OOP PHP的新手,需要对第一个Geo RSS Class进行批评

我是OOP PHP的新手,目前正在阅读“PHP对象,模式和实践”.我需要开发一些可以生成GeoRSS Feed的东西.这就是我所拥有的(它完美地运作,我只想对我能做什么不同/更有效/更安全的一些批评):

class RSS {
 public $channel_title;
 public $channel_description;
 public $channel_link;
 public $channel_copyright;
 public $channel_lang;
 public $item_count;
 public function __construct ($channel_title, $channel_description, $channel_link, $channel_copyright, $channel_lang) {
  $this->channel_title = $channel_title;
  $this->channel_description = $channel_description;
  $this->channel_link = $channel_link;
  $this->channel_copyright = $channel_copyright;
  $this->channel_lang = $channel_lang;
  $this->items = "";
  $this->item_count = 0;
    }
 public function setItem ($item_pubDate, $item_title, $item_link, $item_description, $item_geolat, $item_geolong) {
  $this->items[$this->item_count]['pubDate'] = date("D, j M Y H:i:s T",$item_pubDate);
  $this->items[$this->item_count]['title'] = $item_title;
  $this->items[$this->item_count]['link'] = $item_link;
  $this->items[$this->item_count]['description'] = $item_description;
  $this->items[$this->item_count]['geo:lat'] = $item_geolat;
  $this->items[$this->item_count]['geo:long'] = $item_geolong;
  $this->items[$this->item_count]['georss:point'] = $item_geolat." ".$item_geolong;
  $this->item_count++;
 }
 public function getFeed () {
  foreach ($this->items as $item => $item_element) {
   $items .= "    <item>\n";
   foreach ($item_element as $element => $value) {
    $items .= "      <$element>$value</$element>\n";
   }
   $items .= "    </item>\n";
  }
  $feed = "<?xml version=\"1.0\" encoding=\"utf-8\"?>\n"
    . "<rss version=\"2.0\" xmlns:geo=\"http://www.w3.org/2003/01/geo/wgs84_pos#\" xmlns:georss=\"http://www.georss.org/georss\">\n"
    . "  <channel>\n"
    . "    <title>".$this->channel_title."</title>\n"
    . "    <description>".$this->channel_description."</description>\n"
    . "    <link>".$this->channel_link."</link>\n"
    . "    <copyright>Copyright ".date("Y")." ".$this->channel_copyright.". All rights reserved.</copyright>\n"
    . "    <lang>".$this->channel_lang."</lang>\n"
    . $items
    . "  </channel>\n"
    . "</rss>";
  return $feed;
 }
}

解决方法:

>如果没有令人信服的理由将属性公开或私有,则应始终保护属性.
>在使用之前声明或启动所有变量:您缺少类体中受保护的$项目和getFeed中的$items =”.
>正确启动变量:$this-> items = array();在__construct中.
>不管理自己的item_count.更好地依赖PHP自己的数组附加功能:

$this->items[] = array(
    'pubDate'      => date("D, j M Y H:i:s T",$item_pubDate),
    'title'        => $item_title,
    'link'         => $item_link,
    'description'  => $item_description,
    'geo:lat'      => $item_geolat,
    'geo:long'     => $item_geolong,
    'georss:point' => $item_geolat." ".$item_geolong,
);

好多了,不是吗?
>不要声明您需要的更多变量:

foreach ($this->items as $item) {
    $items .= "    <item>\n";
    foreach ($item as $element => $value) {
         $items .= "      <$element>$value</$element>\n";
    }
    $items .= "    </item>\n";
}

在这里你不需要数组键.所以不要在foreach循环中获取它;)而是使用$item作为值,这比$item_element更好.

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