07-03 21:12:37 +00:00
Commented Jul 3, 2017 at 21:12 — Heretic Monkey
– Heretic Monkey, Commented Jul 3, 2017 at 21:12

Ori Drori · Accepted Answer · 2017-07-03 21:33:14Z

3

Insert all NodeType values into a set, and check the size. If it's more than 1, return mixed, if not return the single item:

const employees = [{"NodeType":"intern","NodeName":"Node1"},{"NodeType":"intern","NodeName":"Node2"},{"NodeType":"full-time","NodeName":"Node1"},{"NodeType":"contract","NodeName":"Node1"}];

const getEmployeesType = (employees) => {
  const types = new Set(employees.map(({ NodeType }) => NodeType));
  return types.size > 1 ? 'mixed' : [...types][0];
};

console.log('mixed: ', getEmployeesType(employees));

console.log('internes: ', getEmployeesType(employees.slice(0, 2)));

edited Jul 3, 2017 at 21:33

answered Jul 3, 2017 at 21:19

Ori Drori

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