Question

我有一个像这样的对象数组，

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

我需要能够查看数组并查看所有员工是否全职然后返回“全职”或者如果员工名单只是“实习生”返回实习生或者如果其混合，则返回“混合”

我试过

         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";

但有一种方法，我不会多次这样做，而是做一次

Answer 1

将所有NodeType值插入集合中，并检查大小。如果它超过1，则返回混合，如果不返回单个项目：

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)));

检查对象数组中的属性，并根据值返回一个字符串

1 个答案: