When someone opens my jar up I am opening a file selector gui so they can choose where they want to store their jar's files like config files and such. This should only take place the first time they open the jar. However, one issue with this approach is that I would have no way to know if it's their first time opening the jar since I will need to save the selected path somewhere. The best solution to this sounds like saving the selected path inside a file in the resource folder which is what I am having issues with. Reading and writing to this resource file will only need to be done when the program is actually running. These read and write operations need to work for packaged jar files (I use maven) and in the IDE.
I am able to read a resources file inside of the IDE and then save that file to the designated location specified in the file selector by doing this. However, I have not been able to do the same from a jar despite trying multiple other approaches from other threads.
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream is = classloader.getResourceAsStream("config.yml");
try {
if(is != null) {
Files.copy(is, testFile.toPath(), StandardCopyOption.REPLACE_EXISTING);
is.close();
}
} catch (IOException e) {
e.printStackTrace();
}
So just to clarify when my project is loaded I need to listen for the user to select a valid path for files like my config. Then I want to write my config to that path which I can do from my IDE and is shown above but I cant figure this out when I compile my project into a jar file since I always receive a file not found error in my cmd. However, the main point of this post is so that I can figure out how to save that selected path to my resource folder to a file (it can be json, yml or whatever u like). In the code above I was able to read a file but I have no idea how to go from that to get the files path since then reading and writing to it would be trivial. Also keep in mind I need to be able to read and write to a resource folder from both my IDE and from a compiled jar.
The following code shows my attempt at reading a resource from a compiled jar. When I added a print statement above name.startWith(path) I generated a massive list of classes that reference config.yml but I am not sure which one I need. I assume it has to be one of the paths relating to my project or possible the META-INF or META-INF/MANIFEST.MF path. Either way how am I able to copy the file or copy the contents of the file?
final String path = "resources/config.yml";
final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());
if(jarFile.isFile()) { // Run with JAR file
try {
final JarFile jar = new JarFile(jarFile);
final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
while(entries.hasMoreElements()) {
final String name = entries.nextElement().getName();
if (name.startsWith(path)) { //filter according to the path
System.out.println(name);
}
}
jar.close();
} catch (IOException exception) {
exception.printStackTrace();
}
}
Also if you were wondering I got the above code from the following post and my first block of code I pasted above is actually in the else statement since the IDE code from that post also did not work.
How can I access a folder inside of a resource folder from inside my jar File?
You can't write to files inside your JAR file, because they aren't actually files, they are ZIP entries.
The easiest way to store configuration for a Java application is to use Preferences:
Preferences prefs = Preferences.userNodeForPackage(MyApp.class);
Now all you have to do is use any of the get methods to read, and put methods to write.
There is absolutely no need to write files into your resource folder inside a jar. All you need to have is a smart classloader structure. Either there is a classloader that allows changing jars (how difficult is that to implement?), or you just setup a classpath that contains an empty directory before listing all the involved jars.
As soon as you want to change a resource, just store a file in that directory. When loading the resource next time, it will be searched on the classpath and the first match is returned - in that case your changed file.
Now it could be that your application needs to be started with that modified classpath, and on top of that it needs to be aware which directory was placed first. You could still setup that classloader structure within a launcher application, which then transfers control to the real application loaded via the newly defined classloader.
You could also check on application startup whether a directory such as ${user.home}/${application_name}/data exists.
If not, create it by extracting a predefined zip into this location.
Then just run your application which would load/write all the data in this directory.
No need to read/write to the classpath. No need to include 3rd party APIs. And modifying this initial data set would just mean to distribute a new zip to be extracted from.
I have a question regarding file handling.
I automate a page using selenium, and I need to upload a file in this page.
I want to put the file in resource folder and read it's path in the test (since many OS and path will be different to any computer WIN/MAC).
I put the file manually in the resource folder, and it put it in:
X:\Project_11_01_2021\src\test\resources
when I used the ClassLoader and try to find the file it not found it,
I saw that if I manually put it in this path it find it, found.
X:\Project_11_01_2021\out\test\resources
the problem is that I am using git and if I add to the resources it upload to git and every one will get the change, and when I put in out\test\resources it is not displayed in the source tree to commit to git.
is their a way that classLoader will search in the first location? and not in the second?
[][path that worked]
[][when here not worked]
/******* test *******/
public void entertax() throws Exception {
WebDriver deiver2 = getWebDriver();
Thread.sleep(1000);
ClassLoader classLoader = getClass().getClassLoader();
String path = classLoader.getResource("TAX12.pdf").getPath();
System.out.println("\n\n path is " + path);
deiver2.switchTo()
.activeElement();
deiver2.findElement(By.xpath("//input[#type='file']"))
.sendKeys(
"X:\\Project_11_01_2021\\out\\test\\resources\\fw8TAX12.pdf");
System.out.println("END");
}
This issue looks more like an IDE configuration problem.
Your code classLoader.getResource("TAX12.pdf") looks correct (1. using classLoader.getResource() method && 2. specifying the correct relative path within the resources folder).
I assume why the code fails to find the file is due to the fact that the resources folder is not part of your applications classpath (I'm assuming your trying to run the code from your IDE which apparently seems to be IntelliJ).
I'm not regularly using IntelliJ, but you can specify the classpath settings in the Module settings.
In the Module settings specify the resources folder as a resource:
When successfully added the resources folder to the classpath it should display an icon like this in the project explorer:
After adding the resources folder to your classpath everything should work.
EDIT:
In case you're using Maven also make sure to specify the resources folder as such in the pom.xml file correspondingly.
Also make sure to spell the name of the file you're looking for "TAX12.pdf" correctly, since it seems to differ in your code and in your screenshots (not sure if it differs only because posting it here or also in your real code base).
I've got a project to do with 2 other classmates.
We used Dropbox to share the project so we can write from our houses (Isn't a very good choice, but it worked and was easier than using GitHub)
My question is now about sharing the object stream.
I want to put the file of the stream in same dropbox shared directory of the code.
BUT, when i initialize the file
File f = new File(PATH);
i must use the path of my computer (C:\User**Alessandro**\Dropbox)
As you can see it is linked to Alessandro, and so to my computer.
This clearly won't work on another PC.
How can tell the compiler to just look in the same directory of the source code/.class files?
You can use Class#getResource(java.lang.String) to obtain a URL corresponding to the location of the file relative to the classpath of the Java program:
URL url = getClass().getResource("/path/to/the/file");
File file = new File(url.getPath());
Note here that / is the root of your classpath, which is the top of the directory containing your class files. So your resource should be placed inside the classpath somewhere in order for it to work on your friend's computer.
Don't use absolute paths. Use relative paths like ./myfile.txt. Start the program with the project directory as the current dir. (This is the default in Eclipse.) But then you have to ensure that this both works for development and for production use.
As an alternative you can create a properties file and define the path there. Your code then only refers to a property name and each developer can adjust the configuration file. See Properties documentation.
I want to understand the best or the standard technique to write a java project when using internal files. To be precise, I want to develop a project that uses text files and images that are needed when the program runs. My goal is to create a runnable jar from the project in which the user does not need to see all these files. Therefore, I decided to create a package called resources and put it inside the folder that contains the source codes. I.e. it is in the same level as other packages. Now, in my codes when I want to use the images I use the following statement:
URL url = getClass().getResource("/resources/image1.gif");
It is working!
Then, to open a text file for reading/writing, I use the following:
String filename= "/resources/"+file1.txt;
Now, this is not working and it complains that it cannot find the file. I am not sure how to go about this?
A google search suggested that I put the resources folder on the project root directory. It is working then but when I created the runnable jar I had to put the resources folder on the same directory as the jar. This means that the user can have access to all the files in there. Any help is much appreciated.
You can use the getResource method for files too.
URL url = getClass().getResource("/resources/file1.txt");
try {
File f = new File(fileUrl.toURI());
} catch (URISyntaxException e) {
//dealing with the exception
}
I need to read a text file when I start my program. I'm using eclipse and started a new java project. In my project folder I got the "src" folder and the standard "JRE System Library" + staedteliste.txt... I just don't know where to put the text file. I literally tried every folder I could think off....I cannot use a "hard coded" path because the text file needs to be included with my app...
I use the following code to read the file, but I get this error:
Error:java.io.FileNotFoundException:staedteliste.txt(No such file or directory)
public class Test {
ArrayList<String[]> values;
public static void main(String[] args) {
// TODO Auto-generated method stub
URL url = Test.class.getClassLoader().getResource("src/mjb/staedteliste.txt");
System.out.println(url.getPath()); // I get a nullpointerexception here!
loadList();
}
public static void loadList() {
BufferedReader reader;
String zeile = null;
try {
reader = new BufferedReader(new FileReader("src/mjb/staedteliste.txt"));
zeile = reader.readLine();
ArrayList<String[]> values = new ArrayList<String[]>();
while (zeile != null) {
values.add(zeile.split(";"));
zeile = reader.readLine();
}
System.out.println(values.size());
System.out.println(zeile);
} catch (IOException e) {
System.err.println("Error :"+e);
}
}
}
Ask first yourself: Is your file an internal component of your application?
(That usually implies that it's packed inside your JAR, or WAR if it is a web-app; typically, it's some configuration file or static resource, read-only).
If the answer is yes, you don't want to specify an absolute path for the file. But you neither want to access it with a relative path (as your example), because Java assumes that path is relative to the "current directory". Usually the preferred way for this scenario is to load it relatively from the classpath.
Java provides you the classLoader.getResource() method for doing this. And Eclipse (in the normal setup) assumes src/ is to be in the root of your classpath, so that, after compiling, it copies everything to your output directory ( bin/ ), the java files in compiled form ( .class ), the rest as is.
So, for example, if you place your file in src/Files/myfile.txt, it will be copied at compile time to bin/Files/myfile.txt ; and, at runtime, bin/ will be in (the root of) your classpath. So, by calling getResource("/Files/myfile.txt") (in some of its variants) you will be able to read it.
Edited: Further, if your file is conceptually tied to a java class (eg, some com.example.MyClass has a MyClass.cfg associated configuration file), you can use the getResource() method from the class and use a (resource) relative path: MyClass.getResource("MyClass.cfg"). The file then will be searched in the classpath, but with the class package pre-appended. So that, in this scenario, you'll typically place your MyClass.cfg and MyClass.java files in the same directory.
One path to take is to
Add the file you're working with to the classpath
Use the resource loader to locate the file:
URL url = Test.class.getClassLoader().getResource("myfile.txt");
System.out.println(url.getPath());
...
Open it
Suppose you have a project called "TestProject" on Eclipse and your workspace folder is located at E:/eclipse/workspace. When you build an Eclipse project, your classpath is then e:/eclipse/workspace/TestProject. When you try to read "staedteliste.txt", you're trying to access the file at e:/eclipse/workspace/TestProject/staedteliste.txt.
If you want to have a separate folder for your project, then create the Files folder under TestProject and then access the file with (the relative path) /Files/staedteliste.txt. If you put the file under the src folder, then you have to access it using /src/staedteliste.txt. A Files folder inside the src folder would be /src/Files/staedteliste.txt
Instead of using the the relative path you can use the absolute one by adding e:/eclipse/workspace/ at the beginning, but using the relative path is better because you can move the project without worrying about refactoring as long as the project folder structure is the same.
Just create a folder Files under src and put your file there.
This will look like src/Files/myFile.txt
Note:
In your code you need to specify like this Files/myFile.txt
e.g.
getResource("Files/myFile.txt");
So when you build your project and run the .jar file this should be able to work.
Depending on your Java class package name, you're probably 4 or 5 levels down the directory structure.
If your Java class package is, for example, com.stackoverflow.project, then your class is located at src/com/stackoverflow/project.
You can either move up the directory structure with multiple ../, or you can move the text file to the same package as your class. It would be easier to move the text file.
MJB
Please try this
In eclipse "Right click" on the text file u wanna use,
see and copy the complete path stored in HDD like (if in UNIX "/home/sjaisawal/Space-11.4-template/provisioning/devenv/Test/src/testpath/testfile.txt")
put this complete path and try.
if it works then class-path issue else GOK :)
If this is a simple project, you should be able to drag the txt file right into the project folder. Specifically, the "project folder" would be the highest level folder. I tried to do this (for a homework project that I'm doing) by putting the txt file in the src folder, but that didn't work. But finally I figured out to put it in the project file.
A good tutorial for this is http://www.vogella.com/articles/JavaIO/article.html. I used this as an intro to i/o and it helped.
Take a look at this video
All what you have to do is to select your file (assuming it's same simple form of txt file), then drag it to the project in Eclipse and then drop it there. Choose Copy instead of Link as it's more flexible. That's it - I just tried that.
You should probably take a look at the various flavours of getResource in the ClassLoader class: https://docs.oracle.com/javase/1.5.0/docs/api/java/lang/ClassLoader.html.