What is the best way to upload a directory in grails ?
I try this code :
def upload = {
if(request.method == 'POST') {
Iterator itr = request.getFileNames();
while(itr.hasNext()) {
MultipartFile file = request.getFile(itr.next());
File destination = new File(file.getOriginalFilename())
if (!file.isEmpty()) {
file.transferTo(destination)
// success
}
else
{
// failure
}
}
response.sendError(200,'Done');
}
}
Unfortunately, I can only upload file by file.
I would like to define my directory, and upload all files directly.
Any ideas ?
There is one major misconception here. The code which you posted will only work if both the server and the client runs at physically the same machine (which won't occur in real world) and if you're using the MSIE browser which has the misbehaviour to send the full path along the filename.
You should in fact get the contents of the uploaded file as an InputStream and write it to any OutputStream the usual Java IO way. The filename can be used to create a file with the same name at the server side, but you'll ensure that you strip the incorrectly by MSIE sent path from the filename.
As to your actual functional requirement, HTML doesn't provide facilities to upload complete directories or multiple files by a single <input type="file"> element. You'll need to create a client application which is capable of this and serve this from your webpage, like a Java Applet using Swing JFileChooser. There exist 3rd party solutions for this, like JumpLoader.
Related
As part of a search application, I want the user to be able to download a report showing the results in a CSV file. I have the following method:
public void downloadCustomerResults(String customer) {
String output = "";
output += produceCustomerID(customer);
output += produceCustomerAddress(customer);
output += produceCustomerContactDetails(customer);
output += produceOrderHeader(customer);
output += producePayments(customer);
// Writes to server desktop, not user desktop.
try {
Writer fileWriter = new FileWriter("C:\\Users\\username\\Desktop\\SAR" + customer + "C.csv");
fileWriter.write(output);
fileWriter.flush();
fileWriter.close();
} catch (IOException e) {
e.printStackTrace();
}
}
This downloads the file to the desktop of the machine running the server, not the user's desktop (accessing the app via JSP's on Tomcat). How would I change the file path string to make this download to the users' desktop? Or would I have to pass the file to the JSP for the user to download via their browser?
Thanks.
Short answer: The server has no means of accessing the client's filesystem.
Longer answer: You might either provide a service for the client to download the file (e.g. a webservice accessible through a URI, like #Kayaman mentioned) or the client provides you a service to write the file (e.g. a remote file system, an FTP server etc.). For the latter there might be libraries providing a special java.nio.FileSystem extension.
You may also provide an application running on the client to receive the file. This client application will then have acces to the client's file systems (unless it lacks the access rights, of course).
So the answer I found was to use the JavaScript package FileSaver.js.
This accepts a blob created from a string, and then saves it with a filename of your choice to the browsers preferred download folder.
I managed to pass the string from Java to JavaScript, and then pass it through FileSaver.js in the .JSP page.
I have Spring MVC web app running on Tomcat.
I upload a file and save it in the /tmp folder on the file system.
Then I need to show a link to that file in the view (Thymeleaf), so that the user can download the file by clicking on the link. How to do that?
I've heard about configuring Tomcat to allow a specific context to link to a folder on the FS, but not sure how to do that or if that is the only solution. Please help.
The way I approach this is slightly different. Basically I use two controller actions for handling file uploads, one for uploading, and for downloading (viewing) files.
So upload action would save files to some preconfigured directory on the file system, I assume you already have that part working.
Then declare download action similar to this
#Controller
public class FileController {
#RequestMapping("/get-file/{filename}")
public void getFileAction(#RequestParam filename, HttpServletResponse response) {
// Here check if file with given name exists in preconfigured upload folder
// If it does, write it to response's output stream and set correct response headers
// If it doesn't return 404 status code
}
}
If you want to make impossible to download file just by knowing the name, after uploading file, save some meta info to the database (or any other storage) and assign some hash (random id) to it. Then, in getFileAction, use this hash to look for file, not the original filename.
Finally, I would discourage using /tmp for file uploads. It depends on the system/application used, but generally temp directory are meant, as name suggest, for temporary data. Usually it is guaranteed data in the temp directory will stay for "reasonable time", but applications must take into account that content of temp directory can be deleted anytime.
This is the precisely setup that worked for me (Tomcat 8, SpringMVC, boot):
server.xml:
<Context docBase="C:\tmp\" path="/images" />
In the controller:
public String createNewsSource(#ModelAttribute("newsSource") NewsSource source, BindingResult result, Model model,
#RequestParam("attachment") final MultipartFile attachment) {
new NewsSourceValidator().validate(source, result);
if (result.hasErrors()) {
return "source/addNewSource";
}
if (!attachment.isEmpty()) {
try {
byte[] bytes = attachment.getBytes();
BufferedOutputStream stream = new BufferedOutputStream(
new FileOutputStream(new File("/tmp/" + attachment.getOriginalFilename())));
stream.write(bytes);
stream.close();
} catch (Exception e) {
e.printStackTrace();
}
}
source.setLogo("images/" + attachment.getOriginalFilename());
newsSourceService.createNewsSourceIfNotExist(source);
return "redirect:/sources/list";
}
As you can see I am saving the file to /tmp, but in the DB (source.setLogo()), I am pointing to images as mapped in server.xml
Here's where I found about Tomcat config:
If the images are all located outside the webapp and you want to have
Tomcat's DefaultServlet to handle them, then all you basically need to
do in Tomcat is to add the following Context element to
/conf/server.xml inside tag:
This way they'll be accessible through http://example.com/images/....
SO answer to a similar question
Hi I am trying to get the returned file path by my JNLP file chooser. Here's my code.
I don't know how and where to get the file path. is it from fileContents? fileConents.getfilepath something like that?
try {
if (fileOpenService==null) {
fileOpenService = (FileOpenService)ServiceManager.
lookup("javax.jnlp.FileOpenService");
}
fileContents = fileOpenService.openFileDialog(path, xtns);
} catch(UnavailableServiceException use) {
use.printStackTrace();
} catch(IOException ioe) {
ioe.printStackTrace();
}
Thanks in advance!
According to http://docs.oracle.com/javase/tutorial/uiswing/components/filechooser.html
You can call other methods on the File object, such as
getPath, isDirectory, or exists to obtain information about the file.
You can also call other methods such as delete and rename to change
the file in some way. Of course, you might also want to open or save
the file by using one of the reader or writer classes provided by the
Java platform. See Basic I/O for information about using readers and
writers to read and write data to the file system.
It is for security reasons that a FileContents will not return a path. The JRE asked the user if our app. could access the content of that file, not it's path.
It is a bit like the brower/HTML based file upload field. Some browsers provide the entire path, while more typically it is just the content/name.
I am trying to create a method that displays a list of files in a given directory. This works fine for normal directories (on disk) but when I enter a url my list of files is null.
public void getListOfFiles(String folderLocation){
File folder = new File(folderLocation);
File[] listFiles = folder.listFiles();
for(int i = 0; i < 10; i++){
System.out.println(listFiles[i]);
}
}
I think my problem is because the File 'folder' is removing one of the '/' in my folderLocation (http://...)
I have tried using URL and URI but have had no luck! Can anyone help?
First of all, File won't work for this as it's not networking-aware.
Secondly, in general there's no mechanism to list files over plain HTTP. If the HTTP server gives you some kind of a listing page when you present it with the URL, you'll have to download the page using, for example, URLConnection and parse it yourself.
To list files over FTP, you could use FTPClient from Apache Commons Net.
A webpage contains a link to an executable (i.e. If we click on the link, the browser will download the file on your local machine).
Is there any way to achieve the same functionality with Java?
Thank you
Yes there is.
Here a simple example:
You can have a JSF(Java Server Faces) page, with a supporting backing bean that contains a method annotated with #PostConstruct This means that any action(for example downloading), will occur when the page is created.
There is already a question very similar already, have a look at: Invoke JSF managed bean action on page load
You can use Java's, URL class to download a file, but it requires a little work. You will need to do the following:
Create the URL object point at the file
Call openStream() to get an InputStream
Open the file you want to write to (a FileOutputStream)
Read from the InputStream and write to the file, until there is no more data left to read
Close the input and output streams
It doesn't really matter what type of file you are downloading (the fact that it's an executable file is irrelevant) since the process is the same for any type of file.
Update: It sounds like what you actually want is to plug the URL of a webpage into the Java app, and have the Java app find the link in the page and then download that link. If that is the case, the wording of your question is very unclear, but here are the basic steps I would use:
First, use steps 1 and 2 above to get an InputStream for the page
Use something like TagSoup or jsoup to parse the HTML
Find the <a> element that you want and extract its href attribute to get the URL of the file you need to download (if it's a relative URL instead of absolute, you will need to resolve that URL against the URL of the original page)
Use the steps above to download that URL
Here's a slight shortcut, based on jsoup (which I've never used before, I'm just writing this from snippets stolen from their webpage). I've left out a lot of error checking, but hey, I usually charge for this:
Document doc = Jsoup.connect(pageUrl).get();
Element aElement = doc.getElementsByTag("a").first() // Obviously you may need to refine this
String newUrl = aElement.attr("abs:href"); // This is a piece of jsoup magic that ensures that the destination URL is absolute
// assert newUrl != null
URL fileUrl = new URL(newUrl);
String destPath = fileUrl.getPath();
int lastSlash = destPath.lastIndexOf('/');
if (lastSlash != -1) {
destPath = destPath.substring(lastSlash);
}
// Assert that this is really a valid filename
// Now just download fileUrl and save it to destPath
The proper way to determine what the destination filename should be (unless you hardcode it) is actually to look for the Content-Disposition header, and look for the bit after filename=. In that case, you can't use openStream() on the URL, you will need to use openConnection() instead, to get a URLConnection. Then you can use getInputStream() to get your InputStream and getRequestProperty("Content-Disposition") to get the header to figure out your filename. In case that header is missing or malformed, you should then fall-back to using the method above to determine the destination filename.
You can do this using apache commons IO FileUtils
http://commons.apache.org/io/apidocs/org/apache/commons/io/FileUtils.html#copyURLToFile(java.net.URL, java.io.File)
Edit:
I was able to successfully download a zip file from source forge site (it is not empty), It did some thing like this
import java.io.File;
import java.net.URL;
import org.apache.commons.io.FileUtils;
public class Test
{
public static void main(String args[])
{
try {
URL url = new URL("http://sourceforge.net/projects/gallery/files/gallery3/3.0.2/gallery-3.0.2.zip/download");
FileUtils.copyURLToFile(url, new File("test.zip"));
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
I was able successfully download tomcat.exe too
URL url = new URL("http://archive.apache.org/dist/tomcat/tomcat-6/v6.0.16/bin/apache-tomcat-6.0.16.exe");