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Cannot Find Path Specified Creating File Despite Existing Parent Directory

I am trying to export zip files to a directory and running into an IOException stating that the file path cannot be found. I am aware that this means that the parent directory does not exist usually, however debugging the line where the file is being written file.getParentFile().exists() returns true, so this is not my issue. To further complicate matters, this only occurs for approximately half of the files written. It is always the same files that fail when unzipping via java, but unzipping them via windows always successfully works.
Here is the code I am using:
ZipInputStream zis =
new ZipInputStream(new ByteArrayInputStream(zipFile));
ZipEntry ze = zis.getNextEntry();
while (ze != null) {
String fileName = ze.getName();
File newFile = new File(outputFolder + File.separator + fileName);
if(!newFile.isDirectory()) {
newFile.getParentFile().mkdirs();
FileOutputStream fos = new FileOutputStream(newFile); //Exception occurs here
//newFile.getParentFile().exists() returns true
//copying the path for newFile.getParentFile() into my file browser leads me to a valid, existing folder
//I have tried newFile.createNewFile() and that errors with a similar exception
int len;
while ((len = zis.read(buffer)) > 0) {
fos.write(buffer, 0, len);
}
fos.close();
results.add(new Foo());
}
ze = zis.getNextEntry();
}
Example exception:
java.io.FileNotFoundException: \\foo\foo\foo\foo\foo\foo\foo.pdf (The system cannot find the path specified)
Some more notes about the system: the file system is a remote network drive, the system is running windows, and the account has full write access to the drive. I have also verified that naming a file foo.pdf (copy and pasting the name of the file intended to be written) does not cause any issue either.

The issue is that zip files can have trailing spaces in their paths. For example, "Test whatever .zip" could be the file name so java sees the folder as "/Test whatever /" and it tries to create that folder. Windows tells java that it succeeded, but really it created a folder at "/Test Whatever/". When dealing with folders, the file IO has no issue with this, but when writing files, it completely bombs as it's looking for the path explicitly. It does not truncate the extra white space the same way it does when dealing with folders, as you would expect.
Here is the code I used to resolve it:
String path = (outputFolder + File.separator + fileName).replace(" ", "");
File newFile = new File(path);

Read directly a file within a Zip file - Java

My situation is that I have a zip file that contains some files (txt, png, ...) and I want to read it directly by their names, I have tested the following code but no result (NullPointerExcepion):
InputStream in = Main.class.getResourceAsStream("/resouces/zipfile/test.txt");
BufferedReader br = new BufferedReader(new InputStreamReader(in, "UTF-8"));
resources is a package and zipfile is a zip file.

If you can be sure that your zip file will never be packed inside another jar, you can use something like:
URL zipUrl = Main.class.getResource("/resources/zipfile.zip");
URL entryUrl = new URL("jar:" + zipUrl + "!/test.txt");
InputStream is = entryUrl.openStream();
Or:
URL zipUrl = Main.class.getResource("/resources/zipfile.zip");
File zipFile = new File(zipUrl.toURI());
ZipFile zip = new ZipFile(zipFile);
InputStream is = zip.getInputStream(zip.getEntry("test.txt"));
Otherwise, your choices are:
Use a ZipInputStream to scan through the zip file once for each entry that you need to load. This may be slow if you have a lot of resources, unless you can reuse the same ZipInputStream for all your resources.
Don't pack the resources in a nested zip file, and just inline them in the jar with the code.
Copy the nested zip file into a temporary directory, and access it using the ZipFile class.

Your current approach is definitely not going to work. You made up an arbitrary 'access' scheme and used it in a class that has no idea what you are trying to do. What you can do is use a ZipInputStream to read the entry you are looking for:
URL zipFileURL = Thread.currentThread().getContextClassLoader().getResource("zipfile.zip");
InputStream inputStream = zipFileURL.openStream();
ZipInputStream zipInputStream = new ZipInputStream(inputStream);
ZipEntry zipEntry = null;
do {
zipEntry = zipInputStream.getNextEntry();
if(zipEntry == null) break;
}
while(zipEntry != null && (! "textfile".equals(zipEntry.getName()));
if(zipEntry != null ) {
// do some stuff
}
This is adhoc code, fix it up to do what you need. Also, there might be some more efficient classes to handle Zip files, for example in the Apache Commons IO library.

What is the path of the test.txt within the zip file? You need to use the path within the zip file to read this file. Also make sure that your zipfile is in the classpath. In fact, you can bundle this in a jar file.

Java: File output help

Fixed: Instead of calling isFile() I used exists() and it seems to be working fine. If possible could someone explain why this change worked?
I'm attempting to write out to an excel file but am having a problem when trying to create that file if the name already exists.
Basically I am taking a file that is uploaded to a server, reading it, and then outputting a report file in a new location with the same filename. I tried to do this by simply checking if the file already existed and then adding a number onto the filename. My code works if the file doesn't exist or if it exists without a number (e.g. filename.xls). If a file exists with the name "filename1.xls" the server just seems to hang when trying to write the file. What can do to fix this?
Here is my code:
String destination = "c:/apache-tomcat-7.0.8/webapps/reports/" + fileName.substring( fileName.lastIndexOf("\\")+1, fileName.lastIndexOf(".")) + ".xls";
int filenum = 1;
while (new File(destination).isFile()) {
destination = "c:/apache-tomcat-7.0.8/webapps/reports/" + fileName.substring( fileName.lastIndexOf("\\")+1, fileName.lastIndexOf(".")) + filenum + ".xls";
filenum++;
}
WritableWorkbook workbook = Workbook.createWorkbook(new File(destination));

That will happen if some process is still keeping the file open. E.g. you've created a FileInputStream on the file to read it, but are never calling close() on it after reading.
Unrelated to the problem, the expanded WAR folder is not the best place to use as a permanent storage. All those files in the expanded WAR folder will get lost whenever you redeploy the WAR. Also hardcoding a servletcontainer-specific path in the code makes it totally unportable.
If your actual intent is to return the Excel file on a per-request basis to the client using a servlet, then you should be using
WritableWorkbook workBook = Workbook.createWorkbook(response.getOutputStream());
// ...
This way it writes to the response immediately without the need for an intermediate file.

Use the File.createTempFile(prefix, suffix, directory) API:
String localName = new File(fileName).getName();
String nameNoExt = localName.substring(0, fileName.lastIndexOf("."));
String extension = localName.substring(fileName.lastIndexOf(".")); // need to include the .
File directory = new File("c:/apache-tomcat-7.0.8/webapps/reports/");
File destFile = File.createTempFile(nameNoExt, extension, directory)

What is the fastest way to extract 1 file from a zip file which contain a lot of file?

I tried the java.util.zip package, it is too slow.
Then I found LZMA SDK and 7z jbinding but they are also lacking something.
The LZMA SDK does not provide a kind of documentation/tutorial of how-to-use, it is very frustrating. No javadoc.
While the 7z jbinding does not provide a simple way to extract only 1 file, however, it only provide way to extract all the content of the zip file. Moreover, it does not provide a way to specify a location to place the unzipped file.
Any idea please?

What does your code with java.util.zip look like and how big of a zip file are you dealing with?
I'm able to extract a 4MB entry out of a 200MB zip file with 1,800 entries in roughly a second with this:
OutputStream out = new FileOutputStream("your.file");
FileInputStream fin = new FileInputStream("your.zip");
BufferedInputStream bin = new BufferedInputStream(fin);
ZipInputStream zin = new ZipInputStream(bin);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
if (ze.getName().equals("your.file")) {
byte[] buffer = new byte[8192];
int len;
while ((len = zin.read(buffer)) != -1) {
out.write(buffer, 0, len);
}
out.close();
break;
}
}

I have not benchmarked the speed but with java 7 or greater, I extract a file as follows.
I would imagine that it's faster than the ZipFile API:
A short example extracting META-INF/MANIFEST.MF from a zip file test.zip:
// file to extract from zip file
String file = "MANIFEST.MF";
// location to extract the file to
File outputLocation = new File("D:/temp/", file);
// path to the zip file
Path zipFile = Paths.get("D:/temp/test.zip");
// load zip file as filesystem
try (FileSystem fileSystem = FileSystems.newFileSystem(zipFile)) {
// copy file from zip file to output location
Path source = fileSystem.getPath("META-INF/" + file);
Files.copy(source, outputLocation.toPath());
}

Use a ZipFile rather than a ZipInputStream.
Although the documentation does not indicate this (it's in the docs for JarFile), it should use random-access file operations to read the file. Since a ZIPfile contains a directory at a known location, this means a LOT less IO has to happen to find a particular file.
Some caveats: to the best of my knowledge, the Sun implementation uses a memory-mapped file. This means that your virtual address space has to be large enough to hold the file as well as everything else in your JVM. Which may be a problem for a 32-bit server. On the other hand, it may be smart enough to avoid memory-mapping on 32-bit, or memory-map just the directory; I haven't tried.
Also, if you're using multiple files, be sure to use a try/finally to ensure that the file is closed after use.

ZipOutputStream relative paths

I create zip file using ZipOutputStream. I put in the zip one file(both file and zip are in the same dir), however the file is stored with fullpath (C:\TEMP\file.xml), how to store it with relative or no path?

You need to set that in the ZipEntry. For example, if you don't want any path, just use the name of the file in your ZipEntry, like this:
File f = new File("C:\\temp\\file.xml");
ZipEntry entry = new ZipEntry(f.getName());