Jaxb system path

Question

Hi,

I just installed java Web Service Developer Pack 2.0 to use JAXB for xml operations. The tutorial says me add "C:\Sun\jwsdp-2.0\jaxb\bin" to the system path.I select Computer->Properties->Advanced System Settings. I come to the Advanced tab I open the window Environment variables. and i selected PATH from user variables,selected edit and append "C:\Sun\jwsdp-2.0\jaxb\bin" to here. But when i write xjc,it is still not recognized. What should i do more ?

Answer 1

I just checked the release notes for JWSDP 2.0, and they're copy right 2005.

• http://java.sun.com/webservices/docs/2.0/ReleaseNotes.html

If you're just looking to run XJC it is also included as part of Java SE 6. You can run it as:

C:\Program Files\Java\jdk1.6.0_20\bin\xjc

As for JAXB implementations you can choose from:

• EclipseLink JAXB (MOXY)
• Metro JAXB (the reference implementation, included in JAVA SE 6.