Java:遍历目录中的 HTML 文件

Question

我将 HTML 文件和一个 XSLT 文件作为输入并生成 HTML 输出，但在我的文件夹中有多个 HTML 文件，我必须将它们中的每一个作为输入并生成相应的输出文件，同时 XSLT 输入文件仍然存在每次都一样。目前在我的代码中，我每次都在重复代码块以获取输入的 HTML 文件。取而代之的是，我想遍历文件夹中的所有 HTML 文件，并将它们作为输入文件一个一个地生成输出。在我当前的代码中，文件名也像 part_1.html 一样固定，但它可能会有所不同，在这种情况下，这段代码将无法工作，这会产生问题。任何人都可以在这件事上帮忙:谢谢!

当前 Java 代码(两个文件的示例):

public void tranformation() {
        // TODO Auto-generated method stub
        transform1();
        transform2();
    }
    public static void transform1(){
        String inXML = "C:/SCORM_CP/part_1.html";
        String inXSL = "C:/source/xslt/html_new.xsl";
        String outTXT = "C:/SCORM_CP/part1_copy_copy.html";
        String renamedFile = "C:/SCORM_CP/part_1.html";
        File oldfile =new File(outTXT);
        File newfile =new File(renamedFile);
        HTML_Convert hc = new HTML_Convert();
        try {
            hc.transform(inXML,inXSL,outTXT);
            } catch(TransformerConfigurationException e) {
            System.err.println("Invalid factory configuration");
            System.err.println(e);
            } catch(TransformerException e) {
            System.err.println("Error during transformation");
            System.err.println(e);
        }
        try{
            File file = new File(inXML);
            if(file.delete()){
                System.out.println(file.getName() + " is deleted!");
                }else{
                System.out.println("Delete operation is failed.");
            }
            }catch(Exception e){
            e.printStackTrace();
        }
        if(oldfile.renameTo(newfile)){
            System.out.println("Rename succesful");
            }else{
            System.out.println("Rename failed");
        }
    }
    public static void transform2(){
        String inXML = "C:/SCORM_CP/part_2.html";
        String inXSL = "C:/source/xslt/html_new.xsl";
        String outTXT = "C:/SCORM_CP/part2_copy_copy.html";
        String renamedFile = "C:/SCORM_CP/part_2.html";
        File oldfile =new File(outTXT);
        File newfile =new File(renamedFile);
        HTML_Convert hc = new HTML_Convert();
        try {
            hc.transform(inXML,inXSL,outTXT);
            } catch(TransformerConfigurationException e) {
            System.err.println("Invalid factory configuration");
            System.err.println(e);
            } catch(TransformerException e) {
            System.err.println("Error during transformation");
            System.err.println(e);
        }
        try{
            File file = new File(inXML);
            if(file.delete()){
                System.out.println(file.getName() + " is deleted!");
                }else{
                System.out.println("Delete operation is failed.");
            }
            }catch(Exception e){
            e.printStackTrace();
        }
        if(oldfile.renameTo(newfile)){
            System.out.println("Rename succesful");
            }else{
            System.out.println("Rename failed");
        }
    }
    public void transform(String inXML,String inXSL,String outTXT)
    throws TransformerConfigurationException,
    TransformerException {
        TransformerFactory factory = TransformerFactory.newInstance();
        StreamSource xslStream = new StreamSource(inXSL);
        Transformer transformer = factory.newTransformer(xslStream);
        transformer.setErrorListener(new MyErrorListener());
        StreamSource in = new StreamSource(inXML);
        StreamResult out = new StreamResult(outTXT);
        transformer.transform(in,out);
        System.out.println("The generated XML file is:" + outTXT);
    }
}

Answer 1

File dir = new File("/path/to/dir");
File[] htmlFiles = dir.listFiles(new FilenameFilter() {
    @Override
    public boolean accept(File dir, String name) {
        return name.endsWith(".html");
    }
});
if (htmlFiles != null) for (File html: htmlFiles) {
    ...
}