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Can not find file that is in resource folder - java

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).

how to define a file path in java to make it ready for production phase?

I am writing a program in java with netbeans IDE which receives a jasper report *.jrxml and then displays the report for the user. I wrote the following line of code for the file path
String reportSource = "src\\jasper-reports\\report.jrxml";
but when I move the dist folder in some other place and try to run the jar file inside it, my program can not find the report.
my problem is that were should I put the *.jrxml file and how to define it's path in my program so that when I want to give my software to someone else it runs without any errors (e.g. the program can find the file)

avoid using absolute paths. try to include the file as a resource in your netbeans project. then in your code you can search for and load the file as
new InputStreamReader((Main.class.getResourceAsStream("/report.jrxml")))
something like that depending on where the file resides in your project

it's more recommended using one of the two approaches:
pass the locations/paths as a -Dproperty=value in the Java application launcher command line http://www.tutorialspoint.com/unix_commands/java.htm
store it the locations/paths in a configurations file with a unique key and edit the file accordingly for different environments,
e.g.this files always stored in ${HOME}/config_files/ directory
absolute paths considered a bad practice

In netbeans, how to save or access an image after deploying jar file

I am trying to make a mess management application in Java using NetBeans. I want to save images of Members in a specified folder inside my src directory. I just created folder named EmpImgs for storing employees images. Here is my code:
File srcDir = new File(file); // current path of image
File dstDir = new File("src\\J_Mess_Mgnt\\EmpImgs\\"+Txt_C_G_M_M_ID.getText());
objm.copyFile(srcDir, dstDir);` // copy image from srcDir to dstDir
Here I use another class for copying images to predefined folders and renaming the images based on their ID.
Everything is working properly in Java IDE.
But unfortunately after making an executable .jar file, this code will not work. I cannot save or access any image file in that directory.
I just went through this site, but I didn't find a suitable answer.
All I need is saving and editing images inside jar folder

Hehe hi mate you need some help. This is a duplicate but I will cut you some slack and maybe you should delete this later. So back to basics, the jvm runs byte code, which you get from compiling java source code to .class files. Now this is different to C and C++ were you just get a .exe. You don't want to give your users a bunch of .class files in all these folders which they can edit and must run a command on the command line, but instead give them what is known as an 'archive' which is just an imutable file structure so they can't screw up the application, known as a jar in java. They can just double click on the archive (which is a jar), and the jvm will call the main method specified in the MetaInf directory (just some information about the jar, same as a manifest in other programming languages).
Now remember your application is now a jar! It is immutable! for the resasons I explained. You can't save anymore data there! Your program will still work on the command line and in IDEs because it is working as if you used your application is distrubuted as bunch of folders with the .class files, and you can write to this location.
If you want to package resources with your application you need to use streams (google it). BUT REMEMBER! you cant then save more resources into the jar! You need to write somewhere else! Maybe use a user.home directory! or a location specified from the class path and the photos will be right next to the jar! Sometimes you might need an installer for your java application, but usually you don't want to create the extra work if you don't need to.

At last I find an answer suit for my question.It is not possible to copy images or files to a executive jar folder.So I used a different Idea.Create some folders(as per our requirement),Where my executable jar folder is located(No matter which drive or where the location is).The code is..
String PRJT_PATH=""; //variable to store path of working directory.
private void getdire() throws IOException{
File f=new File(".");
File[] f1=f.listFiles();
PRJT_PATH=f.getCanonicalPath(); //get path details.for eg:-E:/java/dist
}
private void new_Doc_folder(){ //function for creating new folders
try{
String strManyDirectories="Docs"+File.separator+"Bil_Img"; //i need to create 2 folders,1st a folder namedDocs and In Docs folder another folder named Bil_Img
String SubDirectories="Docs"+File.separator+"EmpImgs"; //same as above but keep in mind that It will not create a Same folder again if already exists,
// Create one directory
boolean success = (new File(strManyDirectories)).mkdirs(); //create more than one directory
boolean success1 = (new File(SubDirectories)).mkdir(); //Creates a single directory
if (success && success1) {
}
}catch (Exception e){//Catch exception if any
System.err.println("Error: " + e.getMessage());
}
It works Successfully.
Regds

Java get current file name EXE

This is my first Stackoverflow question.
I searched across Google for getting the current file name in Java. Most of the sources tell users how to find the current file name if the file is a JAR file, but I'm asking for if the current file is an EXE file.
I saw one EXE answer in Get name of running Jar or Exe, but I don't think it worked.
I use the JSmooth EXE wrapper (launch4j somehow didn't work for me), and Java 8. Is there a straightforward solution to my question? Also, it's nice to explain how it works, or provide a link to the Java documentary about it.
EDIT: To clarify, let's say that I made a Java program and used a JAR wrapper, and I named the resulting EXE "test.exe". I want the Java program be able to give me the current directory of "test.exe", including the filename itself (test.exe).
ANOTHER EDIT: Just to clarify more, go onto your desktop, create a text file, and put some text in it. Save it, and change the text file to an EXE file. Then, try to open it. Windows will give an error. Notice how the title of the message dialog is the file path of the opened file. That is the type of output I want.
Thanks.

Per the JSmooth documentation,
JSmooth also makes some special variable accessible for your application.
Form Meaning
${EXECUTABLEPATH} Replaced by the path to the executable binary. For
instance, if the executable binary launched is located
at c:/program files/jsmooth/test.exe, this variable
is replaced with c:/program files/jsmooth
${EXECUTABLENAME} Replaced by the name of the executable binary. For
instance, if the executable binary launched is located
at c:/program files/jsmooth/test.exe, this variable is
replaced with test.exe
You set these in JSmooth under the "Environment Settings" (the last panel), which allows you to map the variable name. So,
MY_EXECUTABLEPATH=${EXECUTABLEPATH}
MY_EXECUTABLENAME=${EXECUTABLENAME}
In your application, you can get those with
String execPath = System.getProperty("MY_EXECUTABLEPATH");
String execName = System.getProperty("MY_EXECUTABLENAME");

public class JavaApplication1 {
public static void main(String[] args) {
System.out.println("Working Directory = " +
System.getProperty("user.dir"));
}
}
This will print a complete absolute path from where your application has initialized. It is used to get only the DIRECTORY.
If you are using a .exe why do you not create a installer? You can use Inno Setup, this way you are able to specify where do you want to store your .exe, and get it from your application just passing your custom directory

Opening a xml file from eclipse and from a .jar file in java

Yesterday, I had a problem because I couldn't manage to open a xml file (it owuld give me a FileNotFoundException) located in the ressources folder of my .jar file, which I managed to open on eclipse using the following lines of code. You can see my old problem here. This was my code with the problem :
File xmlFile = new File("ressources/emitter.xml");
ConfigurableEmitter emitter = ParticleIO.loadEmitter(xmlFile);
Someone told me it that one way was to use getClassLoader().getRessourceAsStream method to open a xml file in a .jar file that was exported
InputStream i= this.getClass().getClassLoader().getResourceAsStream("ressources/emitter.xml");
ConfigurableEmitter emitter = ParticleIO.loadEmitter(i);
Unfortunately, that solution only works when I export my project into a .jar file, so if I want to go back debugging my program, I have to take the old code that would only works on eclipse.
My question is: is there any better way to do this without having to change my code if I want to export it or if I want to debug it?
Thank you
edit :
Thank you all, it works perfectly fine now
my problem was that I put my ressources folder like that :
+project
+src
+ressources
+emitter.xml

InputStream i= this.getClass().getClassLoader().getResourceAsStream("/ressources/emitter.xml");
The above should work in both cases (Note is is /resources/.... This is assuming say your directory structure is below:
MyProject
+src
+ressources
emitter.xml

Place the file alongside your source files, then you can use the getResourceAsStream() method in both cases. Don't forget to update the path (which should be the package name of your class, but with slashes instead of dots).

My question is: is there any better way to do this without having to
change my code if I want to export it or if I want to debug it?
Yes, use Maven. Maven will handle that and it hooks into Eclipse beautifully (NetBeans too!) What you do is place the resource in src/main/resources and then you can have Eclipse run the test goal of the Maven project or you can just run mvn test from the command line. Another advantage of using Maven here is that you can also have src/test/resources/emitter.xml which overrides the one in src/main with environment-specific test instructions and it won't affect your deployment.

InputStream i= getClass().getClassLoader().getResourceAsStream("ressources/emitter.xml");
or
InputStream i= getClass().getResourceAsStream("/ressources/emitter.xml");
(note the absolute positioning)
both work when the class is in the same jar, on the same class path.
In the jar the names must be case sensitive, but as the jar already works. Ensure that the ressources directory is on the class path too, or copied to the target directory.
As "ressources" is probably configured yourself (not named "resources" as in English), you probably need to add it to the build somehow.