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
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.
After setting a new system variable named JAVA_HOME and setting its path to
"C:\Program Files\Java\jdk1.6.0_25\bin\", I tried the start up script again and this time I get a new error.
D:\Work\apache-tomcat-6.0.35\bin>startup.bat Files\Java\jdk1.6.0_25"" was unexpected at this time.
Any idea what this error means?
I even tried setting the path to
"C:\Program Files\Java\jdk1.6.0_25\"(that is without bin) but same error occurs.
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~ Asked on 2012-01-13 23:37:48
Try installing java somewhere else - in a directory without spaces. Set again the
JAVA_HOME variable and try again. I remember Tomcat had some problems on Window XP with spaces if any variables it was using while starting contained spaces. Maybe it's similar with Windows 7.
I remember I had to change some lines in Tomcat java classes which were handling Tomcat startup.
@Edit: Luciano beat me to noticing it but you should also remove
@Edit: I also remember that another fix (didn't test it myself, though) was to set
JAVA_HOME to the shorthand version e.g.
~ Answered on 2012-01-13 23:53:32
I think that your JAVA_HOME should point to
That is, without the bin folder.
That new error appears to me if I set the JAVA_HOME with the quotes, like you did. Are you using quotation marks? If so, remove them.
~ Answered on 2012-01-13 23:55:37