这个问题是继续上一个问题Accessing External Files Into Our Web Application，实际上我是使用struts标记上传文件<html:file property="file" />
但现在我想显示该位置上传的图片，但我的src位置为http://localhost:9443/D:/resources/images/img1.jpg，这不是该图片的有效路径。
如何访问我服务器目录之外的图像。
这就是我使用绝对路径图像
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上呈现图像的方式
$(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代码处，我的图片路径为http://localhost:9443/D:/resources/images/img1.jpg
答案 0 :(得分：1)
您无法以这种方式渲染图像。 Web服务器将您的映像路径视为相对路径，并在服务器上添加合格的URL位置。您应该创建一个动作来提供图像，例如
<action path="/image" ... scope="request" validate="false"/>
然后呈现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。在允许您查找二进制输出文件的操作中使用参数path。在刷新输出返回null之后，struts不应该进一步转发动作。您还可以添加标头以关闭Cache-Control以确保从服务器检索图像。
答案 1 :(得分：0)
嗨下面是我的问题的答案，我创建ImageServlet用于显示图像，执行步骤：
<强> 1。您需要在web.xml文件中添加映射：
<servlet-name>ImageServlet</servlet-name>
<url-pattern>/ImageServlet/*</url-pattern>
<强> 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();
}
}
}
}
第3。在jsp方面，您需要在 img 标记的第1步中添加映射，即输入类型='图像'：
<input type="image" alt='No image found' src='../ImageServlet/append image name that you want to display' />
您甚至可以创建Action类并使用execute方法执行此操作。