I have written a code to upload the images(profile picture of an student) in the server running in linux environment.The code is shown below
#RequestMapping(value = "/updatePhoto",method = RequestMethod.POST)
public String handleFormUpload(#RequestParam("id") String id,
#RequestParam("file") MultipartFile file,
HttpServletRequest request,
Model model) throws IOException {
if(!file.isEmpty())
{
try
{
String relativePath="/resources";
String absolutePath=request.getServletContext().getRealPath(relativePath);
System.out.print(absolutePath);
byte[] bytes=file.getBytes();
File dir=new File(absolutePath);
if(!dir.exists())
{
dir.mkdir();
}
File uploadFile=new File(dir.getAbsolutePath()+File.separator+id+".jpg");
BufferedOutputStream outputStream=new BufferedOutputStream(new FileOutputStream(uploadFile));
outputStream.write(bytes);
outputStream.close();
model.addAttribute("uploadMessage","image uploaded for id"+id);
}
catch (Exception e)
{
System.out.print(e);
}
}
return "successFileUpload";
}
i have stored in "/resources" folder.but the problem is, whenever i generate the war file of whole application and deploy in server, it flushes the "/resources" folder and deletes the old uploaded images.Is there any way or the path ,i could upload the images.
The way I do is:
Create a directory in the server. For example: /myImages
Then grant full permissions for tomcat user
You are good to go now. I have read somewhere that you shouldn't save your stuff on /resources folder because it makes your app independent from container you are using: with tomcat you could use catalina.home but what if you shift to another container
I store the images inside my Tomcat home location as it will be outside of my project folder(war) and inside the tomcat.
String rootPath = System.getProperty("catalina.home");
File dir = new File(rootPath + File.separator + "images");
The above lines of code will create a folder in tomcat base directory with name 'images'.
This is the one of the best ways to store images.
Here's simple way
System.out.println(System.getProperty("user.dir"));
I have a resources folder/package in the root of my project, I "don't" want to load a certain File. If I wanted to load a certain File, I would use class.getResourceAsStream and I would be fine!! What I actually want to do is to load a "Folder" within the resources folder, loop on the Files inside that Folder and get a Stream to each file and read in the content... Assume that the File names are not determined before runtime... What should I do? Is there a way to get a list of the files inside a Folder in your jar File?
Notice that the Jar file with the resources is the same jar file from which the code is being run...
Finally, I found the solution:
final String path = "sample/folder";
final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());
if(jarFile.isFile()) { // Run with JAR file
final JarFile jar = new JarFile(jarFile);
final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
while(entries.hasMoreElements()) {
final String name = entries.nextElement().getName();
if (name.startsWith(path + "/")) { //filter according to the path
System.out.println(name);
}
}
jar.close();
} else { // Run with IDE
final URL url = Launcher.class.getResource("/" + path);
if (url != null) {
try {
final File apps = new File(url.toURI());
for (File app : apps.listFiles()) {
System.out.println(app);
}
} catch (URISyntaxException ex) {
// never happens
}
}
}
The second block just work when you run the application on IDE (not with jar file), You can remove it if you don't like that.
Try the following.
Make the resource path "<PathRelativeToThisClassFile>/<ResourceDirectory>" E.g. if your class path is com.abc.package.MyClass and your resoure files are within src/com/abc/package/resources/:
URL url = MyClass.class.getResource("resources/");
if (url == null) {
// error - missing folder
} else {
File dir = new File(url.toURI());
for (File nextFile : dir.listFiles()) {
// Do something with nextFile
}
}
You can also use
URL url = MyClass.class.getResource("/com/abc/package/resources/");
The following code returns the wanted "folder" as Path regardless of if it is inside a jar or not.
private Path getFolderPath() throws URISyntaxException, IOException {
URI uri = getClass().getClassLoader().getResource("folder").toURI();
if ("jar".equals(uri.getScheme())) {
FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.emptyMap(), null);
return fileSystem.getPath("path/to/folder/inside/jar");
} else {
return Paths.get(uri);
}
}
Requires java 7+.
I know this is many years ago . But just for other people come across this topic.
What you could do is to use getResourceAsStream() method with the directory path, and the input Stream will have all the files name from that dir. After that you can concat the dir path with each file name and call getResourceAsStream for each file in a loop.
I had the same problem at hands while i was attempting to load some hadoop configurations from resources packed in the jar... on both the IDE and on jar (release version).
I found java.nio.file.DirectoryStream to work the best to iterate over directory contents over both local filesystem and jar.
String fooFolder = "/foo/folder";
....
ClassLoader classLoader = foofClass.class.getClassLoader();
try {
uri = classLoader.getResource(fooFolder).toURI();
} catch (URISyntaxException e) {
throw new FooException(e.getMessage());
} catch (NullPointerException e){
throw new FooException(e.getMessage());
}
if(uri == null){
throw new FooException("something is wrong directory or files missing");
}
/** i want to know if i am inside the jar or working on the IDE*/
if(uri.getScheme().contains("jar")){
/** jar case */
try{
URL jar = FooClass.class.getProtectionDomain().getCodeSource().getLocation();
//jar.toString() begins with file:
//i want to trim it out...
Path jarFile = Paths.get(jar.toString().substring("file:".length()));
FileSystem fs = FileSystems.newFileSystem(jarFile, null);
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(fs.getPath(fooFolder));
for(Path p: directoryStream){
InputStream is = FooClass.class.getResourceAsStream(p.toString()) ;
performFooOverInputStream(is);
/** your logic here **/
}
}catch(IOException e) {
throw new FooException(e.getMessage());
}
}
else{
/** IDE case */
Path path = Paths.get(uri);
try {
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(path);
for(Path p : directoryStream){
InputStream is = new FileInputStream(p.toFile());
performFooOverInputStream(is);
}
} catch (IOException _e) {
throw new FooException(_e.getMessage());
}
}
Another solution, you can do it using ResourceLoader like this:
import org.springframework.core.io.Resource;
import org.apache.commons.io.FileUtils;
#Autowire
private ResourceLoader resourceLoader;
...
Resource resource = resourceLoader.getResource("classpath:/path/to/you/dir");
File file = resource.getFile();
Iterator<File> fi = FileUtils.iterateFiles(file, null, true);
while(fi.hasNext()) {
load(fi.next())
}
If you are using Spring you can use org.springframework.core.io.support.PathMatchingResourcePatternResolver and deal with Resource objects rather than files. This works when running inside and outside of a Jar file.
PathMatchingResourcePatternResolver r = new PathMatchingResourcePatternResolver();
Resource[] resources = r.getResources("/myfolder/*");
Then you can access the data using getInputStream and the filename from getFilename.
Note that it will still fail if you try to use the getFile while running from a Jar.
As the other answers point out, once the resources are inside a jar file, things get really ugly. In our case, this solution:
https://stackoverflow.com/a/13227570/516188
works very well in the tests (since when the tests are run the code is not packed in a jar file), but doesn't work when the app actually runs normally. So what I've done is... I hardcode the list of the files in the app, but I have a test which reads the actual list from disk (can do it since that works in tests) and fails if the actual list doesn't match with the list the app returns.
That way I have simple code in my app (no tricks), and I'm sure I didn't forget to add a new entry in the list thanks to the test.
Below code gets .yaml files from a custom resource directory.
ClassLoader classLoader = this.getClass().getClassLoader();
URI uri = classLoader.getResource(directoryPath).toURI();
if("jar".equalsIgnoreCase(uri.getScheme())){
Pattern pattern = Pattern.compile("^.+" +"/classes/" + directoryPath + "/.+.yaml$");
log.debug("pattern {} ", pattern.pattern());
ApplicationHome home = new ApplicationHome(SomeApplication.class);
JarFile file = new JarFile(home.getSource());
Enumeration<JarEntry> jarEntries = file.entries() ;
while(jarEntries.hasMoreElements()){
JarEntry entry = jarEntries.nextElement();
Matcher matcher = pattern.matcher(entry.getName());
if(matcher.find()){
InputStream in =
file.getInputStream(entry);
//work on the stream
}
}
}else{
//When Spring boot application executed through Non-Jar strategy like through IDE or as a War.
String path = uri.getPath();
File[] files = new File(path).listFiles();
for(File file: files){
if(file != null){
try {
InputStream is = new FileInputStream(file);
//work on stream
} catch (Exception e) {
log.error("Exception while parsing file yaml file {} : {} " , file.getAbsolutePath(), e.getMessage());
}
}else{
log.warn("File Object is null while parsing yaml file");
}
}
}
Took me 2-3 days to get this working, in order to have the same url that work for both Jar or in local, the url (or path) needs to be a relative path from the repository root.
..meaning, the location of your file or folder from your src folder.
could be "/main/resources/your-folder/" or "/client/notes/somefile.md"
Whatever it is, in order for your JAR file to find it, the url must be a relative path from the repository root.
it must be "src/main/resources/your-folder/" or "src/client/notes/somefile.md"
Now you get the drill, and luckily for Intellij Idea users, you can get the correct path with a right-click on the folder or file -> copy Path/Reference.. -> Path From Repository Root (this is it)
Last, paste it and do your thing.
Simple ... use OSGi. In OSGi you can iterate over your Bundle's entries with findEntries and findPaths.
Inside my jar file I had a folder called Upload, this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file, I used the code below:
URL inputUrl = getClass().getResource("/upload/blabla1.txt");
File dest1 = new File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl, dest1);
URL inputUrl2 = getClass().getResource("/upload/blabla2.txt");
File dest2 = new File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2, dest2);
URL inputUrl3 = getClass().getResource("/upload/blabla3.txt");
File dest3 = new File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3, dest3);
It is fairly common to allow users to download a file via having some path modifier in the URL
//MVC Action to download the correct file From our Content directory
public ActionResult GetFile(string name) {
string path = this.Server.MapPath("~/Content/" + name);
byte[] file = System.IO.File.ReadAllBytes(path);
return this.File(file, "html/text");
}
quoted from http://hugoware.net/blog/dude-for-real-encrypt-your-web-config
An application I'm working with has liberal path downloads ( directory based ) sprinkled throughout the application, hence it is super vulnerable to requests like "http://localhost:1100/Home/GetFile?name=../web.config" or ( ..%2fweb.config )
Is there an easy way to restrict access to the config file - do I need to provide a custom Server.MapPath with whitelisted directories - or is there a better way.
How do you secure your file downloads - are path based downloads inherently insecure?
A simple option, assuming that all files in the ~/Content directory are safe to download would be to verify that the path is actually under (or in) the ~/Content directory and not up from it, as ~/Content/../web.config would be. I might do something like this:
// MVC Action to download the correct file From our Content directory
public ActionResult GetFile(string name) {
// Safe path
var safePath = this.Server.MapPath("~/Content");
// Requested path
string path = this.Server.MapPath("~/Content/" + name);
// Make sure requested path is safe
if (!path.StartsWith(safePath))
// NOT SAFE! Do something here, like show an error message
// Read file and return it
byte[] file = System.IO.File.ReadAllBytes(path);
return this.File(file, "html/text");
}
I need to put a property file named "customer.properties" into the "distribution.jar". Fle paths for above files are;
modules\distribution.jar
modules\distributionaManager\customer.properties
My requirement is to have "modules\distribution.ja\distributionaManager\customer.properties"
ProcessBuilder pb = new ProcessBuilder("jar", "uf", "distribution.jar", "distributionaManager\\customer.properties");
pb.directory(jarFile.getParentFile());
try
{
Process process = pb.start();
process.waitFor();
process.destroy();
}
jarFile.getParentFile() - "modules directory"
But this code creates no property file inside the jar. Any suggestions?
Thanks
I am working on website in which i want to copy the file from my application folder to other folder on same server (But this folder is out of my application folder i.e. my application on C driver and the destination folder is on D drive).Is this possible using any functionality of Asp.Net?
Thanks in advance.
YES it's possible, the only concern that you have to watch for is that the CopyTo path should be the full path, not the relative one (ex: c:\websites\myOtherFolder).
this way, you can successfully copy/move the file from your ASP.NET code.
below is a pseudo code to show you how to get it done (assuming that the file has been placed on the root folder of your ASP.NET Application).
using System.IO;
..
..
..
// Get the current app path:
var currentApplicationPath = HttpContext.Current.Request.PhysicalApplicationPath;
//Get the full path of the file
var fullFilePath = currentApplicationPath + fileNameWithExtension;
// Get the destination path
var copyToPath = "This has to be the full path to your destination directory.
Example d:\myfolder";
// Copy the file
File.Copy(fullFilePath , copyToPath );
use this function:
System.IO.File.Copy(FileToCopy, NewCopy)
It's very easy to move file from one folder to other folder. you can change the file name while moving...
string Tranfiles, ProcessedFiles;
//Tranfiles = Server.MapPath(#"~\godurian\sth100\transfiles\" + Filename);
Tranfiles = Server.MapPath(#"~\transfiles\" + Filename);
if (File.Exists(Server.MapPath(#"~\transfiles\" + Filename)))
{
File.Delete(Server.MapPath(#"~\transfiles\" + Filename));
}
//ProcessedFiles = Server.MapPath(#"~\godurian\sth100\ProcessedFiles");
ProcessedFiles = Server.MapPath(#"~\ProcessedFiles");
File.Move(Tranfiles, ProcessedFiles);
That's it now you can check your application folder to confirm the move process status