Question

我正在尝试设置属性并在另一个函数上使用它。

我有

while($texts->employees()){
      $employee = $employees->get();

      switch($employee->getInfoType()){

        case 'email':
            $this->buildemail($employee);
          break;
        case 'Name':
            $this->buildName($employee);
          break;
        case 'Numbers':
            $this->buildNumbers($employee);
          break;
     }

function buildEmail($employee){
    $this->email=$employee->getEmail();  //get the email.
}

function buildName($employee){
    $this->Name=$this->getName(); //get the name
    $this->employeeInfo=$this->email.$this->name;   //combine the email and numbers.

    //$this->email is '' becasue it's only defined in buildEmail(). 
}

function buildNumbers($employee){
     $this->numbers=$this->getNumbers();
}

我似乎无法在buildName方法中获取$this->email，因为this->email方法中定义了buildemail。我需要使用switch，因为每种方法都有很多代码。反正有没有这样做？

Answer 1

为什么不用$employee->getEmail()方法拨打buildName而不是依赖$email？

此外：

    case 'Name':
        $this->buildName($employee);
    case 'Numbers':
        $this->buildNumbers($employee);
      break;

如果buildName返回“姓名”，则

buildNumbers和$employee->getInfoType()都会运行。你错过了两者之间的break;。

Answer 2

你不能这样做：

function buildName($employee){
    $this->Name=$this->getName(); //get the name

    if(null == $this->email)
        $this->buildEmail($employee);

    $this->employeeInfo= $this->email.$this->name;   //combine the email and numbers.

    //$this->email is '' becasue it's only defined in buildEmail(). 
}

我认为每个员工都必须收到一封电子邮件，这是正确的吗？

PHP开关和类属性问题

2 个答案: