Hi,
below are the files I do have:
1.D:\Console\FileDownlaod.java
2.D:\Console\XmlGenerator.java
3.D:\Console\servlet-api.jar
4.D:\Console\log4j-1.2.8.jar
Now i am using :javac -classpath D:\Console\servlet-api.jar;D:\Console\log4j-1.2.8.jar D:\Console\FileDownlaod.java
So i have
5.D:\Console\FileDownlaod.class
XmlGenerator.java calles for the FileUpload class for a cetain functionality,so iah to compile FileUpload.ajava first.
Now how to compile the XmlGenerator.java:
I used:
1. javac -classpath D:\Console\servlet-api.jar;D:\Console\log4j-1.2.8.jar;D:\Console\FileDownlaod.class D:\Console\XmlGenerator.java
Errors I am getting:
1.error: error reading D:\Console\FileDownlaod.class; error in opening zip file
2.D:\Console\XmlGenerator.java:10:
cannot symbol : class FileDownlaod
location: package com.photo.servlet
import com.photo.servlet.FileDownlaod;
3. D:\Console\XmlGenerator.java:53: cannot find symbol
symbol : variable FileDownlaod
location: class com.photo.servlet.XmlGenerator
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\XmlGenerator.java.
Please tell me the valid syntax.