試圖在OOP中獲取非對象的屬性php

Question

我只想支持我的類屬性，以便可以使用它們。在我的項目中，我想獲取mysql表數據的所有結果並將其返回。那麼它的效果很好，但改變返回錯誤消息來了數據，該數據如下：

Trying to get property of non-object in E:\xampp\htdocs\myProject\new_27_03\login_con.php on line 26 

我的代碼::

nonClass.php

<?php 
     $doc = new Dortor(); // object 

     $doc->result("SELECT * FROM doctor"); // function call 

     foreach($doc as $doctor) { 
     echo $doctor->doc_name;  // error msg: Trying to get property of non-object in 
            //E:\xampp\htdocs\myProject\new_27_03\login_con.php on line 26 
     } 
    ?> 

dortor.php

<?php 
    class Dortor extends DatabaseObject { 
     public $table_name = "doctor"; 
     public $id; 
     public $doc_name; 
     public $doc_pass; 
     public $doc_img;  // image directory 

     public function result($sql="") { 
     $result = $database->query($sql); // run query from mysql database 

     $result_array = array(); 
     while($row = $database->fetch_array($result) ) { // this while loop return actual result 
      $result_array[] = $this->get_table_value($row); 
     } 
     return $result_array; 
     } 
    } 
    ?> 

databaseObject.php

<?php 
     class DatabaseObject { 
      public function get_table_value($record) { 
       $className = get_called_class(); // get the class name 
       $object = new $className;  // object call 

       $arrayValue = array(); 
       $i=0; // inital array position 
       foreach($object as $key => $value) { 
        $arrayValue[$i] = $key; // get the class's object attributes 
        $i++; 
       } 

       $i = 1; 
      foreach ($record as $key => $value) { // fetch the database result 
       if(array_key_exists($key, $arrayValue)) { // check if database's column name is exist on 'arrayValue' array, 
        $object->$arrayValue[$i] = $value; // if exist then, store value in 'Doctor' class's object attributes 
        $i++; 
       } 
       } 
      return $object; 
      } 
     } 
    ?> 

Answer 1

你的錯誤可能是在這裏：

$doc = new Dortor(); // object 
$docs = $doc->result("SELECT * FROM doctor"); // you forgot to assign the result 
foreach ($docs as $doctor) { // was looping over $doc here, instead of the result 
    echo $doctor->doc_name; 
} 

Answer 2

這也將是有意義的foreach循環開始之前檢查正確的變量類型：

if (is_array($doc) ) 
{ 
foreach($doc as $doctor) 
    { 
    // foreach code 
    } 
} 
else 
{ 
    echo '$doc is no valid object'; 
    var_dump($doc); 
} 

所以，你可以找到，這就是問題。