File from resources folder with getResource Class. Appreciate all the great replies. The implementation of all these examples and code snippets can be found — this is a Maven-based project, so it should be easy to import and run as it is. A comparison of different ways of resources loading in Java Followings are the preferred ways to load resources in classpath. This method is not appropriate if we want to load resources inside the packages unless we use complete paths starting from root.
After getting the resource bundle, the ResourceBundleTest class iterates over all of the keys in mybundle. How to change maven default 'resources' directory? I believe the above piece will get the Filewhne it is either placed inside the application or if the code is run on local machine. This is useful any time you pack files and other resources into Jar files to distribute your Java application. So, to extend our example: package my. This makes it clear what you're doing. In the followings, we are specifying 'java' and 'resources' directories as resources. My mistake was I had forgotten that to find a resource in a jar file you had to specifically mention the jar file, by name, in the classpath.
The path will be appended if needed. That makes sense because our resources are just under 'classes' in the target folder. Also note that I'm throwing any exceptions that occur rather than handling them. You can access these files and folders from your java code as shown below. You don't need to mess with class loaders. If you do put the code into the jar then when you run the code in the jar the jar is on the classpath so that will also work. File instead of reading it as a stream.
Java: How to list the contents of a getResource directory Greg Briggs' Technical Articles Java: Listing the contents of a resource directory The ClassLoader. Notice how resources are copied in output folder. To call getResourceAsStream in a static method, we use ClassName. I've tried, for trying to figure out what's going on, to use both a text file and an image file. If you select the wrong choice I did , then you'll still, in my case get the NullPointerException. Compiling with javac using the output folder option The resources in this case are not copied at all.
There are some cases in which you want to get a file from the resources folder, in particular when you want to unit test certain code that does file manipulations e. Calling getAbsolutePath on it returns its absolute Path. Hi, I am relatively new to this so was trying to understand. Sorry for my apparent confusion. If not set, the default is the ClassLoader context of the parent Thread.
You just add the library and viola, it woks. If this object was loaded by the bootstrap loader, the call is delegated to ClassLoader. No, getResources does not do it. I'll try to get back to that later. Convert using Java We can use the of java to convert a File to different InputStreams.
Tony Docherty wrote:As Paul has said if the code is running from the src folder and not from the jar then the contents of the jar will not be available unless you have explicitly added the jar file to the classpath used when the program is launched. Notice that the package of the properties file is specified using. In my case I was using a for geolocation retrieval. Project Structure Below image describes the folder structure used in this example. This method delegates the call to its class loader. .
Still trying to figure that one out. © 2002-2017 Greg Briggs except where attributed otherwise. And these names are processed by Java ResourceLoader differently. These files may include configuration files, scripts and other resources needed during run time. I know that resource is there. No I'm saying you need to add the jar to the classpath. If resources are somewhere else they are ignored this is default behavior which can be customized, we will explore that later.
The method returns null if the resource cannot be found or loaded. When the software is executed, it may need to load the contents of these files for some kind of processing — may be properties, sql statements, etc. Resources are loaded from output folder during runtime. The way it works is the class delegates the search to its classloader so it's up to the classloader that loaded the class to decide where to look. To read binary resources, you can use directly use the InputStream instance. If it's the former, then you should start looking at how things are arranged in the jar file after you produce it instead. But if my code is in some server and i want to get the file from my local machine in the code.
The ResourceLoader With Java you can use the classLoader of the current thread and try to load the file, but the Spring Framework provides you with much more elegant solution like the. Same big project, just call m3. Very much appreciate your reply and look forward to hearing back. The resource to be loaded doesn't need to be on the classpath as long it is reachable ie if the path is relative then there is a relative path from somewhere on the classpath that locates a resource with the given name. Therefore I needed to load the file and create a DatabaseReader object, which is stored in the server memory. Thanks very much for any suggestions! Or does it mean that you're talking about how things are arranged in your? Please let me know how this can be possible. Paul Clapham wrote:If you want to use that code as is, then you put the jpg file into the same folder inside the jar as the file which is trying to use the jpg.