通过getRealPath()将上传的文件写入磁盘时出现java.io.FileNotFoundException

时间:2013-02-12 23:12:48

标签: servlets file-upload filenotfoundexception realpath

Glassfish似乎在我想要保存我的图像文件的路径上添加了额外的东西,有没有办法只使用我的servlet获取的绝对路径

String appPath = request.getServletContext().getRealPath("");

我花了几天时间尝试使用不同的方法上传图像文件并让servlet将其保存到磁盘。

我用过这个例子: http://www.codejava.net/java-ee/servlet/how-to-write-upload-file-servlet-with-servlet-30-api

使用debug我可以看到文件名和路径信息被正确收集但代码在`part.write(savePath + File.separator + fileName);``

失败

而glassfish的例外报告是:

exception

java.io.FileNotFoundException: C:\Program Files\glassfish-3.1.2.2\glassfish\domains\domain1\generated\jsp\com.onemore_onemore-web_war_1.0-SNAPSHOT\D:\Dropbox\Git\Brian_JEE\onemore\onemore\onemore-web\target\onemore-web-1.0-SNAPSHOT\image\screengrab_all_products.jpg (The filename, directory name, or volume label syntax is incorrect)

我可以看到正确的路径作为此异常D:\Dropbox\Git\Brian_JEE\onemore\onemore\onemore-web\target\onemore-web-1.0-SNAPSHOT\image\screengrab_all_products.jpg

的一部分

JSP

<form action="${pageContext.request.contextPath}/imageupload" method="post" enctype="multipart/form-data" name="productForm" id="productForm">
<input type="file" name="file" id="file">
<input type="submit" name="Submit" value="Submit"></td>
</form>

的Servlet

@WebServlet(name = "UploadImageServlet", urlPatterns = {"/imageupload"})
@MultipartConfig(fileSizeThreshold = 1024 * 1024 * 2, // 2MB 
maxFileSize = 1024 * 1024 * 10, // 10MB
maxRequestSize = 1024 * 1024 * 50)   // 50MB
public class UploadImageServlet extends HttpServlet {

    /**
     * Name of the directory where uploaded files will be saved, relative to the
     * web application directory.
     */
    private static final String SAVE_DIR = "image";

    /**
     * handles file upload
     */
    @Override
    protected void doPost(HttpServletRequest request,
            HttpServletResponse response) throws ServletException, IOException {
      // gets absolute path of the web application
        String appPath = request.getServletContext().getRealPath("");
        // constructs path of the directory to save uploaded file
        String savePath = appPath + File.separator + SAVE_DIR;

        // creates the save directory if it does not exists
        File fileSaveDir = new File(savePath);
        if (!fileSaveDir.exists()) {
            fileSaveDir.mkdir();
        }

        for (Part part : request.getParts()) {
            String fileName = extractFileName(part);
            part.write(savePath + File.separator + fileName);
        }

        request.setAttribute("message", "Upload has been done successfully!");
        getServletContext().getRequestDispatcher("/WEB-INF/jsp/newproduct.jsp").forward(
                request, response);
    }

    /**
     * Extracts file name from HTTP header content-disposition
     */
    private String extractFileName(Part part) {
        String contentDisp = part.getHeader("content-disposition");
        String[] items = contentDisp.split(";");
        for (String s : items) {
            if (s.trim().startsWith("filename")) {
                return s.substring(s.indexOf("=") + 2, s.length() - 1);
            }
        }
        return "";
    }
}

2 个答案:

答案 0 :(得分:4)

永远不要使用getRealPath()

您不应将上传的文件保存在deploy文件夹中。我之前已经解释了很多次。其中一个解释可以在这个答案中找到:Uploaded image only available after refreshing the page

在某些使用getRealPath()方法的博客/文章中找到的任何JSP / Servlet代码片段都应该带有很大的盐。它的质量和作者的知识应该受到质疑。 There is namely no sensible real world use case for that method

将上传的文件保存在deploy文件夹之外的固定路径中。另请参阅How do I set the folder for storing file uploads using Commons FileUpload

答案 1 :(得分:0)

使用以下代码。我使用这些代码解决了此错误。将其写入您的 doPost 方法。

Part filePart = request.getPart("file");
String fileName = 
Paths.get(filePart.getSubmittedFileName()).getFileName().toString(); 
InputStream inputStream = filePart.getInputStream();
String uploadPath = getServletContext().getRealPath("") + File.separator + "uploaded_file";
File uploadDir = new File(uploadPath);
if (!uploadDir.exists()) {
            uploadDir.mkdir();
        }
FileOutputStream outputStream = new FileOutputStream(uploadPath + 
File.separator + fileName);
        int read = 0;
        final byte[] bytes = new byte[1024];
        while ((read = inputStream.read(bytes)) != -1) {
            outputStream.write(bytes, 0, read);
        }

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