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Compiling In console  RSS feed

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below are the files I do have:

Now i am using :javac -classpath D:\Console\servlet-api.jar;D:\Console\log4j-1.2.8.jar D:\Console\
So i have
5.D:\Console\FileDownlaod.class calles for the FileUpload class for a cetain functionality,so iah to compile FileUpload.ajava first.
Now how to compile the
I used:
1. javac -classpath D:\Console\servlet-api.jar;D:\Console\log4j-1.2.8.jar;D:\Console\FileDownlaod.class D:\Console\
Errors I am getting:
1.error: error reading D:\Console\FileDownlaod.class; error in opening zip file
cannot symbol : class FileDownlaod
location: package

3. D:\Console\ cannot find symbol
symbol : variable FileDownlaod
location: class
filelist = new File(FileDownlaod.ServerPath);

I even tried to convert the FileUpliad.class to .jar file and use it like servlet-api and log4j.jars , but it shows the below error
javac -classpath D:\Console\servlet-api.jar;D:\Console\log4j-1.2.8.jar;D:\Console\D:\Console\FileDownlaod.jar D:\Console\

Please tell me the valid syntax.

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IntelliJ IDE Ruby
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Don't specify class files explicitly; specify the class directory only. Continue specifying jar files as you are now.
Posts: 9610
Android Google Web Toolkit Hibernate IntelliJ IDE Java Spring
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If you want to include all the jars in a folder in the classpath, then include <directory path>\* in the classpath.

Also this question seems quite similar to your other topic to me...
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