Develop Reference

How to Write Output Files from a JAR Program to Directory Outside the JAR?

tl;dr I'm more used to writing command-line scripts that can just output based on the current working directory, so I'm unsure what directory to use for output files in a program that will be launched from a JAR.
Program Description:
My program builds an HTML file from data given to it from the rest of the program, and then is supposed to write it to a file that we'll call "Output.html" for simplicity.
Relevant Code:
public void outputHTML()
{
String output = buildHTML();
// Expanded to explain my confusion better
String fileDirectory = ""; // ???
String fileName = "Output.html";
String fullPath = fileDirectory + "\\" + fileName;
try (BufferedWriter writer = new BufferedWriter(new FileWriter(fullPath)))
{
writer.write(output);
writer.close();
} catch (IOException e)
{
System.out.println("File not found.");
e.printStackTrace();
}
}
Problem
I don't know what to put the file directory as. Usually I run my programs from the command line and use ".\\Output.txt" as my output path, but I don't know where to put it if it's being run from a JAR.
The desired file structure is as follows:
Encompassing Folder
Program.jar
output
Output.html
Or alternatively (not sure if this makes it easier to understand or harder):
main\
main\Program.jar
main\output\
main\output\Output.html
Everything I can find on SE only relates to reading files that are both immutable and internal, but I'm trying to output a non-static file to a location outside of my jar.
Can anyone help with this? Thanks!
Misc Details
I'm using Eclipse without Gradle currently, because I don't know what Gradle is and new things are scary. If this particular problem would be easier to solve with Gradle, let me know and I'll look up more about it.
EDIT:
Added syntax highlighting to code block.
Formatted everything a bit better
Changed title to be more descriptive

You can use an absolute path: e.g. fileDirectory = "\\project\\test\\main\\output";
using normal slash should also work even on Windows ("/project/test/main/output")
Or use a relative path - this will start from the current working directory (user directory), the one where the JVM was started in - e.g. fileDirectory = "main\\output";

store file in spring boot resource folder after deployment

I have deployed a spring-boot application JAR file. Now, I want to upload the image from android and store it in the myfolder of resource directory. But unable to get the path of resource directory.
Error is:
java.io.FileNotFoundException: src/main/resources/static/myfolder/myimage.png
(No such file or directory)
This is the code for storing the file in the resource folder
private final String RESOURCE_PATH = "src/main/resources";
String filepath = "/myfolder/";
public String saveFile(byte[] bytes, String filepath, String filename) throws MalformedURLException, IOException {
File file = new File(RESOURCE_PATH + filepath + filename);
OutputStream out = new FileOutputStream(file);
try {
out.write(bytes);
} catch (Exception e) {
e.printStackTrace();
} finally {
out.close();
}
return file.getName();
}
UPDATED:
This is what I have tried
private final String RESOURCE_PATH = "config/";
controller class:
String filepath = "myfolder/";
String filename = "newfile.png"
public String saveFile(byte[] bytes, String filepath, String filename) throws MalformedURLException, IOException {
//reading old file
System.out.println(Files.readAllBytes(Paths.get("config","myfolder","oldfile.png"))); //gives noSuchFileException
//writing new file
File file = new File(RESOURCE_PATH + filepath + filename);
OutputStream out = new FileOutputStream(file); //FileNotFoundException
try {
out.write(bytes);
} catch (Exception e) {
e.printStackTrace();
} finally {
out.close();
}
return file.getName();
}
Project structure:
+springdemo-0.0.1-application.zip
+config
+myfolder
-oldfile.png
-application.properties
+lib
+springdemo-0.0.1.jar
+start.sh
-springdemo-0.0.1.jar //running this jar file

Usually when you deploy an application (or start it using Java), you start a JAR file. You don't have a resource folder. You can have one and access it, too, but it certainly won't be src/main/resources.
When you build your final artifact (your application), it creates a JAR (or EAR or WAR) file and your resources, which you had in your src/main/resources-folder, are copied over to the output directory and included in the final artifact. That folder simply does not exist when the application is run (assuming you are trying to run it standalone).
During the build process target/ is created and contains the classes, resources, test-resources and the likes (assuming you are building with Maven; it is a little different if you build using Gradle or Ant or by hand).
What you can do is create a folder e.g. docs next to your final artifact, give it the appropriate permissions (chmod/chown) and have your application output files into that folder. This folder is then expected to exist on the target machine running your artifact, too, so if it doesn't, it would mean the folder does not exist or the application lacks the proper permissions to read from / write to that folder.
If you need more details, don't hesitate to ask.
Update:
To access a resource, which is bundled and hence inside your artifact (e.g. final.jar), you should be able to retrieve it by using e.g. the following:
testText = new String(ControllerClass.class.getResourceAsStream("/test.txt").readAllBytes());
This is assuming your test.txt file is right under src/main/resources and was bundled to be directly in the root of your JAR-file (or target folder where your application is run from). ControllerClass is the controller, which is accessing the file. readAllBytes just does exactly this: read all the bytes from a text file. For accessing images inside your artifact, you might want to use ImageIO.
IF you however want to access an external file, which is not bundled and hence not inside your artifact, you may use File image = new File(...) where ... would be something like "docs/image.png". This would require you to create a folder called docs next to your JAR-artifact and put a file image.png inside of it.
You of course also may work with streams and there are various helpful libraries for working with input- and output streams.
The following was meant for AWT, but it works in case you really want to access the bytes of your image: ImageIO. In a controller you usually wouldn't want to do that, but rather have your users access (and thus download) it from a given available folder.
I hope this helps :).

Java 7 zipfs: File is also a folder

I've been using Java 7's ZipFS support.
https://gist.github.com/stain/5591420
shows the behaviour, which I find a bit odd. Basically you can create a ZIP file system, make a file with a given name, and then also make a folder with the same name.
The reason for this seems to be that internally the folder gets "/" appended to its name - however this new name is not returned, therefore you end up in a strange situation where Files.isDirectory() returns false immediately after a successful Files.createDirectory().
try (FileSystem fs = tempZipFS()) {
Path folder = fs.getPath("folder");
Files.createFile(folder);
assertTrue(Files.isRegularFile(folder));
assertFalse(Files.isDirectory(folder));
// try {
Files.createDirectory(folder);
// } catch (FileAlreadyExistsException ex) {
// Is not thrown!
// }
// but a second createDirectory() fails correctly
try {
Files.createDirectory(folder);
} catch (FileAlreadyExistsException ex) {
}
// Look, it's both a file and folder!
Path child = folder.resolve("child");
Files.createFile(child);
// Can this be tested?
assertTrue(Files.isRegularFile(folder));
// Yes, if you include the final /
assertTrue(Files.isDirectory(fs.getPath("folder/")));
// But not the parent
// assertTrue(Files.isDirectory(child.getParent()));
// Or the original Path
// assertTrue(Files.isDirectory(folder));
}
So as long as you have the "/" as the suffix, you can even work with both, and that's how they are listed if you do a directory listing of the root.
Now the ZIP format itself allows this as it only deals with entries in a ZIP file (even allowing multiple entries with the same name), however normal use of a "FileSystem" would normally not allow multiple entries with the same name ; as can be seen when I try to create the folder twice.
The produced ZIP file can be browsed correctly with infozip, 7Zip and Windows 8; but trying to unzip will obviously fail because the native file system don't support such duality.
So is this a feature, bug or something in between?

How to specify a directory when creating a File object?

This should be a really simple question but Google + my code isn't working out.
In Eclipse on Windows, I want my program to look inside a certain folder. The folder is directly inside the Project folder, on the same level as .settings, bin, src, etc. My folder is called surveys, and that's the one I want my File object to point at.
I don't want to specify the full path because I want this to run on both of my computers. Just the path immediately inside my Project.
I'm trying this code but it isn't working - names[] is coming back null. And yes I have some folders and test junk inside surveys.
File file = new File("/surveys");
String[] names = file.list();
for(String name : names)
{
if (new File("/surveys/" + name).isDirectory())
{
System.out.println(name);
}
}
I'm sure my mistake is within the String I'm passing to File, but I'm not sure what's wrong?

In your question you didn't specify what platform you are running on. On non-Windows, a leading slash signifies an absolute path. Best to remove the leading slash. Try this:
File file = new File("surveys");
System.out.println("user.dir=" + System.getProperty("user.dir"));
System.out.println("file is at: " + file.getCanonicalPath());
String[] names = file.list();
for(String name : names)
{
if (new File(file, name).isDirectory())
{
System.out.println(name);
}
}
Make sure the in your run configuration, the program is running from the projects directory (user.dir = <projects>)

Make sure that your file is a directory before using file.list() on it, otherwise you will get a nasty NullPointerException.
File file = new File("surveys");
if (file.isDirectory()){
...
}
OR
if (names!=null){
...
}

If you checked the full path of your file with
System.out.println(file.getCanonicalPath())
the picture would immediately become clear. File.getCanonicalPath gives you exactly the full path. Note that File normalizes the path, eg on Windows "c:/file" is converted to "C:\file".

How to pass a text file as a argument?

Im trying to write a program to read a text file through args but when i run it, it always says the file can't be found even though i placed it inside the same folder as the main.java that im running.
Does anyone know the solution to my problem or a better way of reading a text file?

Do not use relative paths in java.io.File.
It will become relative to the current working directory which is dependent on the way how you run the application which in turn is not controllable from inside your application. It will only lead to portability trouble. If you run it from inside Eclipse, the path will be relative to /path/to/eclipse/workspace/projectname. If you run it from inside command console, it will be relative to currently opened folder (even though when you run the code by absolute path!). If you run it by doubleclicking the JAR, it will be relative to the root folder of the JAR. If you run it in a webserver, it will be relative to the /path/to/webserver/binaries. Etcetera.
Always use absolute paths in java.io.File, no excuses.
For best portability and less headache with absolute paths, just place the file in a path covered by the runtime classpath (or add its path to the runtime classpath). This way you can get the file by Class#getResource() or its content by Class#getResourceAsStream(). If it's in the same folder (package) as your current class, then it's already in the classpath. To access it, just do:
public MyClass() {
URL url = getClass().getResource("filename.txt");
File file = new File(url.getPath());
InputStream input = new FileInputStream(file);
// ...
}
or
public MyClass() {
InputStream input = getClass().getResourceAsStream("filename.txt");
// ...
}

Try giving an absolute path to the filename.
Also, post the code so that we can see what exactly you're trying.

When you are opening a file with a relative file name in Java (and in general) it opens it relative to the working directory.
you can find the current working directory of your process using
String workindDir = new File(".").getAbsoultePath()
Make sure you are running your program from the correct directory (or change the file name so that it will be relative to where you are running it from).

If you're using Eclipse (or a similar IDE), the problem arises from the fact that your program is run from a few directories above where the actual source is located. Try moving your file up a level or two in the project tree.
Check out this question for more detail.

The simplest solution is to create a new file, then see where the output file is. That is the correct place to put your input file into.

If you put the file and the class working with it under same package can you use this:
Class A {
void readFile (String fileName) {
Url tmp = A.class.getResource (fileName);
// Or Url tmp = this.getClass().getResource (fileName);
File tmpFile = File (tmp);
if (tmpFile.exists())
System.out.print("I found the file.")
}
}
It will help if you read about classloaders.

say I have a text file input.txt which is located on the desktop
and input.txt has the following content
i came
i saw
i left
and below is the java code for reading that text file
public class ReadInputFromTextFile {
public static void main(String[] args) throws Exception
{
File file = new File(
"/Users/viveksingh/desktop/input.txt");
BufferedReader br
= new BufferedReader(new FileReader(file));
String st;
while ((st = br.readLine()) != null)
System.out.println(st);
}
}
output on the console:
i came
i saw
i left