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Tomcat not loading new images from context path after server is already started

Inside server.xml of my tomcat folder, i have a virtual folder under <Host> tag:
<Context docBase="C:\app_files\" path="/app_files"/>
So i can access files from inside this folder through the url: http://localhost:8080/app_files/some_file.jpg
But this only works if the image or file was already there BEFORE the server was started. If i go to a URL pointing to an image created after the server was started, it gives the 404 error. After restarting the server, the image loads correctly.
How to sove this problem?

If you use Tomcat application manager you can undeploy/deploy your single application without restarting the whole server (and without impact on other webapps) or, more brutally, you can replace the desired war from the webapps dir (again undeploy/deploy will ensue). If you have to guarantee uptime for your application even in this case you have to go with parallel deployment (here a guide for tomcat 8)

try to add the attribute autoDeploy="true" to your context configuration, this will tell catalina to monitor your docbase location for changes

I actually managed to do what i wanted without using the Context on the server.xml.
It's based on the BalusC's solution to serve static files through servlets
Here's how:
First, i created an environment variable in my system (can be done in every os, just google for "how to create environment variable on windows, linux, etc"), called MANAGEMENT_FILES, the variable value in my case was c:/management_files/
Then, on the method that creates the image that should be shown to the user, i save the image on this folder (which is the value of the environment variable from the previous step):
public String imageUrl;
public void createAndShowImage() {
try {
String imageName = "/nice_images_folder/cool_image.jpg";
File imageFile = new File(System.getenv("MANAGEMENT_FILES") + imageName);
//Here goes your logic to create the file
createImage(imageFile);
//Here i use a fixed URL, you can do it as you see fit
this.imageUrl = "http://localhost:8080/MyCoolApp/" + CoolFileServlet.BASE_URL + imageName + "?delete=true";
} catch (Exception e) {
e.printStackTrace();
}
}
And this is the servlet you have to create, this servlet returns the image or any other file that you put inside the folder:
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.Closeable;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.net.URLDecoder;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
#WebServlet(name="CoolFileServlet", urlPatterns={CoolFileServlet.BASE_URL + "*"})
public class CoolFileServlet extends HttpServlet {
public static final String BASE_URL = "/shiny_happy_files/";
private static final int DEFAULT_BUFFER_SIZE = 10240;
private String filePath;
public void init() throws ServletException {
this.filePath = System.getenv("MANAGEMENT_FILES");
}
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException
{
String requestedFile = request.getPathInfo();
if (requestedFile == null) {
response.sendError(HttpServletResponse.SC_NOT_FOUND); // 404.
return;
}
File file = new File(filePath, URLDecoder.decode(requestedFile, "UTF-8"));
if (!file.exists()) {
response.sendError(HttpServletResponse.SC_NOT_FOUND); // 404.
return;
}
String contentType = getServletContext().getMimeType(file.getName());
if (contentType == null) {
contentType = "application/octet-stream";
}
response.reset();
response.setBufferSize(DEFAULT_BUFFER_SIZE);
response.setContentType(contentType);
response.setHeader("Content-Length", String.valueOf(file.length()));
response.setHeader("Content-Disposition", "attachment; filename=\"" + file.getName() + "\"");
BufferedInputStream input = null;
BufferedOutputStream output = null;
try {
input = new BufferedInputStream(new FileInputStream(file), DEFAULT_BUFFER_SIZE);
output = new BufferedOutputStream(response.getOutputStream(), DEFAULT_BUFFER_SIZE);
byte[] buffer = new byte[DEFAULT_BUFFER_SIZE];
int length;
while ((length = input.read(buffer)) > 0) {
output.write(buffer, 0, length);
}
} finally {
close(output);
close(input);
try {
if ("true".equals(request.getParameter("delete"))) {
if (!file.delete()) {
throw new RuntimeException("File could not be deleted");
}
}
} catch (Exception e) {
e.printStackTrace();
}
}
}
private static void close(Closeable resource) {
if (resource != null) {
try {
resource.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
Notice that you can pass the parameter delete=true in the url when it is accessed, to delete it right after it is recovered (in cases when it will not be needed anymore).
In my case i needed to show the image on the page after the user performed some action, so all i had to do was show the image url:
<h:graphicImage url="#{myManagedBean.imageUrl}"/>
That's it, you can serve any type of file with this servlet, and it will return the file you want, instantly, and the file will remain active between server restart/redeploy (if it was not deleted through delete=true.

If you care for a different approach, you can also do this by mapping a function in a controller that returns a IOUtils object while specifying the media type, then calling the function's URL in your img's src.
#ResponseBody
#RequestMapping(value="/load_photo", params = {"myPhoto"}, method = RequestMethod.GET, produces = MediaType.IMAGE_PNG_VALUE)
public byte[] loadPhoto(#RequestParam(value = "myPhoto") String myPhoto) throws IOException {
File file = new File(servletContext.getRealPath("")+Constants.PATH_TO_FILE+myPhoto);
FileInputStream fis = new FileInputStream(file);
return IOUtils.toByteArray(fis);
}
Then you call your img in your JSP:
<img class="photo" src="/app/controller/load_photo?myPhoto=${myPhoto}">
With this, you can serve dynamically generated images.

Uploading a file in Java Servlet

I have a Java Dynamic Web Project, and I'm using TomCat v7.0.
I am new to web projects and I didn't quite understand how I can upload a file in one of my jsp pages. Since my project is intended to be only local, I thought I could use a multipart form in which the person would choose the file (and this part goes fine) and later retreive the file path from my Servlet. I can't complete this part though, it appears to only give me the name of the file, not its entire path.
Can anyone point me to the right direction? I've read several posts about Apache File Upload and retreiving information from the multipart form but nothing seems to help me.
How can I get the file path from a form or alternatively how can I get the uploaded file to use in my Java classes?
Thanks in advance.
.jsp:
<form method="post" action="upload" enctype="multipart/form-data">
<input type="file" name="filePath" accept="application/vnd.ms-excel, application/vnd.openxmlformats-officedocument.spreadsheetml.sheet"></input>
<input type="submit" value="Enviar"></input>
</form>
Java Servlet:
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
PrintWriter out = response.getWriter();
out.println("<html><body>");
try
{
List<FileItem> items = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);
for (FileItem item : items)
{
if (item.isFormField())
{
// Process regular form field (input type="text|radio|checkbox|etc", select, etc).
String fieldname = item.getFieldName();
String fieldvalue = item.getString();
out.println("<h1>"+fieldname+" / "+fieldvalue+"</h1>");
}
else
{
// Process form file field (input type="file").
String fieldname = item.getFieldName();
String filename = item.getName();
InputStream filecontent = item.getInputStream();
String s = filecontent.toString();
out.println("<h1>"+s+" / "+filename+"</h1>");
item.write(null);
}
}
}
catch (FileUploadException e)
{
throw new ServletException("Cannot parse multipart request.", e);
}
catch (Exception e)
{
e.printStackTrace();
}
out.println("</body></html>");
}

Not providing the file path is a security feature of the browser.
You have the file contents available in your code (InputStream filecontent) so you could use that or use one of the convenience methods on FileItem, e.g.
item.write(new File("/path/to/myfile.txt"));

how to display image which is outside server directory in strtus

This question is continuation to my previous question Accessing External Files Into Our Web Application, actually I am uploading file using struts tag <html:file property="file" />
But now I wanted to show the uploaded images from that location but I am getting src location as http://localhost:9443/D:/resources/images/img1.jpg which is not a valid path for that image.
How to access that image which is outside my server directory.
This is how I am sending Ajax response with Absolute path of images
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);
}
This is how I am rendering images at JSP
$(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'>";
});
but at img tag I am getting image path as http://localhost:9443/D:/resources/images/img1.jpg

Hi Below is the answer to my question, I have created ImageServlet for displaying image, steps to perform:
1. you need to add mapping in web.xml file:
<servlet-name>ImageServlet</servlet-name>
<url-pattern>/ImageServlet/*</url-pattern>
2. create 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. At jsp side you need to add the mapping in step 1 in your img tag i.e. input type='image':
<input type="image" alt='No image found' src='../ImageServlet/append image name that you want to display' />
You can even create Action class and use execute method for doing the same.

You can't render images in such way. Web server treated your image path as relative and add qualifying url location on the server. You should create an action to serve images, for example
<action path="/image" ... scope="request" validate="false"/>
Then render HTML like
cell.innerHTML="<img src='" + '/image?path=' + imagePath + "' alt='" + imagePath + "' height='42' width='42'>";
Now, create the action that write the binary image data to the response output stream. Take a parameter path in the action that let you find a file for binary output. After the flushing output return null so struts should not forward the action further. You could also add headers to turn off a Cache-Control to make sure the images are retrieved from the server.

How to read images from POST call parameter

I am looking at consuming an image that has been passed along with a post request as a parameter.
I am unable to understand how to consume this image with java restlet.
I want to use this image and send it for printing.
Thanks in advance.

Here is the code to put in a class extending ServerResource called at the POST of the form:
#Post
public Representation upload(Representation entity) throws Exception {
// 1/ Create a factory for disk-based file items
DiskFileItemFactory factory = new DiskFileItemFactory();
// 2/ Create a new file upload handler based on the Restlet
// FileUpload extension that will parse Restlet requests and
// generates FileItems.
RestletFileUpload upload = new RestletFileUpload(factory);
// 3/ Request is parsed by the handler which generates a
// list of FileItems
List<FileItem> items = upload.parseRepresentation(entity);
String filename = "myImage.png";
File file = new File(filename);
for (FileItem fi : items) {
if (fi.getFieldName().equals("datafile")) {
System.out.println("Save image ... " + filename);
fi.write(file);
}
}
// Print your image
...
}

Basic File upload in GWT

I'm trying to figure out how to upload one file using GWTs FileUpload widget. I'm using GWT and Google AppEngine with Java but I would like to upload file to my own Linux server.
I have the following code already but now I can't figure out how to submit my file to the Google AppServer server and save it to another server:
public class FileUploader{
private ControlPanel cp;
private FormPanel form = new FormPanel();
private FileUpload fu = new FileUpload();
public FileUploader(ControlPanel cp) {
this.cp = cp;
this.cp.setPrimaryArea(getFileUploaderWidget());
}
#SuppressWarnings("deprecation")
public Widget getFileUploaderWidget() {
form.setEncoding(FormPanel.ENCODING_MULTIPART);
form.setMethod(FormPanel.METHOD_POST);
// form.setAction(/* WHAT SHOULD I PUT HERE */);
VerticalPanel holder = new VerticalPanel();
fu.setName("upload");
holder.add(fu);
holder.add(new Button("Submit", new ClickHandler() {
public void onClick(ClickEvent event) {
GWT.log("You selected: " + fu.getFilename(), null);
form.submit();
}
}));
form.addSubmitHandler(new FormPanel.SubmitHandler() {
public void onSubmit(SubmitEvent event) {
if (!"".equalsIgnoreCase(fu.getFilename())) {
GWT.log("UPLOADING FILE????", null);
// NOW WHAT????
}
else{
event.cancel(); // cancel the event
}
}
});
form.addSubmitCompleteHandler(new FormPanel.SubmitCompleteHandler() {
public void onSubmitComplete(SubmitCompleteEvent event) {
Window.alert(event.getResults());
}
});
form.add(holder);
return form;
}
}
Now, what do I need to do next? What do i need to put in web.xml and how do I write my servlet so i can store file and return url of that object (if possible)

Here's the code from my app:
1) I created a class to accept http request:
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.InputStream;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.apache.commons.fileupload.FileItemIterator;
import org.apache.commons.fileupload.FileItemStream;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
public class FileUpload extends HttpServlet{
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
ServletFileUpload upload = new ServletFileUpload();
try{
FileItemIterator iter = upload.getItemIterator(request);
while (iter.hasNext()) {
FileItemStream item = iter.next();
String name = item.getFieldName();
InputStream stream = item.openStream();
// Process the input stream
ByteArrayOutputStream out = new ByteArrayOutputStream();
int len;
byte[] buffer = new byte[8192];
while ((len = stream.read(buffer, 0, buffer.length)) != -1) {
out.write(buffer, 0, len);
}
int maxFileSize = 10*(1024*1024); //10 megs max
if (out.size() > maxFileSize) {
throw new RuntimeException("File is > than " + maxFileSize);
}
}
}
catch(Exception e){
throw new RuntimeException(e);
}
}
}
2) Then in my web.xml I've added these lines:
<servlet>
<servlet-name>fileUploaderServlet</servlet-name>
<servlet-class>com.testapp.server.FileUpload</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>fileUploaderServlet</servlet-name>
<url-pattern>/testapp/fileupload</url-pattern>
</servlet-mapping>
3) And for form.action did this:
form.setAction(GWT.getModuleBaseURL()+"fileupload");

I would suggest using GWTUpload because it's dead simple to use and extend. You can add it to your project in less than 10 minutes and it supports GAE right out of the box (using GWTUpload-GAE). See the examples for some common usage scenarios.

In GWT, you can post the file to the server using http form methods, and you must use the supplied HttpServlet to accept and save the data as binary blogs in the Appengine BigTable.
Then, you need a second HttpServlet to read the file from bigtable, SET THE MIME TYPE IN THE HTTP HEADER {and caching options}, and then stream the file to the user.
Although RPC isn't NECESSARILY needed, you must let the client know what the generated fileId is so they can access it {unless you want to let user's supply the id and force them to worry about name overrides... ...ick}. Either you can use rpc to ask for a list of / single id {like "newest file id by user"}, or you can return that id in the body of the UploadServlet's response... but then you must make sure your post target is an in-page iframe, poll to make sure the iframe has a body between the submit event and the actual server response, and then parse and use that id in gwt to create an img or object tag that uses the file.
The key part is having one servlet for upload, and another to download. Remember, BigTable just stores binary blobs, so you also need your data entity to have a mime/content Type that can be read from the input file {never rely on file extensions!}. Also, there's a 1MB per entity in the BigTable, and a 10MB request limit for free accounts. You may wish to have your data entity contain a list of 1-10 blobs, each of which are a max 1024bytes.
Basically, your best bet is to find a working, free copy, like Google File Service, and extend it to learn how the system works.
If you wish, I will be posting my own open-source version of file handling, once I finish the gwt control widgets and can consider it all stable enough to be useful to anyone. Email x AT aiyx DOT info if you want me to send you a jar of betalicious code.

Here you go with a complete GWT fileupload with Progress bar
Here you can DOWNLOAD the source