PHP函数OOP阵列拉

目前我的文件结构看起来像这样，一切工作正常。但在我的home.php中，我想能够拉出数组的一部分，例如用户图像。PHP函数OOP阵列拉

$connect->LoadInformation->image

我该怎么写，或者改变我的代码以使其工作？我真的希望这是有道理的。

home.php

 $connect->LoadInformation($_COOKIE['steamID']);

class.php

 public function LoadInformation($steamID){ 

     $query = "SELECT * FROM user WHERE steamid = '$steamID'"; 
     $results = $this->con->get_results($query); 


     echo "<pre>"; 
     print_r($results); 
     echo "<pre>"; 
    }

电流输出

Array 
(
    [0] => Array 
     (
      [id] => 10 
      [steamid] => 76561198026169223 
      [profilestate] => 0 
      [steam_user] => Dz - Lol u mad? 
      [online] => 0 
      [signup] => 2014-01-05 17:28:15 
      [image] => http://media.steampowered.com/steamcommunity/public/images/avatars/b8/b85a568b25210a27a8413e4dc05c9f42a7cdaf22.jpg 
      [description] => 
      [active] => 0 
      [admin] => 0 
     ) 

)

来源

2014-01-05 Brent

无关，与OOP和开放的SQL注入。 – PeeHaa

更改您的class.php这样的..

public function LoadInformation($steamID){ 
     $query = "SELECT * FROM user WHERE steamid = '$steamID'"; 
     $results = $this->con->get_results($query); 
     return $results; 
    }

和你home.php

$arr = $connect->LoadInformation($_COOKIE['steamID']); 
echo $arr[0]['image']; //<-- prints http://media.steampowered.com/steamcommunity/public/images/avatars/b8/b85a568b25210a27a8413e4dc05c9f42a7cdaf22.jpg

来源

2014-01-05 18:17:00

只是把它转换用它来反对使用(object) ...

public function LoadInformation($steamID){ 
    $query = "SELECT * FROM user WHERE steamid = '$steamID'"; 
    $results = $this->con->get_results($query); 
    return (object) array_shift($results); 
}

兼：$connect->LoadInformation(...)->image;

而且我猜steamid将在表中是唯一，在这种情况下，随便挑从数据库查询返回的第一个结果然后按照你的喜好使用它。

来源

2014-01-05 18:15:32 CentaurWarchief

你能解释一下，我有其他所有工作，只需要这一点排序和即时消失！ – Brent

感谢您的配偶 – Brent

PHP函数OOP阵列拉

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