如何将所有这些包装到 PHP 类中？

Question

$image='xiaofl.jpg';
$img=getimagesize($image);
switch ($img[2])
{
    case 1:
    $im =imagecreatefromgif($image);
    break;
    case 2:
    $im =imagecreatefromjpeg($image);
    break;
    case 3:
    $im =imagecreatefrompng($image);
    break;
}
$word=imagecolorallocate($im,212,0,0);
$str=iconv("gbk","utf-8","php100.com");
imagettftext($im,12,0,20,20,$word,'simkai.ttf',$str);

header("content-type: image/jpeg");
Imagejpeg($im);

Answer 1

有很多方法可以做到这一点。您通常希望为每个类确定一个职责。然后，您将对类进行编码以尽可能一般地满足该职责。在这种情况下，职责可能是显示添加了一些文本的图像（？）。

这是一种方法：

 $image='xiaofl.jpg';
 $imageDisplay = new ImageDisplay($image);
 $imageDisplay->show();


 class ImageDisplay
 {

     private $imageName;

     public function __construct($imageName)
     {
         $this->imageName = $imageName;
     }

     public function show()
     {
         $img=getimagesize($this->imageName);
         switch ($img[2])
         {
             case 1:
             $im =imagecreatefromgif($this->imageName);
             break;
             case 2:
             $im =imagecreatefromjpeg($this->imageName);
             break;
             case 3:
             $im =imagecreatefrompng($this->imageName);
             break;
         }
         $word=imagecolorallocate($im,212,0,0);
         $str=iconv("gbk","utf-8","php100.com");
         imagettftext($im,12,0,20,20,$word,'simkai.ttf',$str);

         header("content-type: image/jpeg");
         Imagejpeg($im);
     }

 }

Answer 2

这是一种方法：

<?php

class ImageWorker{

  private $image = '';
  private $font = 'simkai.ttf';
  private $text = 'php100.com';
  private $imgobj = false;

  function __construct($imgfile){
    $this->image = $imgfile;
    $this->init();
  }

  private function init(){
    $img=getimagesize($this->image);
    switch($img[2]){
      case 1:
        $this->imgobj =imagecreatefromgif($this->image);
        break;
      case 2:
        $this->imgobj =imagecreatefromjpeg($this->image);
        break;
      case 3:
        $this->imgobj =imagecreatefrompng($this->image);
        break;
    }
  }

  public function render(){
    if(!$this->imgobj){return false;}
    $word=imagecolorallocate($this->imgobj,212,0,0);
    $str=iconv("gbk","utf-8",$this->text);
    imagettftext($this->imgobj,12,0,20,20,$word,$this->font,$str);
    return $this;
  }

  public function output(){
    if(!$this->imgobj){return false;}
    imagejpeg($this->imgobj);
    return $this;
  }

}

?>

使用类：

<?php

header("content-type: image/jpeg");
$img = new ImageWorker('xiaofl.jpg');
$img->render()->output();

?>

Answer 3

这就是我认为你的班级的样子：

class ImageThing{
    private $image;

    public function setImage($string){
        $this->image = $string;
    }
    public function process{
        $img=getimagesize($this->image);
        switch ($img[2])
        {
        case 1:
            $im =imagecreatefromgif($image);
            break;
        case 2:
            $im =imagecreatefromjpeg($image);
            break;
        case 3:
            $im =imagecreatefrompng($image);
            break;
        }
        $word=imagecolorallocate($im,212,0,0);
        $str=iconv("gbk","utf-8","php100.com");
        imagettftext($im,12,0,20,20,$word,'simkai.ttf',$str);

        header("content-type: image/jpeg");
        Imagejpeg($im);
    }
};

这有点受限，因为我不确定您是如何处理图像的。

因此，您创建对象，然后设置它指向的对象，然后设置它指向的图像，然后让它处理文件。 驱动程序文件可能如下所示：

<?php
    $img = new ImageThing();
    $image->setImage("images/this.jpg");
    $image->process();
?>