The file or folder 'java.exe. that this shortcut refers to cannot be found

[Niyas Ahmed Sheikh]

I have installed JDK1.5 in C:\JDK1.5.
I have installed Tomcat 4.1 in C:\Tomcat.

Path
-----
SET PATH="C:\jdk1.5\bin;"%PATH%
SET JAVA_HOME="C:\jdk1.5;"%PATH%
SET CATALINA_HOME="C:\Tomcat;"
SET CLASSPATH=".;C:\Tomcat\common\lib\servlet.jar;"%PATH%


While I tried to start the Tomcat thru the Start>>Programs>>Apache Tomcat 4.1>>start tomcat, I am getting this error message.

  
"The file or folder 'java.exe. that this shortcut refers to cannot be found"

While I tried to start te tomcat thru' the MS-Dos
C:\cd Tomcat
C:\Tomcat>cd bin
C:\Tomcat\bin>startup

I am getting the following error message:

  
"The CATALINA_HOME environment variable is not defined correctly
This environment variable is needed to run this program"

What is the problem? Whether any problem in the path setting.

[Bear Bibeault]

Moved to the Tomcat forum.

Why on earth are you appending PATH to JAVA_HOME?

[William Brogden]

You get that message when this test fails:

if exist "%CATALINA_HOME%\bin\catalina.bat" goto okHome  

but if CATALINA_HOME is not defined, startup.bat makes a guess of ".."
which should work in your case. So it is seeing CATALINA_HOME but creating an error. I suspect it is due to the trailing ";" in your SET CATALINA_HOME statement. This variable is not dissassembled using ";"
The batch file will NOT discard that so you get an unusable path.
Bill