php – 这是抽象工厂模式的合法用途吗？

>程序应该打开一个zip文件,其中包含许多数据文件
> zip文件的数据文件格式可能不同(例如csv,制表符分隔,甚至可能是某种需要解码的二进制文件)
>但是,在zip文件中,所有数据文件的类型都相同

我一直在阅读Gamma等人的“设计模式”,并一直在寻找抽象工厂模式来试图解决这个问题.

理想情况下,我希望Zip文件有一个类,它可以读取其中的任何类型的数据文件.我想我会有两个类 – FileTypeA和FileTypeB,它们可以处理不同格式的数据(尽管将来会有更多).我想告诉我的ZipFile类在读取数据时使用哪种类型的文件.

到目前为止,这是我提出的：

<?php

/**
 * An abstract factory used for creating data files of any type
 */
abstract class DataFileFactory{
    abstract function createFile($id);
}

/**
 * A factory for creating and setting up a data file of type 'A'
 */
class FileAFactory extends DataFileFactory{
    public function createFile($id){
        $file = new FileA();
        $file->setSampleId($id);
        return $file;
    }
}

/**
 * A factory for creating and setting up a data file of type 'B'
 */
class FileBFactory extends DataFileFactory{
    public function createFile($id){
        $file = new FileB();
        $file->setSampleId($id);
        return $file;
    }
}

/**
 * An abstract class which defines some functionality of a data file
 */
abstract class DataFile{
    abstract function readData();
    abstract function setSampleId();
}

/**
 * Concrete class that processes a data file of type 'A'
 */
class FileA extends DataFile{
    public function readData(){
        echo "Reading data from a file A<br/>";
    }

    public function setSampleId(){
        echo "Setting sample id of a file A<br/>";
    }
}

/**
 * Concrete class that processes a data file of type 'B'
 */
class FileB extends DataFile{
    public function readData(){
        echo "Reading data from a file B<br/>";
    }

    public function setSampleId(){
        echo "Setting sample id of a file B<br/>";
    }
}

/**
 * Concrete class that reads a zip file and reads each file within the zip
 */
class ZipFile{
    private $files = array("file1.txt","file2.txt","file3.txt","file4.txt");//this would be an array read from the zip file
    private $sampleId = 1;//this would be derived from some other function

    /**
     * Read all the files in a zip archive.
     * $factory can be an instance of any class that extends DataFileFactory,and is used for creating each file
     */
    public function readFiles(DataFileFactory $factory){
        foreach($this->files as $fileName){//loop through each file in the zip
            $file = $factory->createFile($this->sampleId);//use the factory to create the desired file
            $file->readData();//now read the data from the file!
            echo "object created of type: ".get_class($file)."<hr/>";
        }
    }
}

/***********************************************************************************************
 * IMPLEMENTATION
 ***********************************************************************************************/
$zip = new ZipFile();//create a new zip file
$factory = new FileAFactory();//instantiate a new factory,depending on which type of file you want to create
$zip->readFiles($factory);//read the files,passing the correct factory object to it

谁能告诉我：
(A)这是否是实现我正在寻找的好方法,还是有一些更简单的方法呢？
(B)这实际上是抽象工厂模式,还是我完全被误解了？

提前致谢！