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从输入流或字节数组构造数据源

apache-commons servlets java

Casey · 技术社区 · 14 年前

我正在写一个小文件上传实用程序作为一个大项目的一部分。最初,我使用ApacheCommons文件实用程序类从servlet处理这个问题。以下是我为该服务编写的快速测试客户机的一个片段:

public static void main(String[] args) {

  JaxWsProxyFactoryBean factory = new JaxWsProxyFactoryBean();

 factory.getInInterceptors().add(new LoggingInInterceptor());
 factory.getOutInterceptors().add(new LoggingOutInterceptor());
 factory.setServiceClass(FileUploadService.class);
 factory.setAddress("http://localhost:8080/FileUploadService/FileUploadService");
 FileUploadService client = (FileUploadService) factory.create();

 FileType file = new FileType();
 file.setName("statemo_1256144312279");
 file.setType("xls");

 DataSource source = new FileDataSource(new File("c:/development/statemo_1256144312279.xls"));
 file.setHandler(new DataHandler(source));
 Boolean ret = client.uploadFile(file);
 System.out.println (ret);
 System.exit(0);

}

这是绝对可行的。现在,当我试图替换ApacheCommons实用程序时,问题出现了。在上面的代码中,我从一个具有绝对路径名的文件创建一个数据源。但是,在我的servlet中,我无法获得绝对路径名,并且我通过线路发送的文件是空的。

以下是servlet代码:

 @SuppressWarnings("unchecked")
    protected void doPost (final HttpServletRequest request, final HttpServletResponse response) 
        throws ServletException, IOException {

    // form should have enctype="multipart/form-data" as an attribute
 if (!ServletFileUpload.isMultipartContent (request)) {
  LOG.info("Invalid form attribute");
  return;
 }

 //DataInputStream in = new DataInputStream(request.getInputStream());

 final DiskFileItemFactory factory = new DiskFileItemFactory ();
 factory.setSizeThreshold(FILE_THRESHOLD_SIZE);

 final ServletFileUpload sfu = new ServletFileUpload (factory);
 sfu.setSizeMax(MAX_FILE_SIZE);

 final HttpSession session = request.getSession();

 final List<FileItem> files = new ArrayList<FileItem>();

 final List<String> filesToProcess = new ArrayList<String>();

 try {
        final List<FileItem> items = sfu.parseRequest(request);

        for (final FileItem f : items) {
            if (!f.isFormField())
                files.add(f);
        }

        /*for (final FileItem f : files) {
         final String absoluteFileName = UPLOAD_DESTINATION + FilenameUtils.getName(f.getName());

            //f.write(new File (absoluteFileName));
            filesToProcess.add(absoluteFileName);
        }*/

        FileItem f = files.get(0);

        LOG.info("File: " + FilenameUtils.getName(f.getName()));
        LOG.info("FileBaseName: " + FilenameUtils.getBaseName(f.getName()));
        LOG.info("FileExtension: " + FilenameUtils.getExtension(f.getName()));

        FileUploadServiceClient client = new FileUploadServiceClient();

        DataSource source = new FileDataSource(new File(f.getName()));

        FileType file = new FileType();
        file.setHandler(new DataHandler(source));
        file.setName(FilenameUtils.getBaseName(f.getName()));
        file.setType(FilenameUtils.getExtension(f.getName()));

        Boolean ret = client.uploadFile(file);

        LOG.info("File uploaded - " + ret);

        filesToProcess.add(UPLOAD_DESTINATION + FilenameUtils.getName(f.getName()));
        session.setAttribute("filesToProcess", filesToProcess);

  final RequestDispatcher dispatcher = request.getRequestDispatcher("Validate");
        if (null != dispatcher) {
         dispatcher.forward(request, response);
        }
    } catch (FileUploadException e) {
        LOG.info("Exception " + e.getMessage());
        e.printStackTrace();
    } catch (Exception e) {
        LOG.info("Exception " + e.getMessage());
        e.printStackTrace();
    }

}

今天早上我一直在做这件事,但什么也没做。即使我完全摆脱了ApacheCommons文件,自己处理请求的解析,我仍然无法适当地构造数据源。

谢谢!

3 回复 | 直到 8 年前

Casey 14 年前

实际上,这相当简单,我只是将字节从输入流复制到数据源:

FileItem f = files.get(0);

// there is a problem here where the file being created is empty, since we only have a
// partial path:
DataSource source = new FileDataSource(new File(f.getName()));

// because of the above problem, we are going to copy over the data ourselves:
byte[] sourceBytes = f.get();
OutputStream sourceOS = source.getOutputStream();
sourceOS.write(sourceBytes);

Bozho MichaÅ Mech 14 年前

This 是公共电子邮件代码吗 ByteArrayDataSource
尝试替换ApacheCommons听起来很奇怪-不要,除非你有一个很好的理由
你可以获取servlet中的绝对路径。你可以打电话 getServletContext().getRealPath("/") 它将返回应用程序的绝对路径,然后您可以获取与之相关的文件。

Yuriy N. 8 年前

在我们的应用程序中,有一些对象具有inputstream和name属性。我们正在使用下一个类用这些属性构造数据源。

public class InputStreamDataSource implements DataSource {

    ByteArrayOutputStream buffer = new ByteArrayOutputStream();
    private final String name;

    public InputStreamDataSource(InputStream inputStream, String name) {
        this.name = name;
        try {
            int nRead;
            byte[] data = new byte[16384];
            while ((nRead = inputStream.read(data, 0, data.length)) != -1) {
              buffer.write(data, 0, nRead);
            }
            inputStream.close();
            buffer.flush();
        } catch (IOException e) {
            e.printStackTrace();
        }

    }

    @Override
    public String getContentType() {            
        return new MimetypesFileTypeMap().getContentType(name);
    }

    @Override
    public InputStream getInputStream() throws IOException {
            return new ByteArrayInputStream(buffer.toByteArray());
    }

    @Override
    public String getName() {
       return name;
    }

    @Override
    public OutputStream getOutputStream() throws IOException {
        throw new IOException("Read-only data");
    }

}