JavaExec workingDir doesn't seem to actually change the working directory. What am I doing wrong?

I’m running an executable jar file using the JavaExec task. The code essentially looks in the working directory for some config files and I have no way of changing this code to use the classpath to find these config files. However, setting the workingDir property in the javaexec config block doesn’t seem to affect the working dir (as determined by runtime errors and --debug output).

Can you please help me understand what I’m doing wrong here?

I’ve included a simplified example of the command I’m invoking, which seems to exhibit the same behavior, based on the debug output.

task javaVersion() << {
 def workingDir = file('c:/')
 println workingDir.absolutePath
     javaexec {
  main '-version'
  workingDir = workingDir

Relevant debug output: bvanoss@STS123 /cygdrive/c/code/workspaces/tetra-workspace/gradle-example $ gradle :javaVersion --debug

11:43:42.590 [INFO] [org.gradle.process.internal.DefaultExecHandle] Starting process ‘command ‘C:\Program Files\Java\jdk1.8.0_25\bin\java.exe’’. Working directory: C:\code\workspaces\tetra-workspace\gradle-example Command: C:\Program Files\Java\jdk1.8.0_25\bin\java.exe -Dfile.encoding=windows-1252 -Duser.language=en -Duser.variant -version

------------------------------------------------------------ Gradle 2.1 ------------------------------------------------------------

Build time:

2014-09-08 10:40:39 UTC Build number: none Revision:



2.3.6 Ant:

Apache Ant™ version 1.9.3 compiled on December 23 2013 JVM:

1.8.0_25 (Oracle Corporation 25.25-b02) OS:

Windows 7 6.1 amd64

Name your variable something other than ‘workingDir’. Within the ‘javaexec {…}’ closure, the symbol ‘workingDir’ delegates first to ‘JavaExecSpec.getWorkingDir()’, which by default is the project directory.

That was exactly the problem. Thanks!