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I am uploading a file. I want to get the file and save to my local system.To do this i am using spring surf webscripts in java. Can any one tell me how i can get my file .

This is my ftl file :

<form name="frmUpload" id="frmUpload"action="${url.context}/upload"     enctype="multipart/form-data" method="get">
<input type="file" size="40" id="toBeUploaded" name="toBeUploaded" tabindex="2" onchange = "document.getElementById('frmUpload').submit()"required />
</form>

I am creating a backed java webscript to get this file. Here is my java code.

public class Upload extends DeclarativeWebScript{

    protected ServiceRegistry serviceRegistry;
    private static final long serialVersionUID = 1L;
    private String fileName;
    private String filePath;

    private File toBeUploaded;  
    private String toBeUploadedFileName = "";  
    private String toBeUploadedContentType;  

    /** Multi-part form data, if provided */
    private FormData formData;

    /** Content read from the inputstream */
    private Content content = null;

    // upload settings
    private static final int MEMORY_THRESHOLD   = 1024 * 1024 * 3;  // 3MB
    private static final int MAX_FILE_SIZE      = 1024 * 1024 * 40; // 40MB
    private static final int MAX_REQUEST_SIZE   = 1024 * 1024 * 50; // 50MB

      @Override
      protected Map executeImpl(WebScriptRequest req,Status status) {         
          System.out.println("backed webscript called");


          Boolean isMultipart  = false;


          String fileName  = req.getParameter("toBeUploaded");        
          if(fileName == null){
              System.out.println("File Name is null");
          }else{
              System.out.println("File Name is :" + fileName);
          }       


          HttpServletRequest request = ServletUtil.getRequest();
          String file = request.getParameter("toBeUploaded");
          File file2 = new File(file);
          String filePath = request.getSession().getServletContext().getRealPath("/");        
          File fileToCreate = new File(filePath, this.toBeUploadedFileName);          
          System.out.println("filepath "+filePath);

          try {
                FileUtils.copyFile(file2, fileToCreate);
                //validateBundle(fileToCreate);
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
               System.out.println("filetocreate "+fileToCreate);


        }

}

file name is comming properly but it is throwing FileNotFoundExeption. Here is stacktrace

java.io.FileNotFoundException: Source 'test.jar' does not exist
        at org.apache.commons.io.FileUtils.copyFile(FileUtils.java:637)
        at org.apache.commons.io.FileUtils.copyFile(FileUtils.java:607)
4

1 回答 1

1

要获取上传的表单，您需要通过FormData对象。您的代码将类似于：

    // Get our multipart form
    final ResourceBundle rb = getResources();
    final FormData form = (FormData)req.parseContent();
    if (form == null || !form.getIsMultiPart())
    {
        throw new ResourceBundleWebScriptException(Status.STATUS_BAD_REQUEST, rb, ERROR_BAD_FORM);
    }

    // Find the File Upload file, and process the contents
    boolean processed = false;
    for (FormData.FormField field : form.getFields())
    {
        if (field.getIsFile())
        {
            // Logic to process/save the file data here
            processUpload(
                    field.getInputStream(),
                    field.getFilename());
            processed = true;
            break;
        }
    }

    // Object if we didn't get a file
    if (!processed)
    {
        throw new ResourceBundleWebScriptException(Status.STATUS_BAD_REQUEST, rb, ERROR_NO_FILE);
    }

如果您确定上传的字段名称，您可以将其中一些逻辑位短路

于 2013-10-30T12:02:39.403 回答