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is it possible to 'source' a script before launching \ debugging an application in pycharm \ intelij?

The use case is that we need some environment variable available to the applications (python \ java)
These variables are created by bash script.
While the script is very simple, it does contain some logic (like handling defaults, checking some system settings, etc) - so it can't be simply replaced by static environment configuration
Is there a way in pycharm and \ or intelij to source a shell script before executing the main program?

Specifically for pycharm, it's possible to use a script instead of a "real" python interpreter.
using a script that sets up environment and then launches the "real" python interpreter works great
something like:
#!/bin/sh
. env.sh
python "$#"
didn't find a way to do a similar thing with java

No, it's not possible. Please vote for the related feature request.
There was also a similar question already asked here.

How to use jruby-complete programatically from Java?

I've followed the following article http://seanchenxi.com/java/sass-compass-jruby-single-jar/ on how to get a jar with ruby, sass and compass.
I'd like to use this as part of a Jenkins plugin. Is there a way to call this programatically from Java the same way that you would run from the command line java -jar jcompass.jar -S compass compile --help?

Warbler can do that ... try packaging up with the runnable feature and see how it goes :
java -jar jcompass.jar -S compass compile
... even if it does not work please report with the project - should be "easily" fixable

Java: How to run a Java process in a specific directory?

When I run a jar file, say in /home/jars like such java -jar /home/jars/jarfile.jar, it seems to run in whatever directory I'm in.
How do I make it so java -jar /home/jars/jarfile.jar runs with /home/jars as the current working directory??

The simple answer:
( cd /home/jars; java -jar /home/jars/jarfile.jar )
See also this response.
The analogy for most languages supporting an exec call, is the variant of exec that lets you specify a working directory.
See this for simple PHP solution.

Can java run a compiled scala code?

Can command java run a compiled scala code? If so, why do we have an exclusive command scala?

You can run byte code generated by Scala if you include all necessary runtime libs for Scala (scala-library.jar, scala-swing.jar ...) in the classpath. The scala command does this automatically, and supports Scala specific command line arguments.

Yes, it can. Scala is compiled down to Java bytecode. But remember that it depends on the Scala runtime classes, so you need to still have Scala's jar files on the classpath.
If so, why do we have an exclusive command scala?
Convenience wrapper.

Scala is designed to integrate easily
with applications that run on modern
virtual machines, primarily the Java
virtual machine (JVM). The main Scala
compiler, scalac, generates Java class
files that can be run on the JVM.
-> http://www.artima.com/scalazine/articles/steps.html
As long as you have installed the scala runtime you should be fine: compile classes with scalac and run them with java.

Just want to add my own answer as additional value for the future readers:
scala, if run without parameter, will run an interactive shell
scala, if run with a text file name as parameter, will regard the file as a scala script
those two can't be done using java

If you look closely, the scala command is simply a bash helper-script which summarize to the below command:
$cat /usr/local/Cellar/scala#2.11/2.11.12_1/libexec/bin/scala
execCommand \
"${JAVACMD:=java}" \
$JAVA_OPTS \
"${java_args[#]}" \
"${classpath_args[#]}" \
-Dscala.home="$SCALA_HOME" \
$OVERRIDE_USEJAVACP \
"$EMACS_OPT" \
$WINDOWS_OPT \
scala.tools.nsc.MainGenericRunner "$#"
There are 2 things required to run a .class file compiled using scalac ( the scala compiler) using the java command.
We need to include the scala-library.jar and the location of the .class file in the classpath. To find the location of scala-library.jar, please execute the
below:
which scala /usr/local/opt/scala#2.11/bin/scala
In my case the scala-*.jar files are in :
/usr/local/Cellar/scala#2.11/2.11.12_1/idea/lib on Mac
The location of the Main2.class file which is in /training/example1/scala.
So, to execute the program we could use the below command:
java -cp /usr/local/Cellar/scala#2.11/2.11.12_1/idea/lib/scala-library.jar:/training/example1/scala/ Main2
EDIT-1: If you are using windows, please use semicolon(;) as the separator in java classpath command.

How do I include jars in a groovy script? [duplicate]

This question already has answers here:
How do I auto load a database jar in Groovy without using the -cp switch?
(5 answers)
Closed 6 years ago.
I have a groovy script that needs a library in a jar. How do I add that to the classpath? I want the script to be executable so I'm using #!/usr/bin/env groovy at the top of my script.

Starting a groovy script with #!/usr/bin/env groovy has a very important limitation - No additional arguments can be added. No classpath can be configured, no running groovy with defines or in debug. This is not a groovy issue, but a limitation in how the shebang (#!) works - all additional arguments are treated as single argument so #!/usr/bin/env groovy -d is telling /usr/bin/env to run the command groovy -d rathen then groovy with an argument of d.
There is a workaround for the issue, which involves bootstrapping groovy with bash in the groovy script.
#!/bin/bash
//usr/bin/env groovy -cp extra.jar:spring.jar:etc.jar -d -Dlog4j.configuration=file:/etc/myapp/log4j.xml "$0" $#; exit $?
import org.springframework.class.from.jar
//other groovy code
println 'Hello'
All the magic happens in the first two lines. The first line tells us that this is a bash script. bash starts running and sees the first line. In bash # is for comments and // is collapsed to / which is the root directory. So bash will run /usr/bin/env groovy -cp extra.jar:spring.jar:etc.jar -d -Dlog4j.configuration=file:/etc/myapp/log4j.xml "$0" $# which starts groovy with all our desired arguments. The "$0" is the path to our script, and $# are the arguments. Now groovy runs and it ignores the first two lines and sees our groovy script and then exits back to bash. bash then exits (exit $?1) with status code from groovy.

If you really have to you can also load a JAR at runtime with:
this.getClass().classLoader.rootLoader.addURL(new File("file.jar").toURL())

My Favorite way to do this is with Groovy Grapes. These access the Maven Central Repository, download the referenced jar, and then put it on the classpath. Then you can use the library like any other library. The syntax is really simple:
#Grab(group='com.google.collections', module='google-collections', version='1.0')
You can read more details here. One major advantage here is that you don't need to distribute your dependencies when you distribute your script. The only drawback to this method is that the Jar has to be in the Maven repository.

You can add the jars to $HOME/.groovy/lib

You can also try out Groovy Grape. It lets you use annotations to modify the classpath. Its experimental right now, but pretty cool. See docs.groovy-lang.org/.../grape

The same as you would in Java.
This is an example of running a MySQL status monitoring script. mysql.jar contains the MySQL connector that I call from script status.groovy.
groovy -cp mysql.jar status.groovy ct1

Below is a combination of Patrick's solution, Maarteen Boekhold's solution, and foozbar's comment that works with both Linux and Cygwin:
#!/bin/bash
// 2>/dev/null; SCRIPT_DIR="$( cd "$( dirname "$0" )" && pwd )"
// 2>/dev/null; OPTS="-cp $SCRIPT_DIR/lib/extra.jar:$SCRIPT_DIR/lib/spring.jar"
// 2>/dev/null; OPTS="$OPTS -d"
// 2>/dev/null; OPTS="$OPTS -Dlog4j.configuration=file:/etc/myapp/log4j.xml"
// 2>/dev/null; exec groovy $OPTS "$0" "$#"; exit $?
import org.springframework.class.from.jar
//other groovy code
println 'Hello'
How it works:
// is a valid groovy comment, so all of the bash commands are ignored by Groovy.
// will return an error, but the error output is redirected to /dev/null and is therefore not displayed.
bash executes commands following a semicolon even though the previous command failed.
exec replaces the current program in the current process without forking a new process. Thus, groovy runs within the original script process (ps shows the process as the script rather than the groovy executable)
The exit $? statement following exec groovy prevents bash from trying to interpret the rest of the script as a bash script, and also preserves the return code from the groovy script.
The above bash trick is more convenient in some cases than the RootLoader trick because you can use regular import statements within the script. Using the RootLoader trick forces you to load all of the classes using reflection. This is fine in some situations (such as when you need to load a JDBC driver), but inconvenient in others.
If you know your script will never be executed on Cygwin, then using Patrick's or Maarteen's solution will likely result in slightly better performance because they avoid the overhead of generating and throwing away an error.

Adding to #Patrick his answer, which helped me a lot, I recently discovered another trick.
If you add lots of jars to the classpath all on one line, things can become quite unreadable. But you can do the following!
#!/bin/bash
//bin/true && OPTS="-cp blah.jar -Dmyopt=value"
//bin/true && OPTS="$OPTS -Dmoreopts=value2"
//usr/bin/env groovy $OPTS "$0" $#; exit $?
println "inside my groovy script"
Let your imagination run wild on how complex a command line you can break down this way into manageable pieces
Maarten

If you want to use it right away before the import declarations it is possible like this :) :
// printEmployees.groovy
this.class.classLoader.rootLoader.addURL(
new URL("file:///C:/app/Dustin/product/11.1.0/db_1/jdbc/lib/ojdbc6.jar"))
import groovy.sql.Sql
sql = Sql.newInstance("jdbc:oracle:thin:#localhost:1521:orcl", "hr", "hr",
"oracle.jdbc.pool.OracleDataSource")
sql.eachRow("SELECT employee_id, last_name, first_name FROM employees")
{
println "The employee's name is ${it.first_name} ${it.last_name}."
}
Taken from this javaworld.com article.