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File (.txt) cannot be found

I am creating a stock market simulator (beginner) and I made a .txt file to save the stock symbol and name within a file. I am having an issue where my code is unable to find the file on my desktop.
public static void load() throws FileNotFoundException {
File file = new File("/Users/dhruvchaudhari/Desktop/stocks.txt");
Scanner scan = new Scanner(file);
while ((scan.hasNextLine())) {
System.out.println(scan.nextLine());
}
}
The error it is throwing is as such
java.io.FileNotFoundException: /Users/*username*/Desktop/stocks.txt (No such file or directory)
I'm on Mac and I checked the directory for the file directory and it should be correct. Any suggestions?

You can check the current working directory to confirm the path of your file by adding System.out.println("Working Directory = " + System.getProperty("user.dir"));
this will return the path you can debug this to get the idea of path in your application.
Also you can add read the file if its not there the code will create it for so you can add your metadata into the generated file.
By using this approach I hope you can move ahead.
public static void load() throws IOException {
File yourFile = new File(path);
yourFile.createNewFile(); // if file already exists will do nothing
Scanner scan = new Scanner(yourFile);
while ((scan.hasNextLine())) {
System.out.println(scan.nextLine());
}
}

Obviously your path is wrong or the file doesn't exist. You can use the if statement to determine whether the file exists first, and create the file when it does not exist.

Zip4j can't remove or overwrite files inside file

I'm trying to edit the contents of an odt file using zip4j (I tried using java ZipEntries but I couldn't even delete the entries from the file itself that's why I chose to use a library instead). I can confirm that the file I am trying to overwrite exits I can even read from it and tell when it was created so that part works. Now when I'm trying to edit the odt contents (removing or overwriting) Zip4j throws a ZipException which says: cannot rename modified zip file. What am I doing wrong?
try
{
File temp = new File(f.getParent()+"/tmp/content.xml");
new File(temp.getParent()).mkdirs();
FileUtils.write(temp, "", encoding);
net.lingala.zip4j.ZipFile zf = new net.lingala.zip4j.ZipFile(f.getPath());
ZipParameters p = new ZipParameters();
p.setEncryptionMethod(EncryptionMethod.NONE);
p.setOverrideExistingFilesInZip(true);
p.setFileNameInZip("content.xml");
p.setCompressionMethod(CompressionMethod.DEFLATE);
zf.addFile(temp, p);
}
catch (Exception e)
{
e.printStackTrace();
}

The zip file system with its jar:file: protocol is supported by Path & Files. A Path maintains its FileSystem, so one can use all operations.
Path osPath = Paths.get("C:/ ... .odt");
URI uri = URI.create("jar:" + osPath.toUri());
Map<String, String> env = new HashMap<>();
env.put("create", "true");
try (FileSystem zipFS = FileSystems.newFileSystem(uri, env)) {
Files.copy(zipFS.getPath("/media/image1.png"), osPath.resolveSibling("image1.png"),
StandardCopyOption.REPLACE_EXISTING);
Files.move(zipFS.getPath("/media/image2a.png"), zipFS.getPath("/media/image2.png"));
}

renaming file name inside a zip file

trying to rename internal file within a zip file without having to extract and then re-zip programatically.
example. test.zip contains test.txt, i want to change it so that test.zip will contain newtest.txt(test.txt renamed to newtest.txt, contents remain the same)
came across this link that works but unfortunately it expects test.txt to exist on the system. In the example the srcfile should exist on the server.
Blockquote Rename file in zip with zip4j
Then icame across zipnote on Linux that does the trick but unfortunately the version i have doesnt work for files >4GB.
Any suggestions on how to accomplish this? prefereably in java.

This should be possible using Java 7 Zip FileSystem provider, something like:
// syntax defined in java.net.JarURLConnection
URI uri = URI.create("jar:file:/directoryPath/file.zip");
try (FileSystem zipfs = FileSystems.newFileSystem(uri, Collections.<String, Object>emptyMap())) {
Path sourceURI = zipfs.getPath("/pathToDirectoryInsideZip/file.txt");
Path destinationURI = zipfs.getPath("/pathToDirectoryInsideZip/renamed.txt");
Files.move(sourceURI, destinationURI);
}

Using zip4j, I am modifying and re-writing the file headers inside of the central directory section to avoid rewriting the entire zip file:
ArrayList<FileHeader> FHs = (ArrayList<FileHeader>) zipFile.getFileHeaders();
FHs.get(0).setFileName("namename.mp4");
FHs.get(0).setFileNameLength("namename.mp4".getBytes("UTF-8").length);
zipFile.updateHeaders ();
//where updateHeaders is :
public void updateHeaders() throws ZipException, IOException {
checkZipModel();
if (this.zipModel == null) {
throw new ZipException("internal error: zip model is null");
}
if (Zip4jUtil.checkFileExists(file)) {
if (zipModel.isSplitArchive()) {
throw new ZipException("Zip file already exists. Zip file format does not allow updating split/spanned files");
}
}
long offset = zipModel.getEndCentralDirRecord().getOffsetOfStartOfCentralDir();
HeaderWriter headerWriter = new HeaderWriter();
SplitOutputStream splitOutputStream = new SplitOutputStream(new File(zipModel.getZipFile()), -1);
splitOutputStream.seek(offset);
headerWriter.finalizeZipFile(zipModel, splitOutputStream);
splitOutputStream.close();
}
The name field in the local file header section remains unchanged, so there will be a mismatch exception in this library.
It's tricky but maybe problematic, I don't know..

How to read data in one zip file located in another zipfile using ZipEntry?

I have a zip file (x.zip) in which there is another zipfile (y.zip). I need to read a file in y.zip. How can I iterate through both the zip files to read the file?
The code I am using to iterate x.zip to read y.zip is as below.
In the code, "zipX" represents "x.zip". When "y.zip" is encountered, it satisfies the "if condition" in the code. Here I need to iterate through "zipEntry" and read a file in that.
How can this be achieved?
private void getFileAsBytes(String path, String name) throws IOException {
ZipFile zipX = new ZipFile(path);
Enumeration<? extends ZipEntry> entries = zipX.entries();
while (entries.hasMoreElements())
{
ZipEntry zipEntry = entries.nextElement();
if(zipEntry.getName().contains(name) && zipEntry.getName().endsWith(".zip")) {
InputStream is;
is = zipX.getInputStream(zipEntry);
// Need to iterate through zipEntry here and read data from a file inside it.
break;
}
}
zipX.close();
}

According to the ZipFile docs, you need to pass in a File object or a file path; an InputStream isn't supported.
With that in mind, you could write that InputStream to a temporary file and then pass that file in to your existing method:
...
is = zipX.getInputStream(zipEntry);
File tmpDir = new File(System.getProperty("java.io.tmpdir"));
//For production, generate a unique name for the temp file instead of using "temp"!
File tempFile = createTempFile("temp", "zip", tmpDir);
this.getFileAsBytes(tempFile.getPath(), name);
break;
...

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");
}