如何启动 J2EE

[zyfhfx]

我运行 J222 为什么总报错?
Microsoft Windows XP [版本 5.1.2600]
(C) 版权所有 1985-2001 Microsoft Corp.

C:\Documents and Settings\zyf>j2ee
C:\Program Files\j2sdk_nb\j2sdk1.4.2\bin\java -Xmx128m
'C:\Program' 不是内部或外部命令，也不是可运行的程序
或批处理文件。

就错在：
%JAVACMD% -Djava.security.policy==%J2EE_HOME%\lib\security\server.policy -Dcom.sun.enterprise.home=%J2EE_HOME% -classpath "%CPATH%" com.sun.enterprise.server.J2EEServer %1 %2
JAVACMD是什么路径？？

j2ee.bat 源文件

@echo off
rem
rem Set JAVA_HOME and J2EE_HOME before running this script.
rem
rem set JAVA_HOME to the path where you have Java 2 (JDK1.2) installed.
rem
rem set J2EE_HOME to the path where you have installed this package (EJB server).
rem

if not "%J2EE_HOME%" == "" goto CONT0
echo ERROR: Set J2EE_HOME before running this script.
goto END
:CONT0

if EXIST "%J2EE_HOME%\bin\setenv.bat" goto CONT1
echo ERROR: Set J2EE_HOME to the path of a valid j2sdkee.
goto END
:CONT1

call %J2EE_HOME%\bin\setenv.bat

if not "%JAVA_HOME%" == "" goto CONT2
echo ERROR: Set JAVA_HOME before running this script.
goto END
:CONT2
rem echo %JAVA_HOME%
if EXIST "%JAVA_HOME%\BIN\java.exe" goto CONT3
echo ERROR: Set JAVA_HOME to the path of a valid jdk.
goto END
:CONT3

rem @echo on
echo %javacmd%
%JAVACMD% -Djava.security.policy==%J2EE_HOME%\lib\security\server.policy -Dcom.sun.enterprise.home=%J2EE_HOME% -classpath "%CPATH%" com.sun.enterprise.server.J2EEServer %1 %2

:END

[bobZ]

J2EE_HOME设了没有？

[zyfhfx]

设了，前面都没有问题，只是在%JAVACMD% -Djava.security.policy==%J2EE_HOME%\lib\security\server.policy -Dcom.sun.enterprise.home=%J2EE_HOME% -classpath "%CPATH%" com.sun.enterprise.server.J2EEServer %1 %2
出错了