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Why do I get NullPointerException when trying to access resource file?

I want to make so a file in the program is packaged in the jar file because it is needed in my program. It is the firebase credentials file. I read that I need to add it to the resource folder but now I cannot access it because I get NullPointerException. I think the path is valid and everything seems okay but I get null every time. Here is the code:
ClassLoader classLoader = EmailSender.class.getClassLoader();
File file = new File(Objects.requireNonNull(classLoader.getResource("/parkingsystem-cf164-firebase-adminsdk-5jjuk-2be72bfcce.json")).getFile());
And here is the code structure:
code structure
If there is another method to add file to jar I am open to hear it, it is just what I found on the Internet but it does not work for me. Any help is appreciated!
Edit: So apparently it works when the / is removed but then I need to convert it to FileInputStream and there I get FileNotFoundExcpetion. The rest of the code:
FileInputStream serviceAccount = null;
try {
serviceAccount = new FileInputStream(file);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
FirebaseOptions options = null;
try {
options = new FirebaseOptions.Builder()
.setCredentials(GoogleCredentials.fromStream(serviceAccount))
.build();
} catch (IOException e) {
e.printStackTrace();
}

Properties file not loading and updating for relative path but working fine for absolute path

I am trying to access a properties file from the src/main/resources folder but when I try to load the file using a relative path it is not getting updated. But it is working fine for an absolute path.
I need the dynamic web project to work across all platforms.
public static void loadUsers() {
try(
FileInputStream in = new FileInputStream("C:\\Users\\SohamGuha\\Documents\\work-coding\\work-coding\\src\\main\\resources\\users.properties")) {
// write code to load all the users from the property file
// FileInputStream in = new FileInputStream("classpath:users.properties");
users.load(in);
System.out.println(users);
in.close();
}
catch(Exception e){
e.printStackTrace();
}

First of all you are using Spring, at least that is what the tags at the bottom say. Secondly C:\\Users\\SohamGuha\\Documents\\work-coding\\work-coding\\src\\main\\resources\\users.properties is the root of your classpath. Instead of loading a File use the Spring resource abstraction.
As this is part of the classpath you can simply use the ClassPathResource to obtain a proper InputStream. This will work regardless of which environment you are in.
try( InputStream in = new ClassPathResource("users.properties").getInputStream()) {
//write code to load all the users from the property file
//FileInputStream in = new FileInputStream("classpath:users.properties");
users.load(in);
System.out.println(users);
} catch(Exception e){
e.printStackTrace();
}
NOTE: you are already using a try with resources so you don't need to close the InputStream that is already handled for you.
Changing things inside your application simply won't work, as this would mean you could change resources (read classes) in your jar which would be quite a security risk! If you want something to be changable you will have to make it a file outside of the classpath and directly on the file-system.

Try the following code
import java.io.FileInputStream;
import java.io.IOException;
public class LoadUsers {
public static void main(String[] args) throws IOException {
try(FileInputStream fis=new FileInputStream("src/main/resources/users.properties")){
Properties users=new Properties();
users.load(fis);
System.out.println(users);
}catch(IOException ioe) {
ioe.printStackTrace();
}
}
}

How to get files from resources folder. Spring Framework

I'm trying to unmarshal my xml file:
public Object convertFromXMLToObject(String xmlfile) throws IOException {
FileInputStream is = null;
File file = new File(String.valueOf(this.getClass().getResource("xmlToParse/companies.xml")));
try {
is = new FileInputStream(file);
return getUnmarshaller().unmarshal(new StreamSource(is));
} finally {
if (is != null) {
is.close();
}
}
}
But I get this errors:
java.io.FileNotFoundException: null (No such file or directory)
Here is my structure:
Why I can't get files from resources folder? Thanks.
Update.
After refactoring,
URL url = this.getClass().getResource("/xmlToParse/companies.xml");
File file = new File(url.getPath());
I can see an error more clearly:
java.io.FileNotFoundException: /content/ROOT.war/WEB-INF/classes/xmlToParse/companies.xml (No such file or directory)
It tries to find WEB-INF/classes/
I have added folder there, but still get this error :(

I had the same problem trying to load some XML files into my test classes. If you use Spring, as one can suggest from your question, the easiest way is to use org.springframework.core.io.Resource - the one Raphael Roth already mentioned.
The code is really straight forward. Just declare a field of the type org.springframework.core.io.Resource and annotate it with org.springframework.beans.factory.annotation.Value - like that:
#Value(value = "classpath:xmlToParse/companies.xml")
private Resource companiesXml;
To obtain the needed InputStream, just call
companiesXml.getInputStream()
and you should be okay :)
But forgive me, I have to ask one thing: Why do you want to implement a XML parser with the help of Spring? There are plenty build in :) E.g. for web services there are very good solutions that marshall your XMLs into Java Objects and back...

ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());

you are suppose to give an absolute path (so add a loading ´/´, where resource-folder is the root-folder):
public Object convertFromXMLToObject(String xmlfile) throws IOException {
FileInputStream is = null;
File file = new File(String.valueOf(this.getClass().getResource("/xmlToParse/companies.xml")));
try {
is = new FileInputStream(file);
return getUnmarshaller().unmarshal(new StreamSource(is));
} finally {
if (is != null) {
is.close();
}
}
}

Open a properties file from a class in a Dynamic Web Project

I'm using Eclipse for EE Developer.
I need to access to a properties file (db.properties) from a class's method (DBQuery.java).
The class is located inside a package inside the src folder.
For the properties file i tried almost everything that i could find over the net to make it work, but looks like i can't.
The properties file is located inside the WebContent folder, and i'll add the code with which i'm trying to load this file:
public class DBQuery {
public static String create_DB_string(){
//the db connection string
String connString = "";
try{
Properties props = new Properties();
FileInputStream fis = new FileInputStream("db.properties");
props.load(fis);
fis.close();
/* creating connString using props.getProperty("String"); */
}
catch (Exception e) {
System.out.println(e.getClass());
}
return connString;
}
}
So my question is, where to put the properties file, and which is the correct way to load it?

You can put this propertie file within your java package for example com/test and use following:
getClass().getResourceAsStream( "com/test/myfile.propertie");
Hope it helps.

Reading Properties file inside conf folder

I have a properties file which is located under conf folder. conf folder is under the project root directory. I am using the following code.
public class PropertiesTest {
public static void main(String[] args) {
InputStream inputStream = PropertiesTest.class
.getResourceAsStream("/conf/sampleprop.conf");
Properties prop = new Properties();
try {
prop.load(inputStream);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println(prop.getProperty("TEST"));
}
}
But I get nullpointer exception.
I have tried using
InputStream inputStream = PropertiesTest.class
.getResourceAsStream("./conf/sampleprop.conf");
and
InputStream inputStream = PropertiesTest.class
.getResourceAsStream("conf/sampleprop.conf");
But all result in nullpointer exception.
Can anyone please help.
Thanks in advance

Try to recover your working directory first:
String workingDir = System.getProperty("user.dir");
System.out.println("Current working dir: " + workingDir);
and then is simple:
Properties propertiesFile = new Properties();
propertiesFile.load(new FileInputStream(workingDir+ "/yourFilePath"));
String first= propertiesFile.getProperty("myprop.first");
Regards, fabio

The getResourceAsStream() method tries to locate and load the resource using the ClassLoader of the class it is called on. Ideally it can locate the files only the class folders .. Rather you could use FileInputStream with relative path.
EDIT
if the conf folder is under src, then you still be able to access with getResourceAsStream()
InputStream inputStream = Test.class
.getResourceAsStream("../conf/sampleprop.conf");
the path would be relative to the class from you invoke getRes.. method.
If not
try {
FileInputStream fis = new FileInputStream("conf/sampleprop.conf");
Properties prop = new Properties();
prop.load(fis);
System.out.println(prop.getProperty("TEST"));
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
NOTE: this will only work if it is Stand alone application/in eclipse. This will not work if its web based (as the root will be Tomcat/bin, for eg)
I would suggest to copy the configuration file at designated place, then you can acess at ease. At certain extent 'System.getProperty("user.dir")' can be used if you are always copying the file 'tomcat` root or application root. But if the files to be used by external party, ideal to copy in a configurable folder (C:\appconf)

Your code works like a charm! But you might have to add the project root dir to your classpath.
If you work with Maven, place your configuration in src/main/resources/conf/sampleprop.conf
When invoking java directly add the project root dir with the java -classpath parameter. Something like:
java -classpath /my/classes/dir:/my/project/root/dir my.Main