Question

function makeGroup(param) {
  this.group = param;
}

duplicateEmployee.makeGroup(index2);

做一个简单的duplicateEmployee.group = index2不起作用。尝试一种方法说＆＃34; makeGroup不是一个函数。＆＃34;

以下是完整的参考代码。

var someVar = [];
vm.contacts = ContactsService.query(employees => {
  employees.forEach(employee => {
    employee.groups.forEach((groupMembership, index2) => {
      // getEmployee() works...
      function getEmployee(param) {
        return param;
      }

      // "makeGroup is not a function"
      function makeGroup(param) {
        this.group = param;
      }

      var duplicateEmployee = getEmployee(employee);
      if (groupMembership) {
        duplicateEmployee.makeGroup(index2); // this breaks it. Says makeGroup is not a function
        console.log(groupMembership, index2); // index2 is working... that's the group number
        someVar.push(duplicateEmployee);
      }
    });
  });
});
vm.contacts = someVar;

试过这个并且它不起作用

      function makeGroup(param1, param2) { // console.log(param2) === undefined
        console.log(param1); // group number... not the employee (still useful)
        console.log(this); // 'this' is the employee, so lets try this.group
        this.group = param1; // still overwriting the variable
      }

      makeGroup.call(duplicateEmployee, index2)

Answer 1

您正在调用makeGroup对象的方法duplicateEmployee。此对象中没有此方法。您可以将其添加到prototype或只更改makeGroup，这样它就是一个接收员工和参数的函数，它就是这样的。

function makeGroup(employee, param) {
  employee.group = param;
}

getEmployee的工作原因是因为您是从全局上下文调用它，而不是像duplicateEmployee那样使用makeGroup上下文调用它。

Answer 2

duplicateEmployee.makeGroup(index2);无法从员工处拨打，因为它不是该员工的成员。
您可以像makeGroup(index2)

一样调用它

Javascript方法不是一个函数

2 个答案: