PHP開關和類屬性問題

Question

我想設置屬性並在另一個函數上使用它。

我

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(); 
} 

我似乎無法得到$this->email在buildName方法，因爲this->email在buildemail方法定義。 我需要使用開關，因爲每種方法都有很多代碼。有沒有辦法做到這一點？

Answer 1

爲什麼不在buildName方法中調用$employee->getEmail()而不是依靠它在$email？

另外：

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

buildName和都將運行，如果$employee->getInfoType()回報 「的名字。」你錯過了兩者之間的break;。

Answer 2

不能你喜歡：

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(). 
} 

我認爲每個員工都必須有一個電子郵件，是正確的？