I create a streamHelper function to load any file that is on the classpath, this is the path for my json file, but I got error when I using this function
streamHelper("/Users/my/IdeaProjects/rules-management/data_tacos-sample/post-migration/sourceData/config_-e48059402bd4.json")
java.io.FileNotFoundException: class path resource [/Users/my/IdeaProjects/rules-management/data_tacos-sample/post-migration/sourceData/config_-e48059402bd4.json] cannot be opened because it does not exist
private InputStream streamHelper(String resourcePath) {
DefaultResourceLoader resourceLoader = new DefaultResourceLoader();
Resource resource = resourceLoader.getResource(resourcePath);
return resource.getInputStream();
}
I print the file name on console, when I click that link it can direct me to the file which means the path is good, why I got this error? How to fix that.
The DefaultResourceLoader is a Spring class. According to Spring docs, the DefaultResourceLoader
Will return a UrlResource if the location value is a URL, and a ClassPathResource if it is a non-URL path or a "classpath:" pseudo-URL.
By default, Spring looks for all your resources in a folder called resources.
By default, this handler serves static content from any of the /static, /public, /resources, and /META-INF/resources directories that are on the classpath. Since src/main/resources is typically on the classpath by default, we can place any of these directories there.
Move your json into a folder called resources, make sure that folder is on your classpath, and then call the streamHelper as follows:
streamHelper("classpath:config_-e48059402bd4.json")
I have a config.ini file I need to open that is quite far back in directory so I used File.getAbsolutePath() to set the base directory and concatenated the remainder of the path.
Printing the path, I get the correct path that I can paste into file explorer, but the object returned is null.
So I started by initializing my Properties and ClassLoader as such:
Properties prop = new Properties();
ClassLoader classLoader = Test.class.getClassLoader();
Then I create the path. I tried escaping a back slash(1) as well as forward slash(2), both return null but both paths work in file explorer.
String absPath = new File("").getAbsolutePath();
absPath = absPath.concat("\\resources\\config\\config.ini"); // (1)
absPath = absPath.concat("/resources/config/config.ini"); // (2)
then I try to set the URL to open an InputStream
URL res = Objects.requireNonNull(classLoader.getResource(absPath), "Unable to open config.ini");
InputStream is = new FileInputStream(res.getFile());
However, the following returns null.
classLoader.getResource(absPath)
I expected this to properly open the file because the path was correct. I am using Intelij and I read that I needed to add the .ini resource file under settings > compiler, which I did but that did not resolve my issue.
Thank you!
That's not the way to load resources via class-loaders.
If your classpath is something like the following ...
java -cp resources;lib/my.jar ... org.mypack.MyClass
then you load it with this path
getClassLoader().getResource("/config/config.ini");
Your classpath includes the resources folder, and the class-loader loads from there.
The absolute path of the OS is quite certainly not in classpath.
In any case, you must be certain that the resource folder is in classpath.
If your config file is not in classpath then you can't use classloaders to load that file.
Just one more thing, if your configuration is not in classpath, but in a child directory of your working dir, why can't you simply use new FileInputStream("resources/config/config.ini");
This is my current database path
"jdbc:ucanaccess://D:/Java/TransactionProcessingSystem/src/transactionprocessingsystem/Resources/TPSystem.accdb"
But I want set the database path in the same compilation folder location, Like this:
"jdbc:ucanaccess://../Resources/TPSystem.accdb"
I am using Netbeans to develop the program.
It just must be a valid path(relative or absolute). Therefore, if you're using a relative path, it must be relative to the current working directory.
You have to use ClassLoader class.
Firstly put the resource file(TPSystem.accdb) in src/main/resource of your maven project.
ClassLoader classLoader = getClass().getClassLoader();
File file = new
File(classLoader.getResource("TPSystem.accdb").getFile());
and now the file Path will be
String filePath = "jdbc:ucanaccess://"+file.getAbsolutePath();
In my Maven project, I have the following code in the main method:
FileInputStream in = new FileInputStream("database.properties");
but always get a file not found error.
I have put the file in src/main/resources and it is properly copied to the target/classes directory (I believe that is the expected behavior for Maven resources) but when actually running the program it seems it can never find the file. I've tried various other paths:
FileInputStream in = new FileInputStream("./database.properties");
FileInputStream in = new FileInputStream("resources/database.properties");
etc. but nothing seems to work.
So what is the proper path to use?
Based on "disown's" answer below, here was what I needed:
InputStream in = TestDB.class.getResourceAsStream("/database.properties")
where TestDB is the name of the class.
Thanks for your help, disown!
You cannot load the file directly like that, you need to use the resource abstraction (a resource could not only be in the file system, but on any place on the classpath - in a jar file or otherwise). This abstraction is what you need to use when loading resources. Resource paths are relative to the location of your class file, so you need to prepend a slash to get to the 'root':
InputStream in = getClass().getResourceAsStream("/database.properties");
In my Java app I need to get some files and directories.
This is the program structure:
./main.java
./package1/guiclass.java
./package1/resources/resourcesloader.java
./package1/resources/repository/modules/ -> this is the dir I need to get
./package1/resources/repository/SSL-Key/cert.jks -> this is the file I need to get
guiclass loads the resourcesloader class which will load my resources (directory and file).
As to the file, I tried
resourcesloader.class.getClass().getResource("repository/SSL-Key/cert.jks").toString()
in order to get the real path, but this way does not work.
I have no idea which path to use for the directory.
I had problems with using the getClass().getResource("filename.txt") method.
Upon reading the Java docs instructions, if your resource is not in the same package as the class you are trying to access the resource from, then you have to give it relative path starting with '/'. The recommended strategy is to put your resource files under a "resources" folder in the root directory. So for example if you have the structure:
src/main/com/mycompany/myapp
then you can add a resources folder as recommended by maven in:
src/main/resources
furthermore you can add subfolders in the resources folder
src/main/resources/textfiles
and say that your file is called myfile.txt so you have
src/main/resources/textfiles/myfile.txt
Now here is where the stupid path problem comes in. Say you have a class in your com.mycompany.myapp package, and you want to access the myfile.txt file from your resource folder. Some say you need to give the:
"/main/resources/textfiles/myfile.txt" path
or
"/resources/textfiles/myfile.txt"
both of these are wrong. After I ran mvn clean compile, the files and folders are copied in the:
myapp/target/classes
folder. But the resources folder is not there, just the folders in the resources folder. So you have:
myapp/target/classes/textfiles/myfile.txt
myapp/target/classes/com/mycompany/myapp/*
so the correct path to give to the getClass().getResource("") method is:
"/textfiles/myfile.txt"
here it is:
getClass().getResource("/textfiles/myfile.txt")
This will no longer return null, but will return your class.
It is strange to me, that the "resources" folder is not copied as well, but only the subfolders and files directly in the "resources" folder. It would seem logical to me that the "resources" folder would also be found under `"myapp/target/classes"
Supply the path relative to the classloader, not the class you're getting the loader from. For instance:
resourcesloader.class.getClassLoader().getResource("package1/resources/repository/SSL-Key/cert.jks").toString();
In the hopes of providing additional information for those who don't pick this up as quickly as others, I'd like to provide my scenario as it has a slightly different setup. My project was setup with the following directory structure (using Eclipse):
Project/
src/ // application source code
org/
myproject/
MyClass.java
test/ // unit tests
res/ // resources
images/ // PNG images for icons
my-image.png
xml/ // XSD files for validating XML files with JAXB
my-schema.xsd
conf/ // default .conf file for Log4j
log4j.conf
lib/ // libraries added to build-path via project settings
I was having issues loading my resources from the res directory. I wanted all my resources separate from my source code (simply for managment/organization purposes). So, what I had to do was add the res directory to the build-path and then access the resource via:
static final ClassLoader loader = MyClass.class.getClassLoader();
// in some function
loader.getResource("images/my-image.png");
loader.getResource("xml/my-schema.xsd");
loader.getResource("conf/log4j.conf");
NOTE: The / is omitted from the beginning of the resource string because I am using ClassLoader.getResource(String) instead of Class.getResource(String).
When you use 'getResource' on a Class, a relative path is resolved based on the package the Class is in. When you use 'getResource' on a ClassLoader, a relative path is resolved based on the root folder.
If you use an absolute path, both 'getResource' methods will start at the root folder.
#GianCarlo:
You can try calling System property user.dir that will give you root of your java project and then do append this path to your relative path for example:
String root = System.getProperty("user.dir");
String filepath = "/path/to/yourfile.txt"; // in case of Windows: "\\path \\to\\yourfile.txt
String abspath = root+filepath;
// using above path read your file into byte []
File file = new File(abspath);
FileInputStream fis = new FileInputStream(file);
byte []filebytes = new byte[(int)file.length()];
fis.read(filebytes);
For those using eclipse + maven. Say you try to access the file images/pic.jpg in src/main/resources. Doing it this way :
ClassLoader loader = MyClass.class.getClassLoader();
File file = new File(loader.getResource("images/pic.jpg").getFile());
is perfectly correct, but may result in a null pointer exception. Seems like eclipse doesn't recognize the folders in the maven directory structure as source folders right away. By removing and the src/main/resources folder from the project's source folders list and putting it back (project>properties>java build path> source>remove/add Folder), I was able to solve this.
resourcesloader.class.getClass()
Can be broken down to:
Class<resourcesloader> clazz = resourceloader.class;
Class<Class> classClass = clazz.getClass();
Which means you're trying to load the resource using a bootstrap class.
Instead you probably want something like:
resourcesloader.class.getResource("repository/SSL-Key/cert.jks").toString()
If only javac warned about calling static methods on non-static contexts...
Doe the following work?
resourcesloader.class.getClass().getResource("/package1/resources/repository/SSL-Key/cert.jks")
Is there a reason you can't specify the full path including the package?
Going with the two answers as mentioned above. The first one
resourcesloader.class.getClassLoader().getResource("package1/resources/repository/SSL-Key/cert.jks").toString();
resourcesloader.class.getResource("repository/SSL-Key/cert.jks").toString()
Should be one and same thing?
In Order to obtain real path to the file you can try this:
URL fileUrl = Resourceloader.class.getResource("resources/repository/SSL-Key/cert.jks");
String pathToClass = fileUrl.getPath;
Resourceloader is classname here.
"resources/repository/SSL-Key/cert.jks" is relative path to the file. If you had your guiclass in ./package1/java with rest of folder structure remaining, you would take "../resources/repository/SSL-Key/cert.jks" as relative path because of rules defining relative path.
This way you can read your file with BufferedReader. DO NOT USE THE STRING to identify the path to the file, because if you have spaces or some characters from not english alphabet in your path, you will get problems and the file will not be found.
BufferedReader bufferedReader = new BufferedReader(
new InputStreamReader(fileUrl.openStream()));
I made a small modification on #jonathan.cone's one liner ( by adding .getFile() ) to avoid null pointer exception, and setting the path to data directory. Here's what worked for me :
String realmID = new java.util.Scanner(new java.io.File(RandomDataGenerator.class.getClassLoader().getResource("data/aa-qa-id.csv").getFile().toString())).next();
Use this:
resourcesloader.class.getClassLoader().getResource("/path/to/file").**getPath();**
One of the stable way to work across all OS would be toget System.getProperty("user.dir")
String filePath = System.getProperty("user.dir") + "/path/to/file.extension";
Path path = Paths.get(filePath);
if (Files.exists(path)) {
return true;
}