的foreach（）错误警告：（）提供的foreach无效参数

Question

我不断收到这个错误对于错误的阵列，我的方法中

这里有设置是代码

public function showerrors() { 
     echo "<h3> ERRORS!!</h3>"; 
     foreach ($this->errors as $key => $value) 
     { 
      echo $value; 
     } 
    } 

我保持得到这个“警告：提供的foreach（）无效参数”当我运行程序 我设置了错误阵列在构造这样

$this->errors = array(); 

所以我不会完全知道为什么它不会打印错误！

public function validdata() { 
     if (!isset($this->email)) { 
      $this->errors[] = "email address is empty and is a required field"; 
     } 

     if ($this->password1 !== $this->password2) { 
      $this->errors[] = "passwords are not equal "; 
     } 
     if (!isset($this->password1) || !isset($this->password2)) { 
      $this->errors[] = "password fields cannot be empty "; 
     } 
     if (!isset($this->firstname)) { 
      $this->errors[] = "firstname field is empty and is a required field"; 
     } 
     if (!isset($this->secondname)) { 
      $this->errors[] = "second name field is empty and is a required field"; 
     } 

     if (!isset($this->city)) { 
      $this->errors[] = "city field is empty and is a required field"; 
     } 


     return count($this->errors) ? 0 : 1; 
    } 

这里是我如何添加数据到数组本身！感谢您的帮助！

好吧，我加入这

public function showerrors() { 
     echo "<h3> ERRORS!!</h3>"; 
     echo "<p>" . var_dump($this->errors) . "</p>"; 
     foreach ($this->errors as $key => $value) 
     { 
      echo $value; 
     } 

然后将其输出我的网页上这

错误的方法！ 字符串（20）“无效提交!!”如果我没有输入任何东西到我的文本框，所以它说一个字符串？

这里是我的构造函数也soory关于这个即时通讯新的PHP！

public function __construct() { 

     $this->submit = isset($_GET['submit'])? 1 : 0; 
     $this->errors = array(); 
     $this->firstname = $this->filter($_GET['firstname']); 
     $this->secondname = $this->filter($_GET['surname']); 
     $this->email = $this->filter($_GET['email']); 
     $this->password1 = $this->filter($_GET['password']); 
     $this->password2 = $this->filter($_GET['renter']); 
     $this->address1 = $this->filter($_GET['address1']); 
     $this->address2 = $this->filter($_GET['address2']); 

     $this->city = $this->filter($_GET['city']); 
     $this->country = $this->filter($_GET['country']); 
     $this->postcode = $this->filter($_GET['postcode']); 


     $this->token = $_GET['token']; 
    } 

Answer 1

上的默认（没有填写）验证，以您的形式，其中消息"invalid submission"设置，你离开了括号[]，造成$this->errors与普通字符串，而不是追加到数组被覆盖。

// Change 
$this->errors = "invalid submission"; 

//...to... 
$this->errors[] = "invalid submission";