The javafx.properties file which contains the JavaFX version information is still in JAVA_HOME/jre/lib. In version 1.8.0, the jfxrt.jar file is in JAVA_HOME/jre/lib/ext (which automatically makes it part of the classpath). Library/Java/JavaVirtualMachines/jdk1.7.0_45.jdk =2.2.45īeginning with JRE 1.8.0, JavaFX was included with the JRE and was placed on the classpath so it's effectively a full part of the core libraries. Just inspecting the JDK 1.7.0 versions I have installed on my machine, the mapping from JDK version to JavaFX version is: /Library/Java/JavaVirtualMachines/jdk1.7.0_06.jdk =2.2.0 The javafx.properties file which contains the JavaFX version information is in the same directory, though in theory at least the JRE version will determine the JavaFX version (since they were shipped together). So for JRE 1.7.0, update 6 and later, the jfxrt.jar file is included in JAVA_HOME/jre/lib. The relationship between JDK/JRE version and JavaFX version is as follows.īeginning with JRE 1.7.0 update 6, JavaFX was included with the JRE, but was not on the classpath. As dfeuer commented, it would help to know your JRE version. ![]() However, if the System property you get at runtime is blank, then I suspect this file doesn't exist. Or whatever the Windows equivalent is, if you're running on Windows. It looks like the System property that you can read at runtime is kept in jre/lib/javafx.properties, so you can do something like cat $JAVA_HOME/jre/lib/javafx.properties JavaFX is just a set of library classes it's not really an executable in the same sense. The difference is that java is a machine executable that launches the JVM. ![]() You can't really do this the same way you do with the Java runtime.
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