PHP - 語言新手

Question

嗨，我只是學習PHP，我正在嘗試各種代碼示例/練習。下面是一個

<?php 
// class definition 
class Bear 
    { 
    // define properties 

    public $name; 
    public $weight; 
    public $age; 
    public $colour; 
    public $sex; 

    // constructor 
    public function _construct() 
     { 
      $this->age = 0; 
      $this->weight = 100; 
      $this->colour = "Brown"; 
      }  
    } 

    // create instance 

    $baby = new Bear; 
    $baby->name = "Baby bear"; 
    $sex->sex = "Male";  

    echo $baby->name." is ".$baby->colour." and weighs ".$baby->weight." units at birth and his sex is ".$baby->sex; 
?> 

當我運行這個文件，它返回

寶寶熊和出生時體重的單位和他的性是

它不拿起變量。 任何建議都會很棒。 謝謝

Answer 1

該程序是非常簡單，它並不複雜。這裏的問題是

1）構造一個構造函數的方式，你創建了一個類Bear。使構造函數代碼中的

public Bear() { //Your code here } //Deprecated in PHP7 
       or 
public function __construct() 

2）您創建了一個名爲$嬰兒和$性實例是該對象的變量。但是你寫了

$sex->sex = "Male"; //This is wrong as $sex is a variable not an object 
$baby->sex = "Male"; //Correct way of representation 

在你的代碼中做這些改變，它肯定會起作用。

希望這可以幫助你。

Answer 2

PHP中的構造函數需要下劃線，你只有一個在你的原代碼。這裏是你的代碼與正確聲明的構造函數：

<?php 
// class definition 
class Bear 
{ 
// define properties 

public $name; 
public $weight; 
public $age; 
public $colour; 
public $sex; 

// constructor 
public function __construct() 
    { 
     $this->age = 0; 
     $this->weight = 100; 
     $this->colour = "Brown"; 
     }  
} 

// create instance 

$baby = new Bear; 
$baby->name = "Baby bear"; 
$baby->sex = "Male";  

echo $baby->name." is ".$baby->colour." and weighs ".$baby->weight." units at birth and his sex is ".$baby->sex; 
?> 

Answer 3

你缺少的構造下劃線，它應該是

public function __construct() {} 

還行$sex->sex = "Male";應該$baby->sex = "Male";