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JAXB generate java class from xsd

开发者 https://www.devze.com 2022-12-17 09:36 出处:网络
JAXB 1.5 installed under C:\\Sun\\jwsdp-1.5 J2SE 1.4.2 installed under C:\\j2sdk1.4.2_08 copied sample.xsd file to C:\\Sun\\jwsdp-1.5\\jaxb\\bin
  1. JAXB 1.5 installed under C:\Sun\jwsdp-1.5
  2. J2SE 1.4.2 installed under C:\j2sdk1.4.2_08
  3. copied sample.xsd file to C:\Sun\jwsdp-1.5\jaxb\bin
  4. went to 开发者_Go百科C:\Sun\jwsdp-1.5\jaxb\bin and ran xjc.bat -p com.package sample.xsd
  5. got error message: Unrecognized option: -p Could not create the Java virtual machine.

Please help me out, thanks a lot


This page seems to indicate tha xjc.bat needs Java 1.5+ :

http://forums.sun.com/thread.jspa?threadID=5359378


The last JAXB version that is compatible with java 1.4.2 is JAXB 1.0.6. I've never heard of a version 1.5... Where did you get it?

Edit

This error message is not generated by jaxb but by the JVM. Looks like, the jvm thinks, the '-p' parameter is a jvm parameter. Here's another page where the same error message was generated at a simple java --version call.

And another explanation for the error.


I faced the similar problem and I resolved it using the following approach. I specified Jdk1.5 as the JDK while installing jwsdp1.5. But the system had jdk1.4 also installed. So I created the following batch file:

set JAVA_HOME=D:\apps\BEA\Weblogic\jdk150_04
set ANT_HOME=H:\Sun\jwsdp-1.5\apache-ant
set JWSDP_HOME=H:\Sun\jwsdp-1.5
set PATH=%JAVA_HOME%\bin;%PATH%;
%JWSDP_HOME%\jaxb\bin\xjc -p package -dtd sample.dtd

This resolved the error. Alternatively we can remove the Jdk1.4 path in the PATH environment variable, in that case, we need to write the batch file.

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