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create zip file without writing to disk

I am working on a Springboot application that has to return a zip file to a frontend when the user downloads some report. I want to create a zip file without writing the zip file or the original files to disk.
The directory I want to zip contains other directories, that contain the actual files. For example, dir1 has subDir1 and subDir2 inside, subDir1 will have two file subDir1File1.pdf and subDir1File2.pdf. subDir2 will also have files inside.
I can do this easily by creating the physical files on the disk. However, I feel it will be more elegant to return these files without writing to disk.

You would use ByteArrayOutputStream if the scope was to write to memory. In essence, the zip file would be entirely contained in memory, so be sure that you don't risk to have too many requests at once and that the file size is reasonable in size! Otherwise this approach can seriously backfire!

You can use following snippet :
public static byte[] zip(final String str) throws IOException {
if (StringUtils.isEmpty(str)) {
throw new IllegalArgumentException("Cannot zip null or empty string");
}
ByteArrayOutputStream bos = new ByteArrayOutputStream();
try (GZIPOutputStream gos = new GZIPOutputStream(bos)) {
gos.write(str.getBytes(StandardCharsets.UTF_8));
}
return bos.toByteArray();
}
But as stated in another answer, make sure you are not risking your program too much by loading everything into your java memory.

Please note that you should stream whenever possible. In your case, you could write your data to https://docs.oracle.com/javase/8/docs/api/index.html?java/util/zip/ZipOutputStream.html.
The only downside of this appproach is: the client won't be able to show a download status bar, because the server will not be able to send the "Content-length" header. That's because the size of a ZIP file can only be known after it has been generated, but the server needs to send the headers first. So - no temporary zip file - no file size beforehand.
You are also talking about subdirectories. This is just a naming issue when dealing with a ZIP stream. Each zip item needs to be named like this: "directory/directory2/file.txt". This will produce subdirectories when unzipping.

NoSuchFileException in Files.newInputStream with StandardOpenOption.CREATE

I am trying to open a file for reading or create the file if it was not there.
I use this code:
String location = "/test1/test2/test3/";
new File(location).mkdirs();
location += "fileName.properties";
Path confDir = Paths.get(location);
InputStream in = Files.newInputStream(confDir, StandardOpenOption.CREATE);
in.close();
And I get java.nio.file.NoSuchFileException
Considering that I am using StandardOpenOption.CREATE option, the file should be created if it is not there.
Any idea why I am getting this exception?

It seems that you want one of two quite separate things to happen:
If the file exists, read it; or
If the file does not exist, create it.
The two things are mutually exclusive but you seem to have confusingly merged them. If the file did not exist and you've just created it, there's no point in reading it. So keep the two things separate:
Path confDir = Paths.get("/test1/test2/test3");
Files.createDirectories(confDir);
Path confFile = confDir.resolve("filename.properties");
if (Files.exists(confFile))
try (InputStream in = Files.newInputStream(confFile)) {
// Use the InputStream...
}
else
Files.createFile(confFile);
Notice also that it's better to use "try-with-resources" instead of manually closing the InputStream.

Accordingly to the JavaDocs you should have used newOutputStream() method instead, and then you will create the file:
OutputStream out = Files.newOutputStream(confDir, StandardOpenOption.CREATE);
out.close();
JavaDocs:
// Opens a file, returning an input stream to read from the file.
static InputStream newInputStream(Path path, OpenOption... options)
// Opens or creates a file, returning an output stream that
// may be used to write bytes to the file.
static OutputStream newOutputStream(Path path, OpenOption... options)
The explanation is that OpenOption constants usage relies on wether you are going to use it within a write(output) stream or a read(input) stream. This explains why OpenOption.CREATE only works deliberatery with the OutputStream but not with InputStream.
NOTE: I agree with #EJP, you should take a look to Oracle's tutorials to create files properly.

I think you intended to create an OutputStream (for writing to) instead of an InputStream (which is for reading)
Another handy way of creating an empty file is using apache-commons FileUtils like this
FileUtils.touch(new File("/test1/test2/test3/fileName.properties"));

How to open a file without saving it to disk

My Question: How do I open a file (in the system default [external] program for the file) without saving the file to disk?
My Situation: I have files in my resources and I want to display those without saving them to disk first. For example, I have an xml file and I want to open it on the user's machine in the default program for reading xml file without saving it to the disk first.
What I have been doing: So far I have just saved the file to a temporary location, but I have no way of knowing when they no longer need the file so I don't know when/if to delete it. Here's my SSCCE code for that (well, it's mostly sscce, except for the resource... You'll have to create that on your own):
package main;
import java.io.*;
public class SOQuestion {
public static void main(String[] args) throws IOException {
new SOQuestion().showTemplate();
}
/** Opens the temporary file */
private void showTemplate() throws IOException {
String tempDir = System.getProperty("java.io.tmpdir") + "\\BONotifier\\";
File parentFile = new File(tempDir);
if (!parentFile.exists()) {
parentFile.mkdirs();
}
File outputFile = new File(parentFile, "template.xml");
InputStream inputStream = getClass().getResourceAsStream("/resources/template.xml");
int size = 4096;
try (OutputStream out = new FileOutputStream(outputFile)) {
byte[] buffer = new byte[size];
int length;
while ((length = inputStream.read(buffer)) > 0) {
out.write(buffer, 0, length);
}
inputStream.close();
}
java.awt.Desktop.getDesktop().open(outputFile);
}
}

Because of this line:
String tempDir = System.getProperty("java.io.tmpdir") + "\\BONotifier\\";
I deduce that you're working on Windows. You can easily make this code multiplatform, you know.
The answer to your question is: no. The Desktop class needs to know where the file is in order to invoke the correct program with a parameter. Note that there is no method in that class accepting an InputStream, which could be a solution.
Anyway, I don't see where the problem is: you create a temporary file, then open it in an editor or whatever. That's fine. In Linux, when the application is exited (normally) all its temporary files are deleted. In Windows, the user will need to trigger the temporary files deletion. However, provided you don't have security constraints, I can't understand where the problem is. After all, temporary files are the operating system's concern.

Depending on how portable your application needs to be, there might be no "one fits all" solution to your problem. However, you can help yourself a bit:
At least under Linux, you can use a pipe (|) to direct the output of one program to the input of another. A simple example for that (using the gedit text editor) might be:
echo "hello world" | gedit
This will (for gedit) open up a new editor window and show the contents "hello world" in a new, unsaved document.
The problem with the above is, that this might not be a platform-independent solution. It will work for Linux and probably OS X, but I don't have a Windows installation here to test it.
Also, you'd need to find out the default editor by yourself. This older question and it's linked article give some ideas on how this might work.

I don't understand your question very well. I can see only two possibilities to your question.
Open an existing file, and you wish to operate on its stream but do not want to save any modifications.
Create a file, so that you could use file i/o to operate on the file stream, but you don't wish to save the stream to file.
In either case, your main motivation is to exploit file i/o existingly available to your discretion and programming pleasure, am I correct?
I have feeling that the question is not that simple and this my answer is probably not the answer you seek. However, if my understanding of the question does coincide with your question ...
If you wish to use Stream io, instead of using FileOutputStream or FileInputStream which are consequent to your opening a File object, why not use non-File InputStream or OutputStream? Your file i/o utilities will finally boil down to manipulating i/o streams anyway.
http://docs.oracle.com/javase/7/docs/api/java/io/OutputStream.html
http://docs.oracle.com/javase/7/docs/api/java/io/InputStream.html
No need to involve temp files.

How to move/rename uploaded file?

I followed this tutorial for uploading a file in my JSF2 application.
The application works fine but I am unhappy with one aspect.
While rebuilding the request, the File sent via request is saved somewhere on the disk.
Even though the file is saved I need to rename the file with a name which is available after entering the Managed Bean containing the action method.
Therefore I decided to create a new file with de desired name, copy the already saved file, and then delete the unneeded one.
private File uploadFile;
//...
try {
BufferedWriter bw = new BufferedWriter(new FileWriter(newFile));
BufferedReader br = new BufferedReader(new FileReader(uploadFile));
String line = "";
while ((line = br.readLine()) != null){
bw.write(line);
}
} catch (Exception e){}
The new file appears in the desired location but this error is thrown when I'm trying to open the file: "Invalid or unsupported PNG file"
These are my questions:
Is there a better way to solve this problem?
Is this solution the best way to upload a picture? Is there a reason to save the file before the business logic when there may be need to resize the picture or the desired name is not available yet.
LE:
I know abot this tutorial as well but I'm trying to do this mojarra only.

There is a rename method built into java.io.File object already, I'd be surprised if it didn't work for your situation.
public boolean renameTo(File dest)
Renames the file denoted by this abstract pathname.
Many aspects of the behavior of this method are inherently platform-dependent:
The rename operation might not be able to move a file from one filesystem to
another, it might not be atomic, and it might not succeed if a file with the
destination abstract pathname already exists. The return value should always
be checked to make sure that the rename operation was successful.
You can also check if a file exists before saving it, and you can use the ImageIO class to do validations on the uploaded file before performing the initial save.

Don't use Reader and Writer when you deal with binary files like images. Use streams: FileInputStream and FileOutputStream. And the best variant is to use #Perception solution with renameTo method.
Readers read file as if it consists of characters (e.g. txt, properties, yaml files). Image files are not characters, they are binary and you must use streams for that.

How to print the content of a tar.gz file with Java?

I have to implement an application that permits printing the content of all files within a tar.gz file.
For Example:
if I have three files like this in a folder called testx:
A.txt contains the words "God Save The queen"
B.txt contains the words "Ubi maior, minor cessat"
C.txt.gz is a file compressed with gzip that contain the file c.txt with the words "Hello America!!"
So I compress testx, obtain the compressed tar file: testx.tar.gz.
So with my Java application I would like to print in the console:
"God Save The queen"
"Ubi maior, minor cessat"
"Hello America!!"
I have implemented the ZIP version and it works well, but keeping tar library from apache ant http://commons.apache.org/compress/, I noticed that it is not easy like ZIP java utils.
Could someone help me?
I have started looking on the net to understand how to accomplish my aim, so I have the following code:
GZIPInputStream gzipInputStream=null;
gzipInputStream = new GZIPInputStream( new FileInputStream(fileName));
TarInputStream is = new TarInputStream(gzipInputStream);
TarEntry entryx = null;
while((entryx = is.getNextEntry()) != null) {
if (entryx.isDirectory()) continue;
else {
System.out.println(entryx.getName());
if ( entryx.getName().endsWith("txt.gz")){
is.copyEntryContents(out);
// out is a OutputStream!!
}
}
}
So in the line is.copyEntryContents(out), it is possible to save on a file the stream passing an OutputStream, but I don't want it! In the zip version after keeping the first entry, ZipEntry, we can extract the stream from the compressed root folder, testx.tar.gz, and then create a new ZipInputStream and play with it to obtain the content.
Is it possible to do this with the tar.gz file?
Thanks.

surfing the net, i have encountered an interesting idea at : http://hype-free.blogspot.com/2009/10/using-tarinputstream-from-java.html.
After converting ours TarEntry to Stream, we can adopt the same idea used with Zip Files like:
InputStream tmpIn = new StreamingTarEntry(is, entryx.getSize());
// use BufferedReader to get one line at a time
BufferedReader gzipReader = new BufferedReader(
new InputStreamReader(
new GZIPInputStream(
inputZip )));
while (gzipReader.ready()) { System.out.println(gzipReader.readLine()); }
gzipReader.close();
SO with this code you could print the content of the file testx.tar.gz ^_^

To not have to write to a File you should use a ByteArrayOutputStream and use the public String toString(String charsetName)
with the correct encoding.