URL root = Thread.currentThread().getContextClassLoader().getResource(packageName.replace(".", "/"));
i use the above statement to access .class files given a package name. But in some cases i need to access the files from \build\test\classes instead of \build\classes. How do i modify the above statement?
You are having problems with your classpath.
${project}/build/test/classes and ${project}/build/classes folders suggest that you have either used Ant to compile your code, or checked "Allow output folders for source folders" option in Eclipse's "New Java Project" dialog.
You should configure you project to compile all source files into the same output folder, in your case /build/classes.
If you use classloader to load a resource the path you give it is not absolute to entire filesystem even if it starts with /.
InputStream is = this.getClass().getResourceAsStream("/some/resource");
This actually means find this file ${project}/build/classes/some/resource.
And in order to get that resource at the target location the easiest way is to place it in ${project}/src/some/resource.
Although you should consider using Maven and m2eclipse if your using Eclipse.
Related
I am trying to read a Properties file in a maven nature project using the Properties.load(); I am specifying a path as a string ex. "./someFolder/file.properties",
but when I try to use my project as dependency in other projects I am forced to copy those files to the other project , simply because the "." means current directory.
Is there a way to specify a path so it will always be valid despite where I am calling it from ? ,
I have tried using the MyClass.class.getClassLoader().getResourceAsStream() but I am having trouble using it , it worked sometimes and failed other times.
There are lots of misconceptions in your question.
"." means classPath
No. When used inside a filesystem path (i.e. a path passed to the constructor of a File, or FileReader, or FileInputStream), "." means the current directory.
When used in a resource path (i.e. passed to Class[Loader].getResource[AsStream]()), it's invalid.
The trick is to carefully read the documentation.
getResourceAsStream() expects a /-separated path.
When using ClassLoader.getResource[AsStream](), this path always starts from the root of the classpath. So you would pass a path looking exactly like a fully qualified class name, except the dots would be replaced by slashes. So, com/foo/bar.properties looks for a resource named bar.properties, in the package com.foo.
When using SomeClass.class.getResource[AsStream](), either the path starts with a /, and the path starts from the root of the classpath, or it doesn't, and it starts from the package of SomeClass. So, if SomeClass is in the package com.foo, using /com/foo/bar.properties is equivalent to using bar.properties.
It's hard to tell what you're doing wrong, since you're not providing any detail. But you really need to understand the difference between opening a file on the file system, and reading a resource loaded by the class loader. Sometimes, the resources just happen to be loaded by the class loader from the filesystem, because the classpath happens to contain directories, and not just jar files.
I noticed that my problem was that I had my properties files in the project path itself, and that the ClassLoader.getResource[AsStream](); looks is the target/classes folder, and that I didn't have the resources folder in my project.
I solved it my adding the resources folder to my build path and adding my files in the src/main/resources as the following src/main/resources/foo/bar.properties and loading it by SomeClass.class.getClassLoader().loadResourceAsStream("foo/bar.properties");.
My code cannot locate the .properties file where i have stored login information.
I have put the file in the src folder to make sure it compiles, and it does correctly.
below is the current location of the file and how i am trying to access it.
I have tried various different paths but no luck.
Change your code;
ResourceBundle bundle = ResourceBundle.getBundle("Selenium/readme");
to
ResourceBundle bundle = ResourceBundle.getBundle("readme");
You don't compile a .properties file, you use it as is.
If you use a FileInputStream it will use the working directory which is set in your Run configuration (most likely the top directory)
But you are loading it as a resource which means it must be in your class path. The simplest thing to do is to create a sub-directory for your configuration and add this to your programs class path.
Read properties file like:
ResourceBundle.getBundle("src/properties/readme.properties"); //Or simply "properties/readme.properties"
Put readme.properties under src/properties directory.
you can use this.getClass().getResourceAsStream("readme.properties");
for more information read:
I see that the .properties file is not inside the src folder. Also check the build path of your project.It will show you the src folders and the output folders location. Once you build the project using eclipse build project option, make sure your properties file is now available in the output folder.
This is the well known problem of loading resources from a jar file. This is not the first time I've tried to do this, but now it doesn't work the way I expect it to.
Normally I try to load the Resources with this.getClass.getResource("foo.png"), or getResourceAsStream()and it works. Now however it does not. The Resource is always null.
If I let System.out.println(this.getClass.getResource("")) print me the path (from eclipse) it shows /path/to/eclipseproject/package/structure/. Running this from a jar it just shows rsrc:package/structure
If I recall correctly this should print the path to the jar. Furthermore I thought this would print the package structure in both cases. Am I doing something wrong?
Here is the thing...
When Extracting the file from the Jar use:
this.getClass.getResource("/foo.png")
When running from a runnable Jar use, to reference an external file in the Jar folder path:
this.getClass.getResource("foo.png")
// When running this from Eclipse, it would refer to files in project root!
I have a code in the lower level determining where I'm running from to determine the correct path.
Doe this get the path you need?
this.getClass().getClassLoader().getResource("<your class name>.class").getPath();
See also this question for more on this issue.
Unless you prepend the path to the resources with '/', Class.getResource() will search for the resource in class package. E.g.: tld.domain.Foo.class.getResource("Bar.txt") will search for tld/domain/Bar.txt
Check the URLClassLoader for all the gory details, but it really depends on whether you are trying to access a ressource in the jar,
using a class loaded inside the same jar, in this case your file 'root' is the root of the jar
using a class loaded outside the jar (your eclipse case) where the root is your 'working directory'
To access resources inside a jar from outside, you should use something like
URL url = new URL( "jar", "", "file:" + jar.getCanonicalPath( ) + "!/" + localPathResource );
url.openStream(...)
This answer provides an explanation of how to load class resources from JAR files, even when the class is not in the JAR file and not in the Class-Path specified in the JAR file's manifest. There are also links to code.
I'm working with a project that is setup using the standard Maven directory structure so I have a folder called "resources" and within this I have made a folder called "fonts" and then put a file in it. I need to pass in the full String file path (of a file that is located, within my project structure, at resources/fonts/somefont.ttf) to an object I am using, from a 3rd party library, as below, I have searched on this for a while but have become a bit confused as to the proper way to do this. I have tried as below but it isn't able to find it. I looked at using ResourceBundle but that seemed to involve making an actual File object when I just need the path to pass into a method like the one below (don't have the actual method call in front of me so just giving an example from my memory):
FontFactory.somemethod("resources/fonts/somefont.ttf");
I had thought there was a way, with a project with standard Maven directory structure to get a file from the resource folder without having to use the full relative path from the class / package. Any advice on this is greatly appreciated.
I don't want to use a hard-coded path since different developers who work on the project have different setups and I want to include this as part of the project so that they get it directly when they checkout the project source.
This is for a web application (Struts 1.3 app) and when I look into the exploded WAR file (which I am running the project off of through Tomcat), the file is at:
<Exploded war dir>/resources/fonts/somefont.ttf
Code:
import java.io.File;
import org.springframework.core.io.*;
public String getFontFilePath(String classpathRelativePath) {
Resource rsrc = new ClassPathResource(classpathRelativePath);
return rsrc.getFile().getAbsolutePath();
}
In your case, classpathRelativePath would be something like "/resources/fonts/somefont.ttf".
You can use the below mentioned to get the path of the file:
String fileName = "/filename.extension"; //use forward slash to recognize your file
String path = this.getClass().getResource(fileName).toString();
use/pass the path to your methods.
If your resources directory is in the root of your war, that means resources/fonts/somefont.ttf would be a "virtual path" where that file is available. You can get the "real path"--the absolute file system path--from the ServletContext. Note (in the docs) that this only works if the WAR is exploded. If your container runs the app from the war file without expanding it, this method won't work.
You can look up the answer to the question on similar lines which I had
Loading XML Files during Maven Test run
The answer given by BobG should work. Though you need to keep in mind that path for the resource file is relative to path of the current class. Both resources and java source files are in classpath
Lots of confusion in this topic. Several Questions have been asked. Things still seem unclear.
ClassLoader, Absolute File Paths etc etc
Suppose I have a project directory structure as,
MyProject--
--dist
--lib
--src
--test
I have a resource say "txtfile.txt" in "lib/txt" directory. I want to access it in a system independent way. I need the absolute path of the project.
So I can code the path as abspath+"/lib/Dictionary/txtfile.txt"
Suppose I do this
java.io.File file = new java.io.File(""); //Dummy file
String abspath=file.getAbsolutePath();
I get the current working directory which is not necessarily project root.
Suppose I execute the final 'prj.jar' from the 'dist' folder which also contains "lib/txt/txtfile.txt" directory structure and resource,It should work here too. I should absolute path of dist folder.
Hope the problem is clear.
You should really be using getResource() or getResourceAsStream() using your class loader for this sort of thing. In particular, these methods use your ClassLoader to determine the search context for resources within your project.
Specify something like getClass().getResource("lib/txtfile.txt") in order to pick up the text file.
To clarify: instead of thinking about how to get the path of the resource you ought to be thinking about getting the resource -- in this case a file in a directory somewhere (possibly inside your JAR). It's not necessary to know some absolute path in this case, only some URL to get at the file, and the ClassLoader will return this URL for you. If you want to open a stream to the file you can do this directly without messing around with a URL using getResourceAsStream.
The resources you're trying to access through the ClassLoader need to be on the Class-Path (configured in the Manifest of your JAR file). This is critical! The ClassLoader uses the Class-Path to find the resources, so if you don't provide enough context in the Class-Path it won't be able to find anything. If you add . the ClassLoader should resolve anything inside or outside of the JAR depending on how you refer to the resource, though you can certainly be more specific.
Referring to the resource prefixed with a . will cause the ClassLoader to also look for files outside of the JAR, while not prefixing the resource path with a period will direct the ClassLoader to look only inside the JAR file.
That means if you have some file inside the JAR in a directory lib with name foo.txt and you want to get the resource then you'd run getResource("lib/foo.txt");
If the same resource were outside the JAR you'd run getResource("./lib/foo.txt");
First, make sure the lib directory is in your classpath. You can do this by adding the command line parameter in your startup script:
$JAVA_HOME/bin/java -classpath .:lib com.example.MyMainClass
save this as MyProject/start.sh or any os dependent script.
Then you can access the textfile.txt (as rightly mentioned by Mark) as:
// if you want this as a File
URL res = getClass().getClassLoader().getResource("text/textfile.txt");
File f = new File(res.getFile());
// As InputStream
InputStream in = getClass().getClassLoader()
.getResourceAsStream("text/textfile.txt");
#Mark is correct. That is by far the simplest and most robust approach.
However, if you really have to have a File, then your best bet is to try the following:
turn the contents of the System property "java.class.path" into a list of pathnames,
identify the JAR pathname in the list based on its filename,
figure out what "../.." is relative to the JAR pathname to give you the "project" directory, and
build your target path relative to the project directory.
Another alternative is to embed the project directory name in a wrapper script and set it as a system property using a -D option. It is also possible to have a wrapper script figure out its own absolute pathname; e.g. using whence.