As per our implementation we are dynamically creating some image files in the server side and sharing the URLs.
But we are unable to retrieve the image resources using the URLs if the file is created while the JETTY is running.
However we can retrieve it if we stop the jetty and start it again.
I would like to know if there is any configuration which will enable us to retrieve the resources without the jetty being stopped?
Is it necessary to persist the dynamic images? If not, you can just stream the dynamic image like this:
public void sendImage(RenderedImage image, String mimeType, String imageIOType, HttpServletResponse response)
{
if(image != null)
{
OutputStream output = null;
try
{
output = response.getOutputStream();
response.setContentType(mimeType);
ImageIO.write(image, imageIOType, output);
}
catch (IOException e)
{
e.printStackTrace();
}
if(output != null)
{
try
{
output.flush();
output.close();
}
catch (IOException e)
{
e.printStackTrace();
}
}
}
}
HTH
Related
public static void write(byte[] aInput, String aOutputFileName, String dirName) {
(new File(dirName)).mkdir();
try {
OutputStream output = null;
try {
output = new BufferedOutputStream(new FileOutputStream(dirName + "/" + aOutputFileName));
output.write(aInput);
} finally {
output.close();
}
} catch (FileNotFoundException ex) {
System.out.println("File not found.");
} catch (IOException ex) {
System.out.println(ex);
}
}
The code above is from a library i'm using and it is supposed to create an output file and write a byte array to it. I checked logcat and saw the Strict Mode Policy violation Write.toDisk. I understand what it is supposed to be for my questions are: (1) Does Strict mode prevent you from doing disk reads and write on the main thread? (2) Does that mean that the file or folder were not actually created? (3) How then would I go about creating a folder or file within my App that doesn't trigger this? (4) What is the recommended way to handle disk read / write off the main ui thread, a real world example would be appreciated
Thanks in Advance
(1) It turns out that Strict mode doesn't actually prevent you from making writes to the disk it just gives a warning. From Android Developer "StrictMode is a developer tool which detects things you might be doing by accident and brings them to your attention so you can fix them". https://developer.android.com/reference/android/os/StrictMode
(2) The files were actually being created it's just that I was just not familiar with writing and reading from disk
(3) There are numerous ways to go about creating files (i) first you get a hold of a file directory to write the file to:
context.getFilesDir()
(ii) then you get an outputstream writer (iii) then you write out the data with the writer
public void makeFile(String filename){
//Create temp file for filename
FileOutputStream fos = null;
try {
fos = new FileOutputStream(new File(filename));
fos.write(filename.getBytes());//Write the contents of the file to app folder
fos.flush();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}finally {
if(fos != null) {
try {
fos.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
(iv) finally you close the outputstream writer
(4) The recommended way is to use either an AsyncTask or some other background running class like FutureTask or to use Threads or Runnable:
public class DownloadFileThread implements Runnable{
public void run(){
//your code here
}
}
according to azure media services - The request body is too large and exceeds the maximum permissible limit large size video works well, but after uploaded ,asset can not been encoded,anyone can help me ?
my code like this:
#Override
protected void createBlobWriter(LocatorInfo uploadLocator, String fileName, InputStream input)
throws ServiceException {
try {
CloudBlobContainer container = new CloudBlobContainer(URI.create(uploadLocator.getPath()));
CloudBlockBlob blob = container.getBlockBlobReference(fileName);
blob.upload(input, input.available());
}
catch (StorageException e) {
e.printStackTrace();
}
catch (IOException e) {
e.printStackTrace();
}
catch (URISyntaxException e) {
e.printStackTrace();
}
}
do encode in https://portal.azure.cn,
and encode run error:
enter image description here
This error occurred because the fileName had a character (in this case, "?") which is not supported. If you can update your code to only allow alphanumeric characters(to be safe), then encoding should work. See the Note in https://learn.microsoft.com/en-us/azure/media-services/media-services-rest-upload-files for a list of unsupported characters.
When /api/upload REST endpoint is accessed I have a UploadController that uses a service UploadService to upload a file to an ftp server with org.apache.commons.net.ftp.FTPClient. I would like to be able to send information back to the user if the ftp client was unable to connect or timed out, or successfully sent the file. I have some IOException handling, but I don't know how to turn that around and send it back to the front-end. Any help appreciated, thanks!
public void upload(InputStream inputStream) {
String filename = "file.txt"
client = new FTPClient();
try {
client.connect("ftpsite");
client.login("username", "password");
client.storeFile(filename, inputStream);
} catch (IOException ioe) {
ioe.printStackTrace();
} finally {
try {
if (inputStream!= null) {
inputStream.close();
}
} catch (IOException ioe) {
ioe.printStackTrace();
}
}
return null;
}
You should throw a new Exception in your catch statement.
For example, you could create a RequestTimeoutException class:
#ResponseStatus(HttpStatus.REQUEST_TIMEOUT)
public class RequestTimeoutException extends RuntimeException { }
and then throw it when need be:
catch (IOException ioe) {
//do some logging while you're at it
throw new RequestTimeoutException();
}
I use a servlet to access a folder outside the web container to load some graphics to web application by using GWT. I use the following snippet in servlet to test the idea:
String s = null;
File inputFile = new File("C:\\Documents and Settings\\User\\My Documents\\My Pictures\\megan-fox.jpg");
FileInputStream fin = null;
try {
fin = new FileInputStream(inputFile);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
byte c[] = new byte[(int) inputFile.length()];
try {
fin.read(c);
} catch (IOException e) {
e.printStackTrace();
}
try {
fin.close();
} catch (IOException e1) {
e1.printStackTrace();
}
String imgFolderPath = getServletContext().getRealPath("/")+"img";
File imgFolder = new File(imgFolderPath);
imgFolder.mkdir();
File newImage = new File("megan-fox.jpg");
FileOutputStream fout = null;
try {
fout = new FileOutputStream(newImage);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
try {
fout.write(c);
} catch (IOException e) {
e.printStackTrace();
}
try {
fout.close();
} catch (IOException e) {
e.printStackTrace();
}
boolean success = newImage.renameTo(new File(imgFolderPath, newImage.getName()));
The code in servlet reads the image file from the specified folder in hard disk, creates a new folder called 'img' in war folder and copies to it the jpg file. Then it returns to the client the path to the image (for now hardcoded as) '/img/megan-fox.jpg'.
The client then uses the Image class in GWT with the returned path-string to display the image, like in the following snippet:
public void onSuccess(String result) {
String myImage = result;
image = new Image(myImage);
RootPanel.get().add(image);
closeButton.setFocus(true);
}
I need to know if there is a way to achieve the same result without using the 'intermediate' step of creating a folder in the web container root (optional) and copying the file there in order to access it with Image GWT class and display it?
updated: The original servlet class.
public class GreetingServiceImpl extends RemoteServiceServlet implements
GreetingService {
// This method is called by the servlet container to process a GET request.
public void doGet(HttpServletRequest req, HttpServletResponse resp) throws IOException {
// Get the absolute path of the image
ServletContext sc = getServletContext();
// i want to load the image in the specified folder (outside the web container)
String filename = sc.getRealPath("C:\\Documents and Settings\\User\\My Documents\\My Pictures\\megan-fox.jpg");
// Get the MIME type of the image
String mimeType = sc.getMimeType(filename);
if (mimeType == null) {
sc.log("Could not get MIME type of "+filename);
resp.setStatus(HttpServletResponse.SC_INTERNAL_SERVER_ERROR);
return;
}
// Set content type
resp.setContentType(mimeType);
// Set content size
File file = new File(filename);
resp.setContentLength((int)file.length());
// Open the file and output streams
FileInputStream in = new FileInputStream(file);
OutputStream out = resp.getOutputStream();
// Copy the contents of the file to the output stream
byte[] buf = new byte[1024];
int count = 0;
while ((count = in.read(buf)) >= 0) {
out.write(buf, 0, count);
}
in.close();
out.close();
}
// This is the method that is called from the client using GWT-RPC
public String greetServer(String input) throws IllegalArgumentException {
HttpServletRequest req = this.getThreadLocalRequest();
HttpServletResponse res = this.getThreadLocalResponse();
try {
doGet(req, res);
} catch (IOException e) {
e.printStackTrace();
}
// actually i dont know what that means but i thought i would have to returned something like the image's url?
return res.encodeURL("/img/image0.png");
}
}
I logically misused the method that was proposed to solve my problem. What is the correct way?
Sure, just have your servlet serve the image directly:
Set the Content-Type header to image/jpeg.
Write out image file contents to servlet response writer.
Here is an example.
I' m looking for a code to save the files created in a applet normally text files i want to save them on a server directory how can i do so.
Here is an example of how to send a String. In fact any Object can be sent this method so long as it's serializable and the same version of the Object exists on both the applet and the servlet.
To send from the applet
public void sendSomeString(String someString) {
ObjectOutputStream request = null;
try {
URL servletURL = new URL(getCodeBase().getProtocol(),
getCodeBase().getHost(),
getCodeBase().getPort(),
"/servletName");
// open the connection
URLConnection con = servletURL.openConnection();
con.setDoOutput(true);
con.setUseCaches(false);
con.setRequestProperty("Content-Type", "application/octet-stream");
// send the data
request =
new ObjectOutputStream(
new BufferedOutputStream(con.getOutputStream()));
request.writeObject(someString);
request.flush();
// performs the connection
new ObjectInputStream(new BufferedInputStream(con.getInputStream()));
} catch (Exception e) {
System.err.println("" + e);
} finally {
if (request != null) {
try {
request.close();
} catch (Exception e) {
System.err.println("" + e);
};
}
}
}
To retrieve on the server side
#Override
public void doPost(HttpServletRequest request, HttpServletResponse response) {
try {
// get the input stream
ObjectInputStream inputStream = new ObjectInputStream(
new BufferedInputStream(request.getInputStream()));
String someString = (String)inputStream.readObject();
ObjectOutputStream oos = new ObjectOutputStream(
new BufferedOutputStream(response.getOutputStream()));
oos.flush();
// handle someString....
} catch (SocketException e) {
// ignored, occurs when connection is terminated
} catch (IOException e) {
// ignored, occurs when connection is terminated
} catch (Exception e) {
log.error("Exception", e);
}
}
No one is going to hand you this on a plate. You have to write code in your applet to make a socket connection back to your server and send the data. One way to approach this is to push the data via HTTP, and use a library such as commons-httpclient. That requires your server to handle the appropriate HTTP verb.
There are many other options, and the right one will depend on the fine details of the problem you are trying to solve.