这是我在我的程序中尝试实现的内容:
>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!
解决方法:
这是一个很好的实现,但如果你使用接口可以稍微调整一下.
一个带有所有虚方法的abtract类它只是一个接口所以不要使用抽象类,使用interfaces.
interface IDataFileFactory{
public function createFile($id);
}
class FileAFactory implements IDataFileFactory
class FileBFactory implements IDataFileFactory
如果您在FileAFactory和FileBFactory方法中找到重复的代码,那么是时候重构您的类并创建继承.
interface IDataFileFactory{
public function createFile($id);
}
abstract class BaseFileFactory implements IDataFileFactory {
//some methods implementation with common features to avoid repeating code
//some abstract methods to be implemented for A and B FileFactories
//absolute abstract base class has no sense because in PHP you can use interfaces.
//...
}
class FileAFactory extends BaseFileFactory
class FileBFactory extends BaseFileFactory
然后使用throug接口:
public function readFiles(IDataFileFactory $factory){
//create a file using factory
return IDataFile; //return Interface implemented by all DataFile types.
}
您可以使用DataFile基类执行相同的操作,依此类推.
我还建议不要在参数中传递工厂,因为工厂不在上下文中.尽量不要将架构实现与数据和信息处理工作流混合在一起.您可以在可供其他类访问的作用域中创建容器来解析工厂.
例如,容器可以读取配置文件以在应用程序引导程序中创建具体工厂;读取某个值,由用户在用户案例的先前步骤中选择,存储在类实例中或在运行时接受参数以解析工厂.它是关于实现某种简单的依赖关系.
无论如何,这只是我的观点,可能是一个不同意见.
我希望它有所帮助.
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