I'm currently developing a scala library and wanted to get a file that is inside it, event when compiled as a Jar dependency. The problem is that when executed from another project where the library is imported, the path is relative to that project. Here is the code to get the file :
private val pathToDocker: Path = Paths.get("src", "main", "resources", "docker-compose")
What can I do to look for my file inside the imported dependency ?
The file won't be in a file system -- it'll be in the compiled JAR.
You can get a JAR resource using a class loader. Here's an example of how to do that in another codebase.
Utility function:
https://github.com/hail-is/hail/blob/6db198ae06/hail/src/main/scala/is/hail/utils/package.scala#L468
Usage to load version info:
https://github.com/hail-is/hail/blob/6db198ae06/hail/src/main/scala/is/hail/package.scala#L21
There is a JarUtil - from an answer of access-file-in-jar-file translate to Scala:
import java.util.jar.JarFile
val jar = new JarFile("path_to_jar/shapeless_2.12-2.3.3.jar")
val entry = jar.getEntry("shapeless/Annotation.class")
val inStream = jar.getInputStream(entry)
As you mentioned scala.io.Source.fromResource works only within your project:
Source.fromResource("tstdir/source.txt")
Make sure the file is in the resources directory, like:
The way I found to achieve getting the path inside a jar depedency was creating a singleton (scala object) that has a method that loads the files. Since it is inside the Jar, the resulting path is relative to the jar itself. Here is the code :
object ClassLoaderTest {
def dockerFile: File =
new File(getClass.getClassLoader.getResource("docker/docker-compose.yml").getPath)
}
When called from a Trait (or an interface), the resulting path is :
jar:file:/home/myname/.ivy2/cache/com.mypackage/libraryname/jars/libraryname.libraryversion.jar!/docker/docker-compose.yml
I am working with scala and want to use SimplexSolver (https://opensource.googleblog.com/2009/06/introducing-apache-commons-math.html). I am using sbt, hence downloaded the .jar file for commons-math-2.2 and imported it in the code. Then I tried to use it, but it showed not found: type SimplexSolver. Following is the code snippet.
import org.apache.commons.math.optimization.linear
class MySimplexSolver{
val solver = new SimplexSolver()
}
I have a lib folder where all the .jar files are kept. Any help regarding this is appreciated. :)
When I try to run my groovy script in Eclipse, I get a cast exception like:
org.codehaus.groovy.runtime.typehandling.GroovyCastException: Cannot cast object 'package.Config#6babd36b' with class 'package.Config' to class 'package.Config'
The exception occurs when I instantiate another object using the Config as parameter.
clazz.newInstance(config: config)
To me, this seems very odd. But perhaps someone out there can bring me an explanation?
I just want to add some information for this topic. Maybe somebody will find it useful.
I received same error then was trying to compile groovy script for using in multithread environment:
GroovyClassLoader groovyClassLoader = new GroovyClassLoader();
Class<? extends Script> clazz = groovyClassLoader.parseClass(groovyCode);
return clazz.newInstance();
But by default GroovyClassLoader are using ClassLoader of current thread:
Thread.currentThread().getContextClassLoader())
So, you can pickup one class loader and use it for all threads. Or avoid shared groovy stuff :)
Solved! :)
In the Eclipse run configuration, I looked at the argument tab and noticed that both source files and compiled class files was part of the classpath parameter:
--classpath "${workspace_loc:/groovyscripts}/src/main/groovy:...
...:${workspace_loc:/groovyscripts}/classes"
the above setting gives different classloaders for my config object (loaded from src/) and the target field (loaded from classes/)
groovy.lang.GroovyClassLoader$InnerLoader#7a06cf15
org.codehaus.groovy.tools.RootLoader#32728d
Removing the first reference to the source files resulted in a successful run. Classloaders after removal is ONLY RootLoader.
org.codehaus.groovy.tools.RootLoader#32728d
The answer is to remove the source from the classpath in arguments tab in Eclipse.
New to Scala and having problems reading an XML file in a Scala worksheet. So far I have:
downloaded the Scala IDE (for Windows) and unzipped it to my C:\ drive
created a Scala project with the following file path: C:\eclipse\workspace\xml_data
created the xml file ...\xml_data\music.xml using the following data
created a package sample_data and create the following object (with file path: ...\xml_data\src\sample_data\SampleData.scala):
package sample_data
import scala.xml.XML
object SampleData {
val data = XML.loadFile("music.xml")
}
object PrintSampleData extends Application {
println(SampleData.data)
}
This runs OK, however, when I create the Scala worksheet test_sample_data.sc:
import sample_data.SampleData
object test {
println(SampleData.data)
}
I get a java.lang.ExceptionInInitializerError which includes: Caused by: java.io.FileNotFoundException: music.xml (The system cannot find the file specified).
The workspace is C:\eclipse\workspace. Any help or insight much appreciated. Cheers!
UPDATE:
Following aepurniet's advice, I ran new java.io.File(".").getAbsolutePath() and got the following respectively:
SampleData.scala: C:\eclipse\workspace\xml_data\.
test_sample_data.sc: C:\eclipse\.
So this is what is causing the problem. Does anyone know why this occurs? Absolute file paths resolve the problem. Is this the best solution?
Regarding what is causing different user directory between the scala class and worksheet:
You are likely hitting the Eclipse IDE issue listed here
https://github.com/scala-ide/scala-worksheet/issues/102
Jfyi, I used Intellij and the issue is not reproducible there.
Regarding using absolute paths:
Using absolute path works fine for quick testing, but would NOT be a good practice for the actual implementation. You can consider passing the path along with the filename as input to SampleData.
Some hack mentioned here to get the base path of the workspace from the scala worksheet: Configure working directory of Scala worksheet
If this is just for your testing, hacking the absolute path of workspace inside the worksheets might be the easiest for you.
SampleData.scala
package sample_data
import scala.xml.XML
object SampleData {
def data(filename: String) = XML.loadFile(filename)
}
object PrintSampleData extends Application {
println(SampleData.data(System.getProperty("user.dir") + "/music.xml")
}
Scala worksheet:
import sample_data.SampleData
object test {
val workDir = ... // Using the hack or hardcoding
println(SampleData.data(workDir + "/music.xml"))
}
I have installed Eclipse 3.5.2 and today's Scala plugin from /update-current (that's Scala 2.8 final.) I can compile and run Scala projects consisting of a single singleton object that implements main().
But, if a project contains more classes, I receive the "Could not find the main class" error.
I have tried searching for the solution and I discovered:
Eclipse is correctly looking for the Main$ class, not the Main class
* under Debug Configurations, my main class is correctly identified as mypackage.Main
* my plugin is up to date and recommended for my version of Eclipse
* cleaning, restarting etc. doesn't help.
The same project will compile with scalac.
Thanks for any ideas on how to solve this.
EDIT: MatthieuF suggested I should post the code.
This snippet produces an error. It's not the most idiomatic code, but I wrote it that way to test my environment. I tried it as a single file and as separate files. It DOES work with scalac.
import swing._
class HelloFrame extends Frame {
title = "First program"
contents = new Label("Hello, world!")
}
object Hello {
val frame = new HelloFrame
def main(args : Array[String]) : Unit = {
frame.visible = true
}
}
BUT, if I nest the definition of HelloFrame within Hello, it works. This snippet runs perfectly:
import swing._
object Hello {
class HelloFrame extends Frame {
title = "First program"
contents = new Label("Hello, world!")
}
val frame = new HelloFrame
def main(args : Array[String]) : Unit = {
frame.visible = true
}
}
For me, the problem was that there was a build error (see Problems tab) which was preventing compilation; oops! The reason you see the error is that the run macro proceeds despite the failed compilation step, and attempts to run class files it expects to be there; they don't exist because there was a build error preventing compilation, so it says it can't find Main (not compiled).
Problem goes away when build can complete successfully, i.e. errors are fixed.
I guess, theoretically, there may be more complicated reasons your build is not completing successfully that are not listed in Problems.
One possibility is that you are trying to launch using ctrl-F11, but from a different class.
The Scala Eclipse plugin does not obey the defaults for Java launching. In Preferences->Run/Debug->Launching, there are some options Launch Operation->Always Launch the previously selected application, etc. This currently does not work in the Scala eclipse plugin. To launch a specified main, you need to launch it from the editor for the class.
There has been a bug raised for this. http://scala-ide.assembla.com/spaces/scala-ide/tickets/1000023-scala-launch--does-not-follow-jdt-behaviour
EDIT: This is now (mostly) fixed.
For me it was Eclipse specific problem. I noticed that .class file wasn't built at all. So bin directory doesn't have compiled classes.
When I manually compiled *.scala file using *.sbt and copied it to bin directory it was working as expected.
I tried different tips and tricks and it wasn't worked until I reinstalled Scala plugin in Eclipse .
I'd solve similar problem by executig "Project->Clean.." with next automatically building.
I had the same error message with a Java application made by myself.
The problem was that I deleted (though inside Eclipse) a jar that belonged to the Java build path, without deleting it from the Java build path (project's Properties window). When I did it the class could compile and run again.
Make sure that the .class files exist, usually below the bin directory.
In particular, if you have errors in unrelated files in the same project then the compilation may fail, and no .class files will be produced.
There can be the case of projects, containing errors, added to the build path of the application which prevents the completion of successful compilation. Make sure you remove any such project from the build path before running the application.
Removing these projects solved the problem for me.
Do you have a proper build tool setup? Like sbt have you installed it?
You can check its version by $sbt --version
If it is not setup you can download from here http://www.scala-sbt.org/download.html
You might have to restart your eclipse after installation.
Just copy your XXX.scala file code. Remove the package and create a new Scala Class. Paste your XXX.scala code. (If you are using maven, do a maven clean and build.) Run configuration again. This works for me.
I have faced this issue. I have just deleted the package name, created scala class, Written the same code, Set Build to "Build Automatically". Finally, It works perfectly fine.
Check scala-ide.log
For me the issue was that there were errors on:
AppData\Local\Temp\sbt_10d322fb\xsbt\ClassName.scala:16: error: not found: value enteringPhase
enteringPhase(currentRun.flattenPhase.next) { s fullName separator }
If you are using Intellij, mark directory as source root