从另一个类中调用的构造返回致命错误

Question

为什么我在尝试调用不相关的构造函数时遇到错误？

Fatal error: Non-static method Employee::__construct() cannot be called statically, assuming $this from incompatible context in.. 

Answer 1

你需要使用它像往常一样构造函数来实例化无关类：

class Employee 
{ 
    function __construct() 
    { 
     echo "<p>Employee construct called!!</p>"; 
    } 
} 

class Manager 
{ 
    function __construct() 
    { 
     Employee::__construct(); 
     echo "<p> Manager Construct Called! </p>"; 
    } 
} 

当我调用它的子类

class Manager extends Employee 

错误工作正常。像这样的： -

class Employee 
{ 
    function __construct() 
    { 
     echo "<p>Employee construct called!!</p>\n"; 
    } 
} 

class Manager 
{ 

    protected $employee; 

    function __construct() 
    { 
     $this->employee = new Employee(); 
     echo "<p>Manager Construct Called! </p>\n"; 
    } 
} 

new Manager(); 

这是简单的方式，但它是更好的做法给员工传递给构造函数的依赖性： -

class Employee 
{ 
    function __construct() 
    { 
     echo "<p>Employee construct called!!</p>\n"; 
    } 
} 

class Manager 
{ 

    protected $employee; 

    function __construct(Employee $employee) 
    { 
     $this->employee = $employee; 
     echo "<p>Manager Construct Called! </p>\n"; 
    } 
} 

$manager = new Manager(new Employee()); 

不过不要担心太多很多，如果你现在还不明白。

See it working。