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java.nio.file.NoSuchFileException When File.transferTo() is called

I've recently inherited a Java API and am having trouble with file uploads. Unfortunately, Java isn't a language I have much experience in so I'm a bit stumped by this.
The MultiPartFile is being received ok, and I can find the file in the temp directory, but when I try to use File.transferTo() to create the final file I just get the below error;
java.nio.file.NoSuchFileException: C:\Users\myUser\AppData\Local\Temp\undertow3706399294849267898upload -> S:\Dev\PolicyData\Temp.xlsx
As I mentioned the temp undertow file exists, and the directory on the S drive also exist, (but there's no Temp.xlsx as my understanding is this should be created by transferTo()). Any solutions I've found to this problem so far are resolved using absolute file paths.
This is a simplified version of the code but the error remains the same.
SpringBoot framework is "1.5.3.RELEASE", running Java 1.8.0_131
ResponseEntity handleFileUpload(#RequestPart(name = "file") MultipartFile file, #PathVariable Long stageFileTypeId) {
if (!file.isEmpty()) {
try {
String filePath = "S:\\Dev\\PolicyData\\Temp.xlsx";
log.info("Upload Path = {}", filePath);
File dest = new File(filePath);
file.transferTo(dest);
return ResponseUtil.wrapOrNotFound(Optional.ofNullable(filePath));
}
catch (Exception ex) {
log.error("An error has occurred uploading the file", ex);
return ResponseEntity.status(HttpStatus.INTERNAL_SERVER_ERROR).build();
}
}
else {
log.error("An error has occurred, no file was received");
return ResponseEntity.status(HttpStatus.INTERNAL_SERVER_ERROR).build();
}
}
If you need any more information please let me know.
Thanks,
Neil

The API for MultipartFile is a bit tricky. The transferTo(File) method javadoc states that (bold are mine):
This may either move the file in the filesystem, copy the file in the
filesystem, or save memory-held contents to the destination file. If
the destination file already exists, it will be deleted first.
If the target file has been moved in the filesystem, this operation
cannot be invoked again afterwards. Therefore, call this method just
once in order to work with any storage mechanism.
It seems that the Undertow implementantion already called it to move the in-memory uploaded file to "C:\Users\myUser\AppData\Loca\Temp\undertow3706399294849267898upload" so another transferTo is failing.
I came across the same problem using javax.servlet.http.Part in a Wildfly containter with Undertow.
If you are using Spring framework >= 5.1, you could try the Multipart.transferTo(Path) method, using dest.toPath()
Or you can copy from the inputStream, with something like this:
try (InputStream is = multipartFile.getInputStream()) {
Files.copy(is, dest.toPath());
}

Google Drive API not downloading FIles (Java v3)

I'm using the code from here:
https://developers.google.com/drive/api/v3/manage-downloads#downloading_a_file
The code snippet I'm using is the following and placed in the main method:
String fileId = "some file ID";
OutputStream outputStream = new ByteArrayOutputStream();
driveService.files().get(fileId)
.executeMediaAndDownloadTo(outputStream);
I have found no sign of the code actually downloading the file, nor do I know where the file is IF it actually downloads.
I'm not sure if I am using the proper scope to gain permission to download files. I am able to upload, list, and delete files as long as I know the fileID, but downloading seems to not work.
private static final List<String> SCOPES = Collections.singletonList(DriveScopes.DRIVE);
Alternatively, I'm trying to create a method to enact the download protocol like so:
private static void downloadFile(Drive service, File file (or String fileID)){
}
but am not sure on how to do so. I've tried looking for samples online but most are from v1 or v2 apis and don't seem to work for me.
Also, I've read somewhere that it is not possible to download a Folder. Instead, I have to download each item in the folder one by one.
So do I have to make an Arraylist/list/array of the fileIDs and iterate through it after initializing a variable to represent fileID?
Edit: Some progress has been made, but I still have some problems I'm trying to thrash out.
List<File> files = result.getFiles();
File newFile;
if (files == null || files.isEmpty()) {
System.out.println("No files found.");
} else {
System.out.println("Files:");
for (File file : files) {
System.out.printf("%s (%s)\n", file.getName(), file.getId());
String fileId = file.getId();
//System.out.println(fileId);
String fileName = file.getName();
//System.out.println(fileName);
OutputStream outputstream = new FileOutputStream();
service.files().get(fileId)
.executeMediaAndDownloadTo(outputstream);
outputstream.flush();
outputstream.close();
}
What I want:
The above code is in the main method. I don't know if this is the proper way to do it, but as the program fetches each file and executes the System.out.printf, I also want it to download that file (with the same mimeType, and pref the same name too) into the destination set in the OutputStream constructor (C://User//some name//Downloads).
What I've tried:
From what I've tested, it only downloads the first file exactly the way I want, but only because I specify the name and extension in OutputStream. I've initialized variables 'fileId' and 'fileName' so that they will change according to the info as the program fetches the metadata for the next file, but I don't know how to change or set multiple constructors into this code:
OutputStream outputstream = new FileOutputStream();
service.files().get(fileId)
.executeMediaAndDownloadTo(outputstream);
to download all the files.
My folder hierarchy in Google Drive is like this:
Logs
-- bin (folder)
---- bunch of .bin files
-- .xml file
-- .xml file

You are using a ByteArrayOutputStream object as the output of your download. If your program terminates without having saved the contents of this object somewhere, you will not be able to find this information in your computer's disk, as it is not written to it but rather saved in memory as a buffered byte-array (refer to the previous link for more information).
If you want to save the output of the download to the file, I suggest you use instead a FileOutputStream as the destination of your download. In order to do that, you have to modify your code as follows:
Add the appropriate import declaration:
import java.io.FileOutputStream;
Modify your outputStream variable assignment as follows:
OutputStream outputStream = new FileOutputStream('/tmp/downloadedfile');
Where the parameter passed to FileOutputStream should be the desired destination path of your download.
After writing any contents to your file, add the following lines of code:
outputStream.flush();
outputStream.close();
This will ensure that your file is being written to properly.
In regards to downloading a folder, you are completely right - you will first need to fetch the folder you want to download, and each of their children. In order to better understand how to do it, I suggest you check out the following answer: Download folder with Google Drive API
Edit - example downloading a folder
String destinationFolder = "/tmp/downloadedfiles/";
List<File> files = result.getFiles();
File newFile;
if (files == null || files.isEmpty()) {
System.out.println("No files found.");
} else {
System.out.println("Files:");
for (File file : files) {
System.out.printf("%s (%s)\n", file.getName(), file.getId());
String fileId = file.getId();
String fileName = file.getName();
OutputStream outputstream = new FileOutputStream(destinationFolder + fileName);
service.files().get(fileId)
.executeMediaAndDownloadTo(outputstream);
outputstream.flush();
outputstream.close();
}
}

Trying to create a new file throws FileNotFoundException but file exists in the same package [duplicate]

This question already has answers here:
java.io.FileNotFoundException: the system cannot find the file specified
(8 answers)
Closed 4 years ago.
I have a csv file in the same path as everything else. Now, when I try to create a File object:
public void getMenu() {
File fileMenu = new File("FastFoodMenu.csv");
try {
Scanner inputStream = new Scanner(fileMenu);
while (inputStream.hasNext()) {
String data = inputStream.next();
System.out.println(data);
}
} catch (FileNotFoundException ex) {
Logger.getLogger(FileHandler.class.getName()).log(Level.SEVERE, null, ex);
}
}
it throws a FileNotFoundException.
the absolute path to all files in the project is:
C:\Users\kenyo\Documents\NetBeansProjects\OrderFastFood\src\fastfoodorderingsystem
I also checked the name a couple of times. fileMenu.exists() returns false.

First, in your root/working directory (in your case it's the folder containing your project), create a folder called 'menus', here you can store all your menus (so you can play around with multi-file input).
Second, move your FastFoodMenu.csv file to that menus folder.
The FastFoodMenu.csv relative path should now look like this: OrderFastFood\menus\FastFoodMenu.csv.
Third, get your working directory from the System properties. This is the folder in which your program is working in. Then, get a reference (File object) to the menus folder.
Lastly, get a reference to the file in question inside the menu folder. When you get to multi-file reading (and at some point, multi-folder reading), you're gonna want to get the files inside the menu folder as a list so that's why I say to just get the menus folder as it's own reference (or just get the file without the isolated reference to the parent aka '\menus\').
So your code should really look like this:
public void getMenu() {
final File workingDir = File(System.getProperty("user.dir"));
final File menusDir = File(workingDir, "menus");
final File fastFoodMenu = File(menusDir, "FastFoodMenu.csv");
try {
final FileInputStream fis = new FileInputStream(fastFoodMenu);
final BufferedInputStream bs = new BufferedInputStream(fis);
while((l = bs.readLine()) != null) {
System.out.println(l);
}
} catch(FileNotFoundException e) {
System.out.println(e.getMessage());
e.printStackTrace()
}
}
This is all psuedocode but that should at least get you started. Make sure to use BufferedInputStream for efficiency, and when reading files, always pass them into FileInputStream's. It's much better than using the Scanner class. I should also mention that when creating a File object, you're not actually creating a file. What you're doing is your're creating an object, giving it the data you want it to have (such as whether it's a folder, and if it is, what child files/folders do you want it to have, whether it's protected or not, hidden or not, etc) before actually telling the system to create the file with everything else.

Your csv file is probably at the wrong place. You're just specifying the file name, which is a relative path.
Relative paths are always resolved against the working directory of your application, not against the directory where your source file(s) are.
To solve the issue, you can
move the files to the real working directory.
use an absolute path (not advisable!)
specify the folder of your data files as program argument or in a config file (in your working directory)
put the files somewhere into the classpath of your application and load them from there via classloader. Note that files that are in your classpath are usually packed with your application and hence not easily modifiable by the user, so this solution doesn't work if the file must be changed by the user.

How to write to a file in Android (without FileNotFoundException Error)

I am having some issues writing to a file in my internal storage in Android. What I am wanting to achieve, is it first of all check whether there is a file (of specified name) already existing in the internal storage. If there isn't, I would then like it to create a file with that name, and write specified content to it. If there is already a file, I would simply like it to write specified content appended to what is already stored within the file. I think I have found how I can check if the file is in existence, and also create a new blank file with the specified name if it isn't (see code below). However, when coming to write to the file I get an error message java.io.FileNotFoundException error, even though my code should ensure that there will always be a file in existence to open with the FileOutputStream. Please could anyone tell me what I am doing wrong. Thanks in advance.
public boolean isFilePresent() {
String strASANo = getIntent().getStringExtra("strASA"); //Part of File Name imported passed from parameter
String logsFile = strASANo + "logs";
File file = getBaseContext().getFileStreamPath(logsFile);
return file.exists();
}
//////////////////////////////////////////////////////////////////////////////
public void addSaveClick(View view) {
if (!isFilePresent()) {
File file = new File(getApplicationContext().getFilesDir(), logsFile);
}
FileOutputStream logWriter;
try {
logWriter = openFileOutput(logsFile, MODE_APPEND); //Unhandled Exception: java.io.FileNotFoundException
}
}

How can I access a non-class file inside a jar FROM 'within' the jar

This should be simple but it has cost me hours. Everything I find on this site indicates I am doing it right but the file still cannot be found.
Inside a jar file I have two files 'CDAkeystore.jks' and 'CDAtruststore.jks' at top level.
Yet when I call
securityProps.setProperty("javax.net.ssl.keyStore","CDAkeystore.jks");
I get a system cannot find the file requested error.
The class file calling this method is inside the same jar in the usual package arrangement.
The jar file is as follows:
com ..... (a lot more class files)
org ..... (lots of class files)
META-INF
CDAtruststore.jks
CDAkeystore.jks
How can this be SOOO difficult?!!
---------- Added INfo ------n
Since the object using the path is open source I found the routine they are using to load the file. It is:
InputStream keystoreInputStream = preBufferInputStream(new FileInputStream(keyStoreName));
which according to the documentation of FileInputStream(String name) is
Creates a FileInputStream by opening a connection to an actual file, the file named by the path name 'name' in the file system. So how should this path be expressed?

Use YourClass.class.getResourceAsStream() or this.getClass().getResourceAsStream(). You can also use class loader if you are in multiple class loaders environment.

The answer is, in short, that you can't. At least in this situation. I am stuck with passing a path to a file to a library implementation that I have no control over. So the library method accesses the file on the assumption that the file exists in unzipped form in the OS's file system. It is getting the path from a Property.setProperty(stringKey, stringPath)
So the only solution I found was an ugly hack. I need to take the resource in my jar and copy it to a file on the system. Then I would pass the path to that file in the above setProperty() method. The ugly hack is implemented as follows (if anyone else can come up with a nicer solution I would be happy). It does solve the problem. The library routine is able to find my newly created file.
/* This evil reads a file as a resource inside a jar and dumps the file where ever
* the loader of this jar/application defines as the current directory. This pain is done
* so the SecurityDomian class can load the file; it cannot access the file from
* the jar. This means the 'name' passed in contains the file name inside the jar
* prefixed with "/" so it is not read with an assumed package extension.
*/
private boolean createFileFromResource(String name)
{
// Dont bother if the file already exists
if(new File(name.replace("/", "")).exists())
{
return true;
}
byte[] keystoreFile = new byte[2048];
ByteArrayOutputStream byteArrayOut = new ByteArrayOutputStream(2048);
// Get the resource
InputStream inputStream = this.getClass().getResourceAsStream(name);
try
{
int bytesRead = 0;
while(true)
{
// Read the resource into the buffer keystoreFile in 2048 byte chunks
bytesRead = inputStream.read(keystoreFile);
if(bytesRead < 0)
{
break;
}
// Copy and append the chunks to the ByteArrayOutputStream (this class
// does the auto-extending of the output array as more chunks are
// added so you don't have to do it.
byteArrayOut.write(keystoreFile, 0, bytesRead);
}
inputStream.close();
// Now create a file at the root of where ever the loader happens to think
// the root is. So remove the "/" in front of the file name
FileOutputStream outputStream = new FileOutputStream(name.replace("/", ""));
// Write out the file. Note you will be left with garbage at that location.
byteArrayOut.writeTo(outputStream);
outputStream.flush();
outputStream.close();
}
catch (IOException e)
{
e.printStackTrace();
return false;
}
return true;
}