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Classpath in Command line

 
Nelo Angelo
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Hello everyone,

I was going through Head First Servlets book and found myself stuck in the BeerSelect.java example.

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PROBLEM
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To compile the servlet following code is being used:
% javac -classpath /Users/bert/Applications2/tomcat/common/lib/servlet-api.jar:classes:. -d classes src/com/example/web/BeerSelect.java

I am using Windows XP and my Tomcat directory's path is this:
C:/Program Files/Apache Software Foundation/Tomcat 5.5/common/lib;


So I am using this code in the command prompt to compile, but it generates an error:
C:\beerV1>javac -classpath /Program Files/Apache Software Foundation/Tomcat 5.5/common/lib//servlet-api.jar:classes:. -d classes src/com/example/web/BeerSelect.java

Error:
javac: invalid flag: Files/Apache
Usage: javac <options> <source files>
 
arvind kushwaha
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set the classpath for your program. For eg.

c:> set classpath=.;Here provide the path of your servlet-api-jar; %z%;
 
Nelo Angelo
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Thanks for the replies, but I want to know how the above given code can be modified to be used in Windows. The compound statement that sets the classpath as well as compiles the code.
 
Campbell Ritchie
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You would appear to have misread the previous post, which uses a Windows® idiom.
Yoiur question is not at all clear. Do you mean this
javac -cp .;c:\MyFolder\MyOtherFolder\MyJar.jar MyClass.java
 
Shankh Pani Parimal
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shankh
 
Nelo Angelo
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Thanks for the reply guys, the problem is resolved.

I used the following code:

C:\beerV1>javac -cp C:\Tomcat_6.0\lib\servlet-api.jar; -d classes src/com/example/web/BeerSelect.java
 
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