Problem statement:
I have a jar file with a set of configuration files in a package mycompany/configuration/file/.
I don't know the file names.
My intention is to load the file names at runtime from the jar file in which they are packaged and then use it in my application.
As far as I understood:
When I use the ClassLoader.getResources("mycompany/configuration/file/") I should be getting all the configuration files as URLs.
However, this is what is happening:
I get one URL object with URL like jar:file:/C:/myJarName.jar!mycompany/configuration/file/
Could you please let me know what I am doing wrong ?
For what you are trying to do I don't think it is possible.
getResource and getResources are about finding named resources within the classloader's classpath, not listing values beneath a directory or folder.
So for example ClassLoader.getResources("mycompany/configuration/file/Config.cfg") would return all the files named Config.cfg that existed in the mycompany/configuration/file path within the class loader's class path (I find this especially useful for loading version information personally).
In your case I think you might almost have half a solution. The URL you are getting back contains the source jar file (jar:file:/C:/myJarName.jar). You could use this information to crack open the jar file a read a listing of the entries, filtering those entries whose name starts with "mycompany/configuration/file/".
From there, you could then fall back on the getResource method to load a reference to each one (now that you have the name and path)
Related
I'm working with a local WebSphere server configured in IntelliJ Idea, and the application I'm working on is using a third-party library that loads a properties file with:
ThirdPartyClass.class.getClassLoader().getResourceAsStream(fileNameParameter);
It uses the default bootstrapClassLoader.
I've been instructed to make sure the properties file is in a config directory so that it can be edited without deploying a code change. My project looks something like this:
ProjectName
Configs
my.properties
src
java (sources root)
packages, .java files, etc
main (resources root)
schemas, web docs, etc
I have tried several of paths to make it work but it always returns null. Since I initially thought it was reaching from within the third party library package, I tried adding several ..\'s to the file path, but then I learned that this method loads from the classpath, so I pulled a
String test = System.getProperty("java.class.path");
and upon inspection, my classpath is all made up of websphere directories and jars within them:
C:\Users\me\Programs\IBM\AppServer\profiles\AppSrv01/properties
C:\Users\me\Programs\IBM\AppServer\AppSrv01/properties
and several jar files in C:\Users\me\Programs\IBM\AppServer/lib/
So just as a test I stuck the file in C:\Users\me\Programs\IBM\AppServer\AppSrv01/properties, then tried to grab it with just its file name (my.properties), but still couldn't reach it. I've also tried moving the file into the src directory and the main directory, but no matter what I do it just can't seem to find the file.
I'm aware that this method is typically used to grab resources from within a jar file, but from my understanding it seems like it should be possible to reach my file from outside of one as long as it's in a directory in the classpath... but apparently not since that didn't work.
I have the absolute path on my hard drive and will have said path on the server; is there a way to derive the path that ClassLoader.getResourceFromStream() wants with that info? Failing that, is there some obvious mistake I'm making with the resource url?
I think your fileNameParameter simply needs to start with / to indicate that it is in the root level of the classpath. Otherwise it will be searched relative to the class it is loaded from, i.e. the package of ThirdPartyClass in your example.
I have a maven project with these standard directory structures:
src/main/java
src/main/java/pdf/Pdf.java
src/test/resources
src/test/resources/files/x.pdf
In my Pdf.java,
File file = new File("../../../test/resources/files/x.pdf");
Why does it report "No such file or dirctory"? The relative path should work. Right?
Relative paths work relative to the current working directory. Maven does not set it, so it is inherited from whatever value it had in the Java process your code is executing in.
The only reliable way is to figure it out in your own code. Depending on how you do things, there are several ways to do so. See How to get the real path of Java application at runtime? for suggestions. You are most likely looking at this.getClass().getProtectionDomain().getCodeSource().getLocation() and then you know where the class file is and can navigate relative to that.
Why does it report "No such file or dirctory"? The relative path should work. Right?
wrong.
Your classes are compiled to $PROJECT_ROOT/target/classes
and your resources are copied to the same folder keeping their relative paths below src/main/resources.
The file will be located relative to the classpath of which the root is $PROJECT_ROOT/target/classes. Therefore you have to write in your Pdf.java:
File file = new File("/files/x.pdf");
Your relative path will be evaluated from the projects current working directory which is $PROJECT_ROOT (AFAIR).
But it does not matter because you want that to work in your final application and not only in your build environment. Therefore you should access the file with getClass().getResource("/path/to/file/within/classpath") which searches the file in the class path of which the root is $PROJECT_ROOT/target/classes.
No the way you are referencing the files is according to your file system. Java knows about the classpath not the file system if you want to reference something like that you have to use the fully qualified name of the file.
Also I do not know if File constructor works with the classpath since it's an abstraction to manage the file system it will depend where the application is run from. Say it is run from the target directory at the same level as source in that case you have to go one directory up and then on src then test the resources the files and finally in x.pdf.
Since you are using a resources folder I think you want the file to be on the classpath and then you can load a resource with:
InputStream in = this.getClass().getClassLoader()
.getResourceAsStream("<path in classpath>");
Then you can create a FileInputStream or something to wrap around the file. Otherwise use the fully qualiefied name and put it somewere like /home/{user}/files/x.pdf.
I have a maven project and want to read file in it form its class path. The code that i am using is
InputStream is = getClass().getResourceAsStream ("filename.json");
But every time i am getting null inputstreams. I am not sure why ?
The file is places under /src/main/resources. The same folder which contains log4j.xml and it is being picked up decently.
Please note, I am trying to run this file from Eclipse i.e., run or debug mode. No vm arguments or whatsoever.
The Class.getResourceAsStream(String) method looks for the given resource within the same namespace (i.e. package) that the given class is in unless you give it an absolute path (see the API documentation); If it can't find the resource on the classpath in this namespace, it returns null. Since your class is likely inside e.g. com.myproject.resourcemanagement, your resource file has to analogously be under src/main/resources/com/myproject/resourcemanagement, similary to how your class source files are organised (under src/main/java/com/myproject/resourcemanagement).
when reading the properties file, iam getting nullpointer exception.
faceConfig.load(ReadPropertyFile.class.getClassLoader().getResourceAsStream("/resources/faces.properties"));
below is the path of properties file facedetections/src/main/resources/faces.properties
i tried in different combinations as my class file that reads in below path /facedetections/src/main/java/com/facial/facedetection/utils/ReadPropertyFile.java
combinations are ../../../../../resource/faces.properties , /resource/faces.properties and
../../../resource/faces.properties
Please suggest what is the correct path i can provide for this.
Edit :
I extracted the war file and providing its path below.
looks your property file under class folder not in resource.
as your screen shot cant find resource folder under class folder.
So just use
ReadPropertyFile.class.getClassLoader().getResourceAsStream("faces.properties")
The path is relative to the point where the object (.class) files are located. Are you sure you have configured your build/test tool to copy the resource file into that structure? Exactly where? That's what counts, not the position of the sources.
Additionally, my understanding is that getResourceAsStream() of most ClassLoaders do not support .. notation.
The position where your resource file is currently located is out of reach of the ClassLoader. If you move your file to /facedetections/src/main/java/resources/faces.properties, then you will be able to use the current code
getResourceAsStream("resources/faces.properties")
I'm making assumptions about your environment. In particular, this is entirely dependent on classloaders. If this doesn't help, please provide object file location, not sources (unless it is the same, but state it).
Since it is unlikely that you get a NPE when the file is not found I assume that faceConfig is null when you execute that line.
So I see there has already been a post very similar to this issue, however I am in a situation where I have no power to specify the location of this file within my jar and so am hoping someone is aware of a solution to get around this.
So I currently use the following snippet to obtain a file as an input stream, the file 'plugin.xml' is located at the root of the jar and I cannot change this location as another piece of software (dynatrace) creates this file and determines its location.
the standard snippet:
InputStream is = JmxPlugin.class.getResourceAsStream("/plugin.xml");
Now I am aware that the issue is that the ClassLoader is picking up the first file which matches the name 'Plugin.xml' along the classpath (one which isn't in my jar, yay).
Can anyone think of a way to ensure I pick up the correct file without having to move it? The relative path of my class in the jar is com/something/jmx/JmxPlugin.class.
(Id rather not have to unpack the jar in memory).
Many thanks for any contributions,
I'm not absolutely sure, but seems like Thread.currentThread().getContextClassLoader().getResourceAsStream("/plugin.xml") may solve your issue. If not, you'll have to create your own ClassLoader and resolve the issue there.
The simplies way is to move your jar in classpath to be the first containing Plugin.xml,
Another approach is to use getResource() to locate your jar file:
URL myJar=JmxPlugin.class.getResource("/"+JmxPlugin.class.getName().replace(".","/")+".class");
then use this URL to open jar file and extract Plugin.xml.