When I am trying to run tomcat using startup.bat
I get the following error,
The JAVA_HOME environment variable is not defined correctly
This environment variable is needed to run this program
NB: JAVA_HOME should point to a JDK not a JRE
But then I try C:>echo %java_home%
and I get the following result
C:Program FilesJavajdk1.6.0_25in
I have even tried setting JAVA_HOME
manually to system variable list, but this issue remains.
What can I do to solve it?
I am using Windows 7.
Update
After setting a new system variable named JAVA_HOME and setting its path to "C:Program FilesJavajdk1.6.0_25in"
, I tried the start up script again and this time I get a new error.
D:Workapache-tomcat-6.0.35in>startup.bat
FilesJavajdk1.6.0_25"" was unexpected at this time.
Any idea what this error means?
I even tried setting the path to "C:Program FilesJavajdk1.6.0_25"
(that is without bin) but same error occurs.
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