I create ImageServlet to refer to videos out of my web application scope.
The location of all of my videos are on a intranet location that could be reached from any computer in the intranet:
String path = "\\myip\storage\ogg\VX-276.ogg"
In my application, when I write it as URL - it can't display it!
If I try to open it with chrome it automatically changes it to file://myip/storage/ogg/VX-276.ogg and the file is being displayed.
I tried to do so: file:////odelyay_test64/storage/ogg/
as well but Java converts the string to: file:\myip\storage\ogg\VX-276.ogg which does not exist!
What is the correct way to refer to it?
EDITED
I create a small test:
String path = "file://myip/storage/ogg/VX-276.ogg";
File file = new File(path);
if (file.exists())
System.out.println("exists");
else {
System.out.println("missing" + file.getPath());
}
and I get:
missing file:\myip\storage\ogg\VX-276.ogg
As you can see the slashes are being switched
As per your previous question, you're referencing the resource in a HTML <video> tag. All URLs in the HTML source code must be http:// URLs (or at least be relative to a http:// URL). Most browsers namely refuse to load resources from file:// URLs when the HTML page is itself been requested by http://. You just need to let the URL point to the servlet. If the servlet's doGet() method get hit, then the URL is fine and you should not change it.
Your concrete problem is in the way how you open and read the desired file in the servlet. You need to ensure that the path in File file = new File(path) points to a valid location before you open a FileInputStream on it.
String path = "file://myip/storage/ogg/VX-276.ogg";
File file = new File(path);
// ...
If the servlet code is well written that it doesn't suppress/swallow exceptions and you have read the server logs, then you should have seen an IOException such as FileNotFoundException along with a self-explaining message in the server logs whenever reading the file fails. Go read the server logs.
Update as per the comments, it turns out that you're using Windows and thus file:// on a network disk isn't going to work for Java without mapping it on a drive letter. You need to map //myip on a drive letter first, for example X:.
String path = "X:/storage/ogg/VX-276.ogg";
File file = new File(path);
// ...
in the end I used VFS library of apache and my code looks like this:
public static void main(String[] args) {
FileSystemManager fsManager = null;
String path = "\\\\myip\\storage\\ogg\\VX-276.ogg";
try {
fsManager = VFS.getManager();
FileObject basePath;
basePath = fsManager.resolveFile("file:" + path);
if (basePath.exists())
System.out.println("exists");
else {
System.out.println("missing" + basePath.getURL());
}
} catch (FileSystemException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
In this way, I don't need to create a driver for each user of the system and it allows me not to depend on operation system!
Related
I read here that one should not save the file in the server anyway as it is not portable, transactional and requires external parameters. However, given that I need a tmp solution for tomcat (7) and that I have (relative) control over the server machine I want to know :
What is the best place to save the file ? Should I save it in /WEB-INF/uploads (advised against here) or someplace under $CATALINA_BASE (see here) or ... ? The JavaEE 6 tutorial gets the path from the user (:wtf:). NB : The file should not be downloadable by any means.
Should I set up a config parameter as detailed here ? I'd appreciate some code (I'd rather give it a relative path - so it is at least Tomcat portable) - Part.write() looks promising - but apparently needs a absolute path
I'd be interested in an exposition of the disadvantages of this approach vs a database/JCR repository one
Unfortunately the FileServlet by #BalusC concentrates on downloading files, while his answer on uploading files skips the part on where to save the file.
A solution easily convertible to use a DB or a JCR implementation (like jackrabbit) would be preferable.
Store it anywhere in an accessible location except of the IDE's project folder aka the server's deploy folder, for reasons mentioned in the answer to Uploaded image only available after refreshing the page:
Changes in the IDE's project folder does not immediately get reflected in the server's work folder. There's kind of a background job in the IDE which takes care that the server's work folder get synced with last updates (this is in IDE terms called "publishing"). This is the main cause of the problem you're seeing.
In real world code there are circumstances where storing uploaded files in the webapp's deploy folder will not work at all. Some servers do (either by default or by configuration) not expand the deployed WAR file into the local disk file system, but instead fully in the memory. You can't create new files in the memory without basically editing the deployed WAR file and redeploying it.
Even when the server expands the deployed WAR file into the local disk file system, all newly created files will get lost on a redeploy or even a simple restart, simply because those new files are not part of the original WAR file.
It really doesn't matter to me or anyone else where exactly on the local disk file system it will be saved, as long as you do not ever use getRealPath() method. Using that method is in any case alarming.
The path to the storage location can in turn be definied in many ways. You have to do it all by yourself. Perhaps this is where your confusion is caused because you somehow expected that the server does that all automagically. Please note that #MultipartConfig(location) does not specify the final upload destination, but the temporary storage location for the case file size exceeds memory storage threshold.
So, the path to the final storage location can be definied in either of the following ways:
Hardcoded:
File uploads = new File("/path/to/uploads");
Environment variable via SET UPLOAD_LOCATION=/path/to/uploads:
File uploads = new File(System.getenv("UPLOAD_LOCATION"));
VM argument during server startup via -Dupload.location="/path/to/uploads":
File uploads = new File(System.getProperty("upload.location"));
*.properties file entry as upload.location=/path/to/uploads:
File uploads = new File(properties.getProperty("upload.location"));
web.xml <context-param> with name upload.location and value /path/to/uploads:
File uploads = new File(getServletContext().getInitParameter("upload.location"));
If any, use the server-provided location, e.g. in JBoss AS/WildFly:
File uploads = new File(System.getProperty("jboss.server.data.dir"), "uploads");
Either way, you can easily reference and save the file as follows:
File file = new File(uploads, "somefilename.ext");
try (InputStream input = part.getInputStream()) {
Files.copy(input, file.toPath());
}
Or, when you want to autogenerate an unique file name to prevent users from overwriting existing files with coincidentally the same name:
File file = File.createTempFile("somefilename-", ".ext", uploads);
try (InputStream input = part.getInputStream()) {
Files.copy(input, file.toPath(), StandardCopyOption.REPLACE_EXISTING);
}
How to obtain part in JSP/Servlet is answered in How to upload files to server using JSP/Servlet? and how to obtain part in JSF is answered in How to upload file using JSF 2.2 <h:inputFile>? Where is the saved File?
Note: do not use Part#write() as it interprets the path relative to the temporary storage location defined in #MultipartConfig(location). Also make absolutely sure that you aren't corrupting binary files such as PDF files or image files by converting bytes to characters during reading/writing by incorrectly using a Reader/Writer instead of InputStream/OutputStream.
See also:
How to save uploaded file in JSF (JSF-targeted, but the principle is pretty much the same)
Simplest way to serve static data from outside the application server in a Java web application (in case you want to serve it back)
How to save generated file temporarily in servlet based web application
I post my final way of doing it based on the accepted answer:
#SuppressWarnings("serial")
#WebServlet("/")
#MultipartConfig
public final class DataCollectionServlet extends Controller {
private static final String UPLOAD_LOCATION_PROPERTY_KEY="upload.location";
private String uploadsDirName;
#Override
public void init() throws ServletException {
super.init();
uploadsDirName = property(UPLOAD_LOCATION_PROPERTY_KEY);
}
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
// ...
}
#Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
Collection<Part> parts = req.getParts();
for (Part part : parts) {
File save = new File(uploadsDirName, getFilename(part) + "_"
+ System.currentTimeMillis());
final String absolutePath = save.getAbsolutePath();
log.debug(absolutePath);
part.write(absolutePath);
sc.getRequestDispatcher(DATA_COLLECTION_JSP).forward(req, resp);
}
}
// helpers
private static String getFilename(Part part) {
// courtesy of BalusC : http://stackoverflow.com/a/2424824/281545
for (String cd : part.getHeader("content-disposition").split(";")) {
if (cd.trim().startsWith("filename")) {
String filename = cd.substring(cd.indexOf('=') + 1).trim()
.replace("\"", "");
return filename.substring(filename.lastIndexOf('/') + 1)
.substring(filename.lastIndexOf('\\') + 1); // MSIE fix.
}
}
return null;
}
}
where :
#SuppressWarnings("serial")
class Controller extends HttpServlet {
static final String DATA_COLLECTION_JSP="/WEB-INF/jsp/data_collection.jsp";
static ServletContext sc;
Logger log;
// private
// "/WEB-INF/app.properties" also works...
private static final String PROPERTIES_PATH = "WEB-INF/app.properties";
private Properties properties;
#Override
public void init() throws ServletException {
super.init();
// synchronize !
if (sc == null) sc = getServletContext();
log = LoggerFactory.getLogger(this.getClass());
try {
loadProperties();
} catch (IOException e) {
throw new RuntimeException("Can't load properties file", e);
}
}
private void loadProperties() throws IOException {
try(InputStream is= sc.getResourceAsStream(PROPERTIES_PATH)) {
if (is == null)
throw new RuntimeException("Can't locate properties file");
properties = new Properties();
properties.load(is);
}
}
String property(final String key) {
return properties.getProperty(key);
}
}
and the /WEB-INF/app.properties :
upload.location=C:/_/
HTH and if you find a bug let me know
I'm trying to get a load a mp4 file from the android file browser. However the path from the following code does not work. It throws an exception
Uri currFileURI = data.getData();
String path=currFileURI.getPath();
MediaExtractor extractor = new MediaExtractor();
try {
extractor.setDataSource(path);
} catch( IOException e) {
e.printStackTrace();
return;
}
Data I have:
data = {Intent#9869} "Intent { dat=content://com.android.providers.downloads.documents/document/44 flg=0x1 }"
currFileURI = {Uri$HierarchicalUri#9870} "content://com.android.providers.downloads.documents/document/44"
path = "/document/44"
I see this code in stackoverflow:
How to get the Full file path from URI
This looks like an overkill
I can see the method takes in a file path.
https://developer.android.com/reference/android/media/MediaExtractor#setDataSource(java.lang.String)
So it looks the path from the android file browser is different from what the method wants. I also see that the path from the file browser is different than the file name. Anybody have an insight into what the path should look like?
So it looks the path from the android file browser is different from what the method wants
It is not a filesystem path, because a Uri is not a file.
I can see the method takes in a file path
There are many forms of setDataSource() on MediaExtractor, including one that takes a Uri. Try using that method with your Uri.
I read here that one should not save the file in the server anyway as it is not portable, transactional and requires external parameters. However, given that I need a tmp solution for tomcat (7) and that I have (relative) control over the server machine I want to know :
What is the best place to save the file ? Should I save it in /WEB-INF/uploads (advised against here) or someplace under $CATALINA_BASE (see here) or ... ? The JavaEE 6 tutorial gets the path from the user (:wtf:). NB : The file should not be downloadable by any means.
Should I set up a config parameter as detailed here ? I'd appreciate some code (I'd rather give it a relative path - so it is at least Tomcat portable) - Part.write() looks promising - but apparently needs a absolute path
I'd be interested in an exposition of the disadvantages of this approach vs a database/JCR repository one
Unfortunately the FileServlet by #BalusC concentrates on downloading files, while his answer on uploading files skips the part on where to save the file.
A solution easily convertible to use a DB or a JCR implementation (like jackrabbit) would be preferable.
Store it anywhere in an accessible location except of the IDE's project folder aka the server's deploy folder, for reasons mentioned in the answer to Uploaded image only available after refreshing the page:
Changes in the IDE's project folder does not immediately get reflected in the server's work folder. There's kind of a background job in the IDE which takes care that the server's work folder get synced with last updates (this is in IDE terms called "publishing"). This is the main cause of the problem you're seeing.
In real world code there are circumstances where storing uploaded files in the webapp's deploy folder will not work at all. Some servers do (either by default or by configuration) not expand the deployed WAR file into the local disk file system, but instead fully in the memory. You can't create new files in the memory without basically editing the deployed WAR file and redeploying it.
Even when the server expands the deployed WAR file into the local disk file system, all newly created files will get lost on a redeploy or even a simple restart, simply because those new files are not part of the original WAR file.
It really doesn't matter to me or anyone else where exactly on the local disk file system it will be saved, as long as you do not ever use getRealPath() method. Using that method is in any case alarming.
The path to the storage location can in turn be definied in many ways. You have to do it all by yourself. Perhaps this is where your confusion is caused because you somehow expected that the server does that all automagically. Please note that #MultipartConfig(location) does not specify the final upload destination, but the temporary storage location for the case file size exceeds memory storage threshold.
So, the path to the final storage location can be definied in either of the following ways:
Hardcoded:
File uploads = new File("/path/to/uploads");
Environment variable via SET UPLOAD_LOCATION=/path/to/uploads:
File uploads = new File(System.getenv("UPLOAD_LOCATION"));
VM argument during server startup via -Dupload.location="/path/to/uploads":
File uploads = new File(System.getProperty("upload.location"));
*.properties file entry as upload.location=/path/to/uploads:
File uploads = new File(properties.getProperty("upload.location"));
web.xml <context-param> with name upload.location and value /path/to/uploads:
File uploads = new File(getServletContext().getInitParameter("upload.location"));
If any, use the server-provided location, e.g. in JBoss AS/WildFly:
File uploads = new File(System.getProperty("jboss.server.data.dir"), "uploads");
Either way, you can easily reference and save the file as follows:
File file = new File(uploads, "somefilename.ext");
try (InputStream input = part.getInputStream()) {
Files.copy(input, file.toPath());
}
Or, when you want to autogenerate an unique file name to prevent users from overwriting existing files with coincidentally the same name:
File file = File.createTempFile("somefilename-", ".ext", uploads);
try (InputStream input = part.getInputStream()) {
Files.copy(input, file.toPath(), StandardCopyOption.REPLACE_EXISTING);
}
How to obtain part in JSP/Servlet is answered in How to upload files to server using JSP/Servlet? and how to obtain part in JSF is answered in How to upload file using JSF 2.2 <h:inputFile>? Where is the saved File?
Note: do not use Part#write() as it interprets the path relative to the temporary storage location defined in #MultipartConfig(location). Also make absolutely sure that you aren't corrupting binary files such as PDF files or image files by converting bytes to characters during reading/writing by incorrectly using a Reader/Writer instead of InputStream/OutputStream.
See also:
How to save uploaded file in JSF (JSF-targeted, but the principle is pretty much the same)
Simplest way to serve static data from outside the application server in a Java web application (in case you want to serve it back)
How to save generated file temporarily in servlet based web application
I post my final way of doing it based on the accepted answer:
#SuppressWarnings("serial")
#WebServlet("/")
#MultipartConfig
public final class DataCollectionServlet extends Controller {
private static final String UPLOAD_LOCATION_PROPERTY_KEY="upload.location";
private String uploadsDirName;
#Override
public void init() throws ServletException {
super.init();
uploadsDirName = property(UPLOAD_LOCATION_PROPERTY_KEY);
}
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
// ...
}
#Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
Collection<Part> parts = req.getParts();
for (Part part : parts) {
File save = new File(uploadsDirName, getFilename(part) + "_"
+ System.currentTimeMillis());
final String absolutePath = save.getAbsolutePath();
log.debug(absolutePath);
part.write(absolutePath);
sc.getRequestDispatcher(DATA_COLLECTION_JSP).forward(req, resp);
}
}
// helpers
private static String getFilename(Part part) {
// courtesy of BalusC : http://stackoverflow.com/a/2424824/281545
for (String cd : part.getHeader("content-disposition").split(";")) {
if (cd.trim().startsWith("filename")) {
String filename = cd.substring(cd.indexOf('=') + 1).trim()
.replace("\"", "");
return filename.substring(filename.lastIndexOf('/') + 1)
.substring(filename.lastIndexOf('\\') + 1); // MSIE fix.
}
}
return null;
}
}
where :
#SuppressWarnings("serial")
class Controller extends HttpServlet {
static final String DATA_COLLECTION_JSP="/WEB-INF/jsp/data_collection.jsp";
static ServletContext sc;
Logger log;
// private
// "/WEB-INF/app.properties" also works...
private static final String PROPERTIES_PATH = "WEB-INF/app.properties";
private Properties properties;
#Override
public void init() throws ServletException {
super.init();
// synchronize !
if (sc == null) sc = getServletContext();
log = LoggerFactory.getLogger(this.getClass());
try {
loadProperties();
} catch (IOException e) {
throw new RuntimeException("Can't load properties file", e);
}
}
private void loadProperties() throws IOException {
try(InputStream is= sc.getResourceAsStream(PROPERTIES_PATH)) {
if (is == null)
throw new RuntimeException("Can't locate properties file");
properties = new Properties();
properties.load(is);
}
}
String property(final String key) {
return properties.getProperty(key);
}
}
and the /WEB-INF/app.properties :
upload.location=C:/_/
HTH and if you find a bug let me know
I read here that one should not save the file in the server anyway as it is not portable, transactional and requires external parameters. However, given that I need a tmp solution for tomcat (7) and that I have (relative) control over the server machine I want to know :
What is the best place to save the file ? Should I save it in /WEB-INF/uploads (advised against here) or someplace under $CATALINA_BASE (see here) or ... ? The JavaEE 6 tutorial gets the path from the user (:wtf:). NB : The file should not be downloadable by any means.
Should I set up a config parameter as detailed here ? I'd appreciate some code (I'd rather give it a relative path - so it is at least Tomcat portable) - Part.write() looks promising - but apparently needs a absolute path
I'd be interested in an exposition of the disadvantages of this approach vs a database/JCR repository one
Unfortunately the FileServlet by #BalusC concentrates on downloading files, while his answer on uploading files skips the part on where to save the file.
A solution easily convertible to use a DB or a JCR implementation (like jackrabbit) would be preferable.
Store it anywhere in an accessible location except of the IDE's project folder aka the server's deploy folder, for reasons mentioned in the answer to Uploaded image only available after refreshing the page:
Changes in the IDE's project folder does not immediately get reflected in the server's work folder. There's kind of a background job in the IDE which takes care that the server's work folder get synced with last updates (this is in IDE terms called "publishing"). This is the main cause of the problem you're seeing.
In real world code there are circumstances where storing uploaded files in the webapp's deploy folder will not work at all. Some servers do (either by default or by configuration) not expand the deployed WAR file into the local disk file system, but instead fully in the memory. You can't create new files in the memory without basically editing the deployed WAR file and redeploying it.
Even when the server expands the deployed WAR file into the local disk file system, all newly created files will get lost on a redeploy or even a simple restart, simply because those new files are not part of the original WAR file.
It really doesn't matter to me or anyone else where exactly on the local disk file system it will be saved, as long as you do not ever use getRealPath() method. Using that method is in any case alarming.
The path to the storage location can in turn be definied in many ways. You have to do it all by yourself. Perhaps this is where your confusion is caused because you somehow expected that the server does that all automagically. Please note that #MultipartConfig(location) does not specify the final upload destination, but the temporary storage location for the case file size exceeds memory storage threshold.
So, the path to the final storage location can be definied in either of the following ways:
Hardcoded:
File uploads = new File("/path/to/uploads");
Environment variable via SET UPLOAD_LOCATION=/path/to/uploads:
File uploads = new File(System.getenv("UPLOAD_LOCATION"));
VM argument during server startup via -Dupload.location="/path/to/uploads":
File uploads = new File(System.getProperty("upload.location"));
*.properties file entry as upload.location=/path/to/uploads:
File uploads = new File(properties.getProperty("upload.location"));
web.xml <context-param> with name upload.location and value /path/to/uploads:
File uploads = new File(getServletContext().getInitParameter("upload.location"));
If any, use the server-provided location, e.g. in JBoss AS/WildFly:
File uploads = new File(System.getProperty("jboss.server.data.dir"), "uploads");
Either way, you can easily reference and save the file as follows:
File file = new File(uploads, "somefilename.ext");
try (InputStream input = part.getInputStream()) {
Files.copy(input, file.toPath());
}
Or, when you want to autogenerate an unique file name to prevent users from overwriting existing files with coincidentally the same name:
File file = File.createTempFile("somefilename-", ".ext", uploads);
try (InputStream input = part.getInputStream()) {
Files.copy(input, file.toPath(), StandardCopyOption.REPLACE_EXISTING);
}
How to obtain part in JSP/Servlet is answered in How to upload files to server using JSP/Servlet? and how to obtain part in JSF is answered in How to upload file using JSF 2.2 <h:inputFile>? Where is the saved File?
Note: do not use Part#write() as it interprets the path relative to the temporary storage location defined in #MultipartConfig(location). Also make absolutely sure that you aren't corrupting binary files such as PDF files or image files by converting bytes to characters during reading/writing by incorrectly using a Reader/Writer instead of InputStream/OutputStream.
See also:
How to save uploaded file in JSF (JSF-targeted, but the principle is pretty much the same)
Simplest way to serve static data from outside the application server in a Java web application (in case you want to serve it back)
How to save generated file temporarily in servlet based web application
I post my final way of doing it based on the accepted answer:
#SuppressWarnings("serial")
#WebServlet("/")
#MultipartConfig
public final class DataCollectionServlet extends Controller {
private static final String UPLOAD_LOCATION_PROPERTY_KEY="upload.location";
private String uploadsDirName;
#Override
public void init() throws ServletException {
super.init();
uploadsDirName = property(UPLOAD_LOCATION_PROPERTY_KEY);
}
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
// ...
}
#Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
Collection<Part> parts = req.getParts();
for (Part part : parts) {
File save = new File(uploadsDirName, getFilename(part) + "_"
+ System.currentTimeMillis());
final String absolutePath = save.getAbsolutePath();
log.debug(absolutePath);
part.write(absolutePath);
sc.getRequestDispatcher(DATA_COLLECTION_JSP).forward(req, resp);
}
}
// helpers
private static String getFilename(Part part) {
// courtesy of BalusC : http://stackoverflow.com/a/2424824/281545
for (String cd : part.getHeader("content-disposition").split(";")) {
if (cd.trim().startsWith("filename")) {
String filename = cd.substring(cd.indexOf('=') + 1).trim()
.replace("\"", "");
return filename.substring(filename.lastIndexOf('/') + 1)
.substring(filename.lastIndexOf('\\') + 1); // MSIE fix.
}
}
return null;
}
}
where :
#SuppressWarnings("serial")
class Controller extends HttpServlet {
static final String DATA_COLLECTION_JSP="/WEB-INF/jsp/data_collection.jsp";
static ServletContext sc;
Logger log;
// private
// "/WEB-INF/app.properties" also works...
private static final String PROPERTIES_PATH = "WEB-INF/app.properties";
private Properties properties;
#Override
public void init() throws ServletException {
super.init();
// synchronize !
if (sc == null) sc = getServletContext();
log = LoggerFactory.getLogger(this.getClass());
try {
loadProperties();
} catch (IOException e) {
throw new RuntimeException("Can't load properties file", e);
}
}
private void loadProperties() throws IOException {
try(InputStream is= sc.getResourceAsStream(PROPERTIES_PATH)) {
if (is == null)
throw new RuntimeException("Can't locate properties file");
properties = new Properties();
properties.load(is);
}
}
String property(final String key) {
return properties.getProperty(key);
}
}
and the /WEB-INF/app.properties :
upload.location=C:/_/
HTH and if you find a bug let me know
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");