I am trying to create a temp file and generate a file name and then save a multipart file. I am using Spring Boot and the following code is working in local. But in heroku or docker it is throwing FileNotFoundException; So how to create a temp directory and save a file inside that temp directory in docker/heroku? Or what is the best way to save multipart file to temp folder in server? Anybody can help me? Thanks in advance.
File tempDirectory = new File(new File(System.getProperty("java.io.tmpdir")), "files");
if (!tempDirectory.exists()) {
tempDirectory.mkdir();
}
String filePath = new File(tempDirectory.getAbsolutePath() + "/temp.pdf").getAbsolutePath();
File tempDirectory = new File(new File(System.getProperty("java.io.tmpdir")), "files");
if(tempDirectory.exists()){
System.out.println("something");
}else{
tempDirectory.mkdirs();
}
File file = new File(tempDirectory.getAbsolutePath()+"/abcd.txt");
if(!file.exists()){
file.createNewFile();
}
String file2= new File(tempDirectory.getAbsolutePath()+"/something.txt").getAbsolutePath();
System.out.println(file2);
Works totally fine at my end. The only problem you might be having is
String filePath = new File(tempDirectory.getAbsolutePath() + "/temp.exe").getAbsolutePath();
This doesn't create the file in the temp directory you have created. It just returns the absolute path if it was to be saved in the directory mentioned. This might be the reason you are getting not found error. Try actually saving by using
file.transferTo(wherefileneedstobesavedlocation);
Below is my project like:
projectName
-package
- Util.java
- Test.json
In Util.java, I need to read the content from Test.json file and parse it.
Thus I use:
File currentfile = new File("");//get the current path
String absJsonPath = currentfile.getAbsolutePath() + "/Test.json";
While it did not work when I use a main method to test it. The thing is that the /src/package is lost in the obtained file path and I just got the path of the project.
And, when I deploy the project to weblogic server, I got another new error, the obtained current path is like:
.../DefaultDomain/.
I just want the file path in the file system, which is not related to the server.
What can I do for this? Thanks!
Put your file in resources folder and get it as following:
//Get file from resources folder
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("Test.json").getFile());
To read the content you can use following:
StringBuilder result = new StringBuilder("");
try (Scanner scanner = new Scanner(file)) {
while (scanner.hasNextLine()) {
String line = scanner.nextLine();
result.append(line).append("\n");
}
scanner.close();
} catch (IOException e) {
e.printStackTrace();
}
It's CanonicalPath() not Absolute. Check this out, let me know if it any helps.
Try this to get current path
String currentPath = System.getProperty("user.dir");
If you want to read a file in a specific package. You can use
File jsonFile = new File(getClass().getResource("/Test.json").getFile());
I am making a program that works with MySQL database,for now i store URL, login, password e.t.c as public static String. Now i need to make it possible to work on another computer, so database adress will vary, so i need a way to edit it inside programm and save. I would like to use just external txt file, but i don't know how to point it's location.
I decided to make it using Property file, i put it in src/res folder. It work correct while i'm trying it inside Intellij Idea, but when i build jar (artifact) i get java.io.FileNotFoundException
I tried two ways:
This one was just copied
private String getFile(String fileName) {
StringBuilder result = new StringBuilder("");
//Get file from resources folder
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource(fileName).getFile());
System.out.println(file.length());
try (Scanner scanner = new Scanner(file)) {
while (scanner.hasNextLine()) {
String line = scanner.nextLine();
result.append(line).append("\n");
}
scanner.close();
} catch (IOException e) {
e.printStackTrace();
}
return result.toString();
}
System.out.println(obj.getFile("res/cfg.txt"));</code>
And second one using Properties class:
try(FileReader reader = new FileReader("src/res/cfg.txt")) {
Properties properties = new Properties();
properties.load(reader);
System.out.println(properties.get("password"));
}catch (Exception e) {
e.printStackTrace();
System.out.println(e);
}
In both ways i get java.io.FileNotFoundException. What is right way to attach config file like that?
Since the file is inside a .JAR, it can't be accessed via new File(), but you can still read it via the ClassLoader:
Properties properties = new Properties();
try (InputStream stream = getClass().getResourceAsStream("/res/cfg.txt")) {
properties.load(stream);
}
Note that a JAR is read-only. So this approach won't work.
If you want to have editable configuration, you should place your cfg.txt outside the JAR and read it from the filesystem. For example like this:
Properties properties = new Properties();
File appPath = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().toURI()).getParentFile();
try (InputStream stream = new FileInputStream(new File(appPath, "cfg.txt"))) {
properties.load(stream);
}
There are multiple places your can place your configuration options, and a robust deployment strategy will utilize some (or all) of the following techniques:
Storing configuration files in a well known location relative to the user's home folder as I mentioned in the comments. This works on Windows (C:\Users\efrisch), Linux (/home/efrisch) and Mac (/Users/efrisch)
File f = new File(System.getProperty("user.home"), "my-settings.txt");
Reading environment variables to control it
File f = new File(System.getenv("DEPLOY_DIR"), "my-settings.txt");
Using a decentralized service such as Apache ZooKeeper to store your database settings
Use Standalone JNDI
(or the JNDI built-in to your deployment target)
Use a Connection Pool
How to write or read any type of file (for example .txt file) to a resources folder with the config.property file, but without using absolute path file.
I tried to solve this like below:
ClassLoader classLoader = Setting.class.getClassLoader();
Setting setting = new Setting();
try (InputStream resourceAsStream = classLoader.getResourceAsStream("config.properties")) {
setting.load(resourceAsStream);
}
String readFileName = setting.getValue("pathSource.txt");
String writeFileName = setting.getValue("outPutPathSourceFile.txt");
String s = System.getProperty("line.separator");
File readFile = new File("./src/main/resources" + File.separator + "pathSource.txt");
File writeFile = new File("./src/main/resources" + File.separator + "outPutPathSourceFile.txt");
However, I don't want using ./src/main/resources prefix.
If your file is located under resources you able to access it like:
File file = new File(classLoader.getResource(fileName).getFile());
if you are using Spring you can use ResourceUtils.getFile():
File file = ResourceUtils.getFile("classpath:fileName")
UPDATE:
If I understand correctly, you want to read file which located under your src/main/resources without relative path usage /src/main/resources.
I created small demo:
public class ReadResourceFile {
public static void main(String[] args) throws IOException {
String fileName = "webAutomationConfig.xml";
ClassLoader classLoader = ReadResourceFile.class.getClassLoader();
File file = new File(classLoader.getResource(fileName).getFile());
System.out.println("File exists: " + file.exists());
String content = new String(Files.readAllBytes(file.toPath()));
System.out.println(content);
}
}
Output:
File exists: true
<?xml version="1.0" encoding="utf-8"?>
<config>
<baseUrl>https://www.gmail.com</baseUrl>
</config>
Project structure:
Resources (files on the class path, possibly inside a jar)can be read, but are not intended to be written to. You can use them as template to create an inital copy on the file system. – Joop Eggen
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);