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Reading file inside Spring boot fat jar

We have a spring boot application which has a legacy jar api that we use that needs to load properties by using InputFileStream. We wrapped the legacy jar in our spring boot fat jar and the properties files are under BOOT-INF/classes folder. I could see spring loading all the relevant properties but when I pass the properties file name to the legacy jar it could not read the properties file as its inside the jar and is not under physical path. In this scenario how do we pass the properties file to the legacy jar?
Please note we cannot change the legacy jar.

Actually you can, using FileSystem. You just have to emulate a filesystem on the folder you need to get the file from. For example if you wanted to get file.properties, which is under src/main/resources you could do something like this:
FileSystem fs = FileSystems.newFileSystem(this.getClass().getResource("").toURI(), Collections.emptyMap());
String pathToMyFile = fs.getPath("file.properties").toString();
yourLegacyClassThatNeedsAndAbsoluteFilePath.execute(pathToMyFile)

I also have faced this issue recently. I have a file I put in the resource file, let's call it path and I was not able to read it. #madoke has given one of the solutions using FileSystem. This is another one, here we are assuming we are in a container, and if not, we use the Java 8 features.
#Component
#Log4j2
public class DataInitializer implements
ApplicationListener<ApplicationReadyEvent> {
private YourRepo yourRepo; // this is your DataSource Repo
#Value(“${path.to.file}”). // the path to file MUST start with a forward slash: /iaka/myfile.csv
private String path;
public DataInitializer(YourRepo yourRepo) {
this.yourRepo = yourRepo;
}
#Override
public void onApplicationEvent(ApplicationReadyEvent event) {
try {
persistInputData();
log.info("Completed loading data");
} catch (IOException e) {
log.error("Error in reading / parsing CSV", e);
}
}
private void persistInputData() throws IOException {
log.info("The path to Customers: "+ path);
Stream<String> lines;
InputStream inputStream = DataInitializer.class.getClassLoader().getResourceAsStream(path);
if (inputStream != null) { // this means you are inside a fat-jar / container
InputStreamReader inputStreamReader = new InputStreamReader(inputStream);
BufferedReader bufferedReader = new BufferedReader(inputStreamReader);
lines = bufferedReader.lines();
} else {
URL resource = this.getClass().getResource(path);
String path = resource.getPath();
lines = Files.lines(Paths.get(path));
}
List<SnapCsvInput> inputs = lines.map(s -> s.split(","))
.skip(1) //the column names
.map(SnapCsvInput::new)
.collect(Collectors.toList());
inputs.forEach(i->log.info(i.toString()));
yourRepo.saveAll(inputs);
}
}

Basically, you can't, because the properties "file" is not a file, it's a resource in a jar file, which is compressed (it's actually a .zip file).
AFAIK, the only way to make this work is to extract the file from the jar, and put it on your server in a well-known location. Then at runtime open that file with an FileInputStream and pass it to the legacy method.

According to #moldovean's solution, you just need to call 'getResourceAsStream(path)' and continue with returned InputStream. Be aware that the 'path' is based on ClassPath. For example:
InputStream stream = this.getClass().getResourceAsStream("/your-file.csv");
In root of your class path, there is a file named 'your-file.csv' that you want to read.

How to link a local file

I'm trying to use a local file in which I've specified my db connection properties which is named dao.properties. And I'm proceeding this way:
InputStream fichierProperties = classLoader.getResourceAsStream( "/src/dao/dao.properties" );
However, when using this path, I'm getting an exception stating that the debugger wasn't able to find that file.
Here are some packages in my project:
The dao.properties is just under the dao package.
How do I resolve this, please?

If you put the file inside the src folder, the IDE probably is packaging, when instructed to compile and build, the file into the bundled generated jar. So you can reach with the method GetResourceAsStream.
So if you put the file (dao.properties) in root folder of your sources files (generally the src folder), just simple referring to dao.properties will refer to the resource.
If you put the file inside a subfolder of src, the correct way to reference it would be subfolder/dao.properties.
The first "/" is not necessary as the getResourceAsStream always search in the classpath, that for default is the root of the sources folder, inside the jar. (where are not talking about external files!)
Updated:
Assuming you place a file name notes.txt inside a folder(package) named ´sub´, this is valid example, only for purporses of how to get a bundled file that is in jar.
public class Main {
public static void main (String[] args) throws IOException {
InputStream is = Main.class.getResourceAsStream("sub/notes.txt");
InputStreamReader isr = new InputStreamReader(is);
BufferedReader br = new BufferedReader(isr);
String s = br.readLine();
while (s != null) {
System.out.println (s);
s = br.readLine();
}
is.close();
}
}
I add more information about this, by referring to this post

How can I get path of resource under WEB-INF/class folder (java ee web dynamic project)

I develop web dynamic project in Eclipse. One file, file named 'users.txt' is located under classes folder (WEB-INF/classes/users.txt).
How I can get relative path of this file in class (base class, not servlet class)? I will use that path for append several lines of text.
Christian this is the code that works for reading. Problem is I don't know how to create object for writing (output) in the same file with relative path.
public Products(){
try{
InputStream in = getClass().getClassLoader().getResourceAsStream("products.txt");
BufferedReader br = new BufferedReader(new InputStreamReader(in));
readProducts(br);
in.close();
br.close();
}catch(Exception e){
System.out.println("File doesn't exists");
}
}
public void readProducts(BufferedReader in) throws NumberFormatException, IOException{
String id="";
String name="";
double price=0;
StringTokenizer st;
String line;
while((line=in.readLine())!=null){
st=new StringTokenizer(line,";");
while(st.hasMoreTokens()){
id=st.nextToken();
name=st.nextToken();
price=Double.parseDouble(st.nextToken());
}
products.put(id,new Product(id,name,price));
}
}

Generally speaking you should not rely on modifying files in your web application directory.
One of the reasons is that servlet container is not obligated to extract your .war file to the fs, it could, in theory, run your web application from memory. Yeah, Tomcat and most of the containers do unpack .war but there is nothing in Java EE specs that says that they must do this.
Anyway, if you think that this worth the risk you can use ServletContext.getRealPath to get location of the file inside your webapp directory.
This should look something like this:
Create a file myfile under your webapp root (should be located in the same folder as WEB-INF in your .war).
do something like this in your servlet (or anywhere where you can access servlet context):
String filePath = getServletContext().getRealPath("myfile");
Note that in order for this to work you should be able to fetch the file with request, like: http://<host>:<port>/<contextPath>/myfile (see method docs for details)

How do I reference a file that is placed in the WEB-INF folder when using Arquillian?

I am using maven and the standard directory layout. So I have added a testdata.xml file in the src/test/resources folder, and I also added it as:
.addAsWebInfResource("testdata.xml", "testdata.xml")
in the deployment method, and I have confirmed that it is there. This will make the file appear in /WEB-INF/testdata.xml. Now I need to have a reference to this file in my code and I tried several different getClass().getResourceAsStream(...) and failing again and again so I need some advise now.
I need it for my DBUnit integration test. Is this not possible?

Option A) Use ServletContext.getResourceXXX()
You should have a Aquillarian MockHttpSession and a MockServletContext. E.g.:
#Test
public void myTest()
{
HttpSession session = new MockHttpSession(new MockServletContext());
ServletLifecycle.beginSession(session);
..testCode..
// You can obtain a ServletContext (will actually be a MockServletContext
// implementation):
ServletContext sc = session.getServletContext();
URL url = sc.getResource("/WEB-INF/testdata.xml")
Path resPath = new Path(url);
File resFile = new File(url);
FileReader resRdr = new FileReader(resFile);
etc...
..testCode..
ServletLifecycle.endSession(session);
}
You can create resource files & subdirectories in:
the web module document root - resources are accessible from the browser and from classes
WEB-INF/classes/ - resources are accessible to classes
WEB-INF/lib/*.jar source jar - accessible to classes
WEB-INF/lib/*.jar dedicated resource-only jar - accessible to classes
WEB-INF/ directly within directory - accessible to classes. This is what you are asking for.
In all cases the resource can be accessed via:
URL url = sc.getResource("/<path from web doc root>/<resourceFileName>");
OR
InputStream resIS = sc.getResourceAsStream("/<path from web doc root>/<resourceFileName>");
>
These will be packaged into the WAR file and may be exploded into directories on the deployed app server OR they may stay within the WAR file on the app server. Either way - same behaviour for accessing resources: use ServletContext.getResourceXXX().
Note that as a general principle, (5) the top-level WEB-INF directory itself is intended for use by the server. It is 'polite' not to put your web resources directly in here or create your own directory directly in here. Instead, better to use (2) above.
JEE5 tutorial web modules
JEE6 tutorial web modules
Option B): Use Class.getResourceXXX()
First move the resource out of WEB-INF folder into WEB-INF/classes (or inside a jar WEB-INF/lib/*.jar).
If your test class is:
com.abc.pkg.test.MyTest in file WEB-INF/classes/com/abc/pkg/test/MyTest.class
And your resource file is
WEB-INF/classes/com/abc/pkg/test/resources/testdata.xml (or equivalent in a jar file)
Access File using Relative File Location, via the Java ClassLoader - finds Folders/Jars relative to Classpath:
java.net.URL resFileURL = MyTest.class.getResource("resources/testdata.xml");
File resFile = new File(fileURL);
OR
InputStream resFileIS =
MyTedy.class.getResourceAsStream("resources/testdata.xml");
Access File Using full Package-like Qualification, Using the Java ClassLoader - finds Folders/Jars relative to Classpath:
java.net.URL resFileURL = MyTest.class.getResource("/com/abc/pkg/test/resources/testdata.xml");
File resFile = new File(fileURL);
OR
InputStream resFileIS =
MyTest.class.getResourceAsStream("/com/abc/pkg/test/resources/testdata.xml");
OR
java.net.URL resFileURL = MyTest.class.getClassLoader().getResource("com/abc/pkg/test/resources/testdata.xml");
File resFile = new File(fileURL);
OR
InputStream resFileIS =
MyTest.class.getClassLoader().getResourceAsStream("com/abc/pkg/test/resources/testdata.xml");
Hope that nails it! #B)

The way to access files under WEB-INF is via three methods of ServletContext:
getResource("/WEB-INF/testdata.xml") gives you a URL
getResourceAsStream gives you an input stream
getRealPath gives you the path on disk of the relevant file.
The first two should always work, the third may fail if there is no direct correspondence between resource paths and files on disk, for example if your web application is being run directly from a WAR file rather than an unpacked directory structure.

Today I was struggling with the same requirement and haven't found any full source sample, so here I go with smallest self contained test I could put together:
#RunWith(Arquillian.class)
public class AttachResourceTest {
#Deployment
public static WebArchive createDeployment() {
WebArchive archive = ShrinkWrap.create(WebArchive.class).addPackages(true, "org.apache.commons.io")
.addAsWebInfResource("hello-kitty.png", "classes/hello-kitty.png");
System.out.println(archive.toString(true));
return archive;
}
#Test
public void attachCatTest() {
InputStream stream = getClass().getResourceAsStream("/hello-kitty.png");
byte[] bytes = null;
try {
bytes = IOUtils.toByteArray(stream);
} catch (IOException e) {
e.printStackTrace();
}
Assert.assertNotNull(bytes);
}
}
In your project hello-kitty.png file goes to src/test/resources. In the test archive it is packed into the /WEB-INF/classes folder which is on classpath and therefore you can load it with the same class loader the container used for your test scenario.
IOUtils is from apache commons-io.
Additional Note:
One thing that got me to scratch my head was related to spaces in path to my server and the way getResourceAsStream() escapes such special characters: sysLoader.getResource() problem in java

Add this class to your project:
import java.io.UnsupportedEncodingException;
import java.net.URLDecoder;
public class Init {
private static final String WEB_INF_DIR_NAME="WEB-INF";
private static String web_inf_path;
public static String getWebInfPath() throws UnsupportedEncodingException {
if (web_inf_path == null) {
web_inf_path = URLDecoder.decode(Init.class.getProtectionDomain().getCodeSource().getLocation().getPath(), "UTF8");
web_inf_path=web_inf_path.substring(0,web_inf_path.lastIndexOf(WEB_INF_DIR_NAME)+WEB_INF_DIR_NAME.length());
}
return web_inf_path;
}
}
Now wherever you want to get the full path of the file "testdata.xml" use this or similar code:
String testdata_file_location = Init.getWebInfPath() + "/testdata.xml";

Create file in resources/source folder in java programmatically?

I have two resources folders.
src - here are my .java files
resources - here are my resources files (images, .properties) organized in folders (packages).
Is there a way to programmatically add another .properties file in that resources folder?
I tried something like this:
public static void savePropertiesToFile(Properties properties, File propertiesFile) throws IOException {
FileOutputStream out = new FileOutputStream(propertiesFile);
properties.store(out, null);
out.close();
}
and before that created:
new File("/folderInResources/newProperties.properties");
But it looks for that path on the file system. How can I force it to look in the resources folder?
EDIT: Let me say what is it about. I have a GUI application and I support 2 languages (2 .properties files in resources folder). Now I added a option that user can easily translate application and when he finishes I save that new .properties on a disk in some hidden folder and read it from there. But I was hoping I could save that new .properties files (new language) next to the current languages (resources folder). I have a static Messages class which knows how to load resources both from the disk and both the default ones in resources folder. But if user takes this .jar file on some other machine, he would't have that new languages since they are on disk on that computer, not inside .jar file.

Java 8 Solution
Path source = Paths.get(this.getClass().getResource("/").getPath());
Path newFolder = Paths.get(source.toAbsolutePath() + "/newFolder/");
Files.createDirectories(newFolder);
This will surely create new folder in resource folder. but you will find new folder in your target runtime.
which will be ProjectName/target/test-classes/newFolder. if you are running this code in test case. Other wise it would be in target/classes
Don't try to find new folder in your src/resources.
it will be surely in target/test-classes or target/classes.

As other people have mentioned, resources are obtained through a ClassLoader. What the two current responses have failed to stress, however, is these points:
ClassLoaders are meant to abstract the process of obtaining classes and other resources. A resource does not have to be a file in a filesystem; it can be a remote URL, or anything at all that you or somebody else might implement by extending java.lang.ClassLoader.
ClassLoaders exist in a child/parent delegation chain. The normal behavior for a ClassLoader is to first attempt to obtain the resource from the parent, and only then search its own resources—but some classloaders do the opposite order (e.g., in servlet containers). In any case, you'd need to identify which classloader's place for getting stuff you'd want to put stuff into, and even then another classloader above or below it might "steal" your client code's resource requests.
As Lionel Port points out, even a single ClassLoader may have multiple locations from which it loads stuff.
ClassLoaders are used to, well, load classes. If your program can write files to a location where classes are loaded from, this can easily become a security risk, because it might be possible for a user to inject code into your running application.
Short version: don't do it. Write a more abstract interface for the concept of "repository of resource-like stuff that I can get stuff from," and subinterface for "repository of resource-like stuff that I can get stuff from, but also add stuff from." Implement the latter in a way that both uses ClassLoader.getContextClassLoader().getResource() (to search the classpath) and, if that fails, uses some other mechanism to get stuff that the program may have added from some location.

Problem would be the classpath can contain multiple root directories so distinguishing which one to store would be hard without an existing file or directory.
If you have an existing file loaded.
File existingFile = ...;
File parentDirectory = existingFile.getParentFile();
new File(parentDirectory, "newProperties.properties");
Otherwise try an get a handle on a directory you know is unique in your resources directory. (Not sure if this works)
URL url = this.getClass().getResource("/parentDirectory");
File parentDirectory = new File(new URI(url.toString()));
new File(parentDirectory, "newProperties.properties");

Cut the main project folder of the compiled subfolders ("/target/classes", "target/test-classes") and you have the basic path to reconstruct your project folders with:
import java.io.File;
import java.io.IOException;
import java.net.URISyntaxException;
public class SubfolderCreator {
public static void main(String... args) throws URISyntaxException, IOException {
File newResourceFolder = createResourceSubFolder("newFolder");
}
private static File createResourceSubFolder(String folderName) throws URISyntaxException, IOException {
java.net.URL url = SubfolderCreator.class.getResource("/EXISTING_SUBFOLDER/");
File fullPathToSubfolder = new File(url.toURI()).getAbsoluteFile();
String projectFolder = fullPathToSubfolder.getAbsolutePath().split("target")[0];
File testResultsFolder = new File(projectFolder + "src/test/resources/" + folderName);
if (!testResultsFolder.exists()) {
testResultsFolder.mkdir();
}
return testResultsFolder;
}
}

The following code writes into the classes directory, along with the class files.
As others have noted, beware of overwriting class files. Best to put your new files into a separate directory; however, that directory needs to already exist. To create it, create a sub-directory within the resources in the source, perhaps containing an empty file. For example src\main\resources\dir\empty.txt.
public class WriteResource {
public static void main(String[] args) throws FileNotFoundException {
String thing = "Text to write to the file";
String dir = WriteResource.class.getResource("/").getFile();
//String dir = WriteResource.class.getResource("/dir").getFile();
OutputStream os = new FileOutputStream(dir + "/file.txt");
final PrintStream printStream = new PrintStream(os);
printStream.println(thing);
printStream.close();
}
}
This does the trick, but I'd be nervous about deploying this outside of a strictly controlled environment. I don't really like the idea of unauthorised persons writing to my classes directory!

Based on the below code you get or create a file and use it as a util method.
public static File getFileFromResource(String filePath) {
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
return new File(Objects.requireNonNull(classloader.getResource(filePath)).getFile());
}
Project structure
You should specify the relative path to file or folder. Usage:
getFileFromResource("application.properties");
Output:
File with path C:\Users\????\IdeaProjects\back-end\target\classes\application.properties