使用Servlet上传文件？

Question

注意:

在阅读此问题及其答案之前,请检查您的输入元素是否具有name属性.

我正在尝试使用servlet上传文件.Eclipse控制台显示没有错误.但是文件没有上传.对我来说,似乎一切都很好.但我在某个地方犯了错误.

在控制台我得到了

Inside Servlet //Printed by code
Items: [] // Printed by Cdoe

Html代码:

<form action="ImageUploadServlet" method="post" enctype="multipart/form-data">
<table>
<tr>
  <td><label>Select Image: </label></td>
  <td><input type="file" id="sourceImage" /></td>
   <tr>
        <td colspan="3">
         <input type="submit" value="Upload"/><span id="result"></span>
        </td>
      </tr> 
</table>
</form>

Servlet代码:

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    boolean isMultiPart = ServletFileUpload.isMultipartContent(request);
    System.out.println("Inside Servlet");
    if(!isMultiPart){
        System.out.println("Form type is not multipart/form-data");
        System.out.println("File Not Uploaded");
    }
    else
    {
        FileItemFactory dfit = new DiskFileItemFactory();
        ServletFileUpload sfu = new ServletFileUpload(dfit);
        List aList = null;

        try {
            aList = sfu.parseRequest(request);
            System.out.println("Items: "+aList);
        } 
        catch (FileUploadException fue) 
        {
            fue.printStackTrace();
        }

        Iterator itr = aList.iterator();
        while(itr.hasNext())
        {
            FileItem fi = (FileItem) itr.next();
            if(fi.isFormField())
            {
                System.out.println("File Name: "+fi.getFieldName());
                System.out.println("File Size: "+fi.getSize());

                try 
                {
                    File f = new File("D:/MyUploads/", fi.getName());
                    fi.write(f);
                } 
                catch (Exception e) 
                {
                    e.printStackTrace();
                }
            }
            else
            {
                System.out.println("It's Not Form Item;");
            }
        }
    }
}
}

任何建议,将不胜感激.

谢谢!

Answer 1

有两个问题:

首先,你需要给这个领域一个name.然后它成为请求参数名称.

<input type="file" id="sourceImage" name="sourceImage" />

其次,你需要处理的文件上传中else的情况下,FileItem#isFormField()如按照通用FileUpload指南.您的代码当前忽略它们并仅执行sysout.

if (item.isFormField()) {
    // Process regular form field (input type="text|radio|checkbox|etc", select, etc).
    String fieldname = item.getFieldName();
    String fieldvalue = item.getString();
    // ... (do your regular form field processing job here)
} else {
    // Process form file field (input type="file").
    String fieldname = item.getFieldName();
    String filename = FilenameUtils.getName(item.getName());
    // ... (do your uploaded file job here)
    File file = new File("D:/MyUploads/", filename);
    item.write(file);
}

请注意,您需要使用FilenameUtils#getName()提取文件名,因为MSIE浏览器错误地沿文件名发送完整的客户端文件路径.另请参阅Commons FileUpload FAQ.

您还需要记住,此方法将覆盖任何以前上传的具有相同名称的文件.您可能希望为文件名添加自动生成的后缀.

也可以看看:

• 如何在JSP/Servlet中上传文件？