We Keep Coding

Cannot be resolved to absolute file path because it does not reside in the file system

My Code:
XWPFDocument doc = new XWPFDocument(OPCPackage.open(ResourceUtils.getFile("classpath:assets/OPTIONS_" + jubilar1.getJubiLanguage().toUpperCase() + ".docx")));
I have already tried instead of .getFile(), extractJarFileFromURL or resource.getInputStream() but all this does not work. When I package my project and run it as a jar file and it tries to open the following file it always returns the following message.
Error:
java.io.FileNotFoundException: class path resource [assets/OPTIONS_DE.
docx] cannot be resolved to absolute file path because it does not
reside in the file system:
jar:file:/home/tkf6y/IdeaProjects/hrapps/backend/target/backend-3.0.0.jar!/BOOT-INF/classes!/assets/OPTIONS_EN.docx

So yes it was the problem, as you are now using an InputStream as I suggested. The problem was (and always has been) the getFile stuff. What I suggest to do is don't use what you have now but rather do a new ClassPathResource(your location).getInputStream()) instead, it is easier, or even use a ResourceLoader (a Spring interface you can inject) and then use the path you had an again use getInputStream(). –

This works for me.
String strJson = null;
ClassPathResource classPathResource = new ClassPathResource("json/data.json");
try {
byte[] binaryData = FileCopyUtils.copyToByteArray(classPathResource.getInputStream());
strJson = new String(binaryData, StandardCharsets.UTF_8);
} catch (IOException e) {
e.printStackTrace();
}

Java 8: How to copy files written in a list to a TXT file from one directory to another directory?

I have a simple text file called small_reports.txt that looks like:
report_2021_05_02.csv
report_2021_05_05.csv
report_2021_06_08.csv
report_2021_06_25.csv
report_2021_07_02.csv
This reported is generated with my java code and takes in each of these files from the directory /work/dir1/reports and writes them into the file combined_reports.txt and then places the txt file back into /work/dir1/reports.
My question is, for each line in small_reports.txt, find that same file (line) in /work/dir1/reports and then COPY them to a new directory called /work/dir1/smallreports?
Using Java 8 & NIO (which is really helpful and good) I have tried:
Path source = Paths.get("/work/dir1/reports/combined_reports.txt");
Path target = Paths.get("/work/dir1/smallreports/", "combined_reports.txt");
if (Files.notExists(target) && target != null) {
Files.createDirectories(Paths.get(target.toString()));
}
Files.copy(source, target, StandardCopyOption.REPLACE_EXISTING);
But this is just copying the actual txt file combined_reports.txt into the new directory and not the contents inside like i thought it would.

final String SOURCE_DIR = "/tmp";
final String TARGET_DIR = "/tmp/root/delme";
List<String> csvFileNames = Files.readAllLines(FileSystems.getDefault().getPath("small_reports.txt"), Charset.forName("UTF-8"));
for (String csvFileName : csvFileNames) {
Path source = Paths.get(SOURCE_DIR, csvFileName);
Path target = Paths.get(TARGET_DIR, csvFileName);
if (Files.notExists(target) && target != null) {
Files.createDirectories(Paths.get(target.toString()));
}
Files.copy(source, target, StandardCopyOption.REPLACE_EXISTING);
}
Should do it for you. Obviously change the constants appropriately

Configuration file not found JAVA EE [duplicate]

This question already has answers here:
Where to place and how to read configuration resource files in servlet based application?
(6 answers)
Closed 6 years ago.
I'm trying to load a configuration file. But it doesn't work my configuration file is placed under WEB-INF folder
and here is my code to load that conf file :
private static final String PROPERTIES_FILE = "/WEB-INF/dao.properties";
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream fichierProperties = classloader.getResourceAsStream(PROPERTIES_FILE);
if (fichierProperties == null) {
throw new DAOConfigurationException("file "+PROPERTIES_FILE+ " not found" );
}
I'm always getting this error file not found, Should make some changes on the build path ??

For simple purpose, try
Put dao.properties inside src folder (where put source code).
Change to
private static final String PROPERTIES_FILE = "dao.properties"; // <-------
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream fichierProperties = classloader.getResourceAsStream(PROPERTIES_FILE);
if (fichierProperties == null) {
throw new DAOConfigurationException("file "+PROPERTIES_FILE+ " not found" );
}

If you put your file inside the WEB-INF directory, You can use context object to read your file as shown if you have access to servlet context
InputStream input = context.getResourceAsStream("/WEB-INF/dao.properties");

Reading a TXT file in Java [duplicate]

This question already has answers here:
What is a NullPointerException, and how do I fix it?
(12 answers)
Closed 7 years ago.
I keep getting a java.lang.NullPointerException when trying to open a txt file in eclipse. Basically, this is a main menu, and when you click the "Rules" button, the rules text file should open. Currently, the txt file is located in a package called "Resources" (which is where all of the other img files I've used in making the game are). Here's the code:
private List<String> readFile(String filename)
{
List<String> records = new ArrayList<String>();
try
{
BufferedReader buff = new BufferedReader(new InputStreamReader(
Configuration.class.getResourceAsStream(filename)));
String line;
while ((line = buff.readLine()) != null)
{
records.add(line);
}
buff.close();
return records;
}
catch (Exception e)
{
System.err.format("Exception occurred trying to read '%s'.", filename);
e.printStackTrace();
return null;
}
}
//action performed
public void actionPerformed(ActionEvent ae) {
JButton b = (JButton)ae.getSource();
if( b.equals(newGameButton) )
{
flag = true;
controller.startGame();
buttonPressed = "newGameBtn";
}
if(b.equals(quitButton))
{
System.exit(0);
}
if(b.equals(ruleButton)){
readFile("../resource/riskRules.txt");
}
}
Appreciate the help!

If "Resources" it's marked as resource in Eclipse. The txt file should be copied to your class path when you build.
As per what I can guess from your code you should be doing something like
Configuration.class.getResourceAsStream("riskRules.txt")
Since your file will be at the root level of your class path.
If for example the file is withing a dir called "text" in your resources you would use something like
Configuration.class.getResourceAsStream("text/riskRules.txt")

There needs to be some level of rudimentary error checking on the result returned from getResourceAsStream before you attempt to use it. Is there a reason you're using getResourceAsStream instead of getResource? If the file exists on disk (I see from your OP that it's because it's in a package, and may not physically exist on the disk), then you can just use that to return the path to it, and create a file object from it.
String path = "/path/to/resource"; // note the leading '/' means "search from root of classpath"
URL fileUrl = getClass().getResource(path);
if (fileUrl != null ) {
File f = new File(fileUrl.toURI());
BufferedReader = new BufferedReader(new FileReader(f));
// do stuff here...
}
else {
// file not found...
}
If you need to pull the file out of the JAR archive, then you can do this:
String path = "/path/to/resource"; // note the leading '/' means "search from root of classpath"
InputStream is = getClass().getResourceAsStream(path);
if (is != null ) {
BufferedReader = new BufferedReader(new InputStreamReader(is));
// do stuff here...
}
else {
// file not found...
}
In the event your resource is not found, you will avoid the NPE and you can properly account for the fact that it's missing.
Note that if you do have your resources in a package (jar), then you cannot use a path to locate it that uses "..", since there is no "relative path" in a jar archive, it's not actually a file on the filesystem.
Your "resources" are located by the relative path you specify in the getResource... method. A leading "/" means to look at the root of your classpath for locating the resource. No leading "/" means to look relative to the location of the class file that you're using to locate the resource.
If your file is in a location called "com.program.resources", and you're trying to locate it from a class called "com.program.someotherpackage.MyClass", then you'd use:
getClass().getResourceAsStream("/com/program/resources/<file.txt>");
to find it.
Here's my example illustrated:
<classpath root>
com
program
resources
file.txt
img.png
someotherpackage
MyClass.class
Generally, it's common practice to leave resources outside your package structure, to avoid confusion when locating them later. Most IDE's have a way to mark your directories as resources, so when the program is compiled, they will be copied to the proper location in the classpath root, and can be found by any class asking for them.

Is it possible to create a NEW zip file using the java FileSystem?

I've successfully modified the contents of a (existing) zip file using the FileSystem provided by java 7, but when I tried to create a NEW zip file by this method it fails, with the error message that says: "zip END header not found", it is logical because of the way I'm doing it, first I create the file (Files.createFile) which is a completely empty file, and then I try to access to its file system , and since the file is empty its impossible to find any header inside the zip, my question is is there any way to create a new zip file completely empty using this method?; the hack that I've considered is adding an empty new ZipEntry to a the zip file and then using that new empty file to crate the file system based on it, but i really want to think that the guys of oracle implemented a better (easier) way to do this with nio and the filesystems...
this is my code (the error appears when creating the file system):
if (!zipLocation.toFile().exists()) {
if (creatingFile) {
Files.createFile(zipLocation);
}else {
return false;
}
} else if (zipLocation.toFile().exists() && !replacing) {
return false;
}
final FileSystem fs = FileSystems.newFileSystem(zipLocation, null);
.
.
.
zipLocation is a Path
creatingFile is a boolean
ANSWER:
in my particular case the answer given didn't work appropriately because of the spaces in the path, therefore i have to do it the way i didn't want to:
Files.createFile(zipLocation);
ZipOutputStream out = new ZipOutputStream(
new FileOutputStream(zipLocation.toFile()));
out.putNextEntry(new ZipEntry(""));
out.closeEntry();
out.close();
it does not mean that the given answer is wrong, it just didn't work for my particular case

As described in The Oracle Site:
public static void createZip(Path zipLocation, Path toBeAdded, String internalPath) throws Throwable {
Map<String, String> env = new HashMap<String, String>();
// check if file exists
env.put("create", String.valueOf(Files.notExists(zipLocation)));
// use a Zip filesystem URI
URI fileUri = zipLocation.toUri(); // here
URI zipUri = new URI("jar:" + fileUri.getScheme(), fileUri.getPath(), null);
System.out.println(zipUri);
// URI uri = URI.create("jar:file:"+zipLocation); // here creates the
// zip
// try with resource
try (FileSystem zipfs = FileSystems.newFileSystem(zipUri, env)) {
// Create internal path in the zipfs
Path internalTargetPath = zipfs.getPath(internalPath);
// Create parent directory
Files.createDirectories(internalTargetPath.getParent());
// copy a file into the zip file
Files.copy(toBeAdded, internalTargetPath, StandardCopyOption.REPLACE_EXISTING);
}
}
public static void main(String[] args) throws Throwable {
Path zipLocation = FileSystems.getDefault().getPath("a.zip").toAbsolutePath();
Path toBeAdded = FileSystems.getDefault().getPath("a.txt").toAbsolutePath();
createZip(zipLocation, toBeAdded, "aa/aa.txt");
}