1

这个问题是我之前的问题Accessing External Files Into Our Web Application的延续，实际上我正在使用 struts 标签上传文件<html:file property="file" />

但是现在我想显示从该位置上传的图像，但我得到的src位置http://localhost:9443/D:/resources/images/img1.jpg不是该图像的有效路径。

如何访问我的服务器目录之外的那个图像。

这就是我使用图像的绝对路径发送 Ajax 响应的方式

public ActionForward getAjaxUploadedFiles(ActionMapping mapping, ActionForm form, HttpServletRequest request, HttpServletResponse response) throws Exception
    {

        String imagePath = "D:/resources/images/";
        ArrayList<String> path = new ArrayList<String>();

        File imageFile = new File(imagePath);
        File imageFiles[] = imageFile.listFiles();

        for (int i = 0; i < imageFiles.length; i++) {
            path.add(imageFiles[i].getAbsolutePath());
        }

        PrintWriter out = response.getWriter();
        response.setContentType("text/xml");
        response.setHeader("Cache-Control", "no-cache");
        response.setStatus(HttpServletResponse.SC_OK);

        StringBuffer strXMl = new StringBuffer();
        strXMl.append("<?xml version=\"1.0\" encoding=\"UTF-8\"?>");
        strXMl.append("<start>"); 


        for (String imagePth : path) {
            strXMl.append("<imagePath>");
            strXMl.append(imagePth);
            strXMl.append("</imagePath>");
        }

        strXMl.append("</start>");

        if(strXMl != null){ 
            String Xml = strXMl.toString();
            out.write(Xml);
            System.err.println("XMl Reponse is: " + Xml);
        }
        else {
            response.setStatus(HttpServletResponse.SC_BAD_REQUEST);
        }
        out.flush();

        return mapping.findForward(null);
    }

这就是我在 JSP 上渲染图像的方式

 $(response).find("imagePath").each(function() {
            row = tblReportList.insertRow(0);
            row.className="TableBordergray";
            row.style.width="100%";

            var imagePath = $(this).text();

            cell = row.insertCell(0);
            cell.innerHTML="<img src='" + imagePath + "' alt='" + imagePath + "' height='42' width='42'>";
        });

但在img标签上我得到的图像路径为http://localhost:9443/D:/resources/images/img1.jpg

4

2 回答 2

2

您好以下是我的问题的答案，我创建了ImageServlet用于显示图像的执行步骤：

1.需要在web.xml文件中添加映射：

    <servlet-name>ImageServlet</servlet-name>
    <url-pattern>/ImageServlet/*</url-pattern>

2. 创建 ImageServlet：

public class ImageServlet extends HttpServlet {

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

        //Setting image path
        ImageLocationService locationService = new ImageLocationService();

        try {
            String imageCategory = request.getParameter("imageCategory");
            if (imageCategory != null) {
                this.imagePath = locationService.getImageLocation(imageCategory);
            }else{
                this.imagePath = ConfigConstants.imageLocation;
            }
        } catch (Exception e) {
            e.printStackTrace();
        }

        // Get requested image by path info.
        String requestedImage = request.getPathInfo();

        // Check if file name is actually supplied to the request URI.
        if (requestedImage == null) {
            // Do your thing if the image is not supplied to the request URI.
            // Throw an exception, or send 404, or show default/warning image, or just ignore it.
            response.sendError(HttpServletResponse.SC_NOT_FOUND); // 404.
            return;
        }

        // Decode the file name (might contain spaces and on) and prepare file object.
        File image = new File(imagePath, URLDecoder.decode(requestedImage, "UTF-8"));

        // Check if file actually exists in filesystem.
        if (!image.exists()) {
            // Do your thing if the file appears to be non-existing.
            // Throw an exception, or send 404, or show default/warning image, or just ignore it.
            response.sendError(HttpServletResponse.SC_NOT_FOUND); // 404.
            return;
        }

        // Get content type by filename.
        String contentType = getServletContext().getMimeType(image.getName());

        // Check if file is actually an image (avoid download of other files by hackers!).
        // For all content types, see: http://www.w3schools.com/media/media_mimeref.asp
        if (contentType == null || !contentType.startsWith("image")) {
            // Do your thing if the file appears not being a real image.
            // Throw an exception, or send 404, or show default/warning image, or just ignore it.
            response.sendError(HttpServletResponse.SC_NOT_FOUND); // 404.
            return;
        }

        // Init servlet response.
        response.reset();
        response.setBufferSize(DEFAULT_BUFFER_SIZE);
        response.setContentType(contentType);
        response.setHeader("Content-Length", String.valueOf(image.length()));
        response.setHeader("Content-Disposition", "inline; filename=\"" + image.getName() + "\"");

        // Prepare streams.
        BufferedInputStream input = null;
        BufferedOutputStream output = null;

        try {
            // Open streams.
            input = new BufferedInputStream(new FileInputStream(image), DEFAULT_BUFFER_SIZE);
            output = new BufferedOutputStream(response.getOutputStream(), DEFAULT_BUFFER_SIZE);

            // Write file contents to response.
            byte[] buffer = new byte[DEFAULT_BUFFER_SIZE];
            int length;
            while ((length = input.read(buffer)) > 0) {
                output.write(buffer, 0, length);
            }
        } finally {
            // Gently close streams.
            close(output);
            close(input);
        }
    }

    private static void close(Closeable resource) {
        if (resource != null) {
            try {
                resource.close();
            } catch (IOException e) {
                // Do your thing with the exception. Print it, log it or mail it.
                e.printStackTrace();
            }
        }
    }
}

3. 在 jsp 端，您需要在 img 标签中添加步骤 1 中的映射，即 input type='image'：

<input  type="image" alt='No image found' src='../ImageServlet/append image name that you want to display' />

你甚至可以创建Action类并使用execute方法来做同样的事情。

于 2013-10-14T06:32:35.847 回答
1

您不能以这种方式渲染图像。Web 服务器将您的图像路径视为相对路径，并在服务器上添加合格的 url 位置。您应该创建一个操作来提供图像，例如

<action path="/image" ... scope="request" validate="false"/>

然后像渲染 HTML

cell.innerHTML="<img src='" + '/image?path=' + imagePath + "' alt='" + imagePath + "' height='42' width='42'>";

现在，创建将二进制图像数据写入响应输出流的操作。在操作中获取一个参数path，让您找到二进制输出的文件。在刷新输出返回后null，struts 不应该进一步转发动作。您还可以添加标题以关闭 aCache-Control以确保从服务器检索图像。

于 2013-10-04T11:11:48.213 回答