Question

I am trying to read a config file based on the code here:

http://www.opencodez.com/java/read-config-file-in-java.htm

So I found that if the config.cfg is in same directory as of where I am running the code, then everything is fine but if the config is at different directory

example: /path/to/config.cfg

I get this error:

java.lang.NullPointerException
at java.util.Properties$LineReader.readLine(Properties.java:418)
at java.util.Properties.load0(Properties.java:337)
at java.util.Properties.load(Properties.java:325)
at packagename.conf.Config.<init>(Config.java:14)
at packagename.conf.Config.main(Config.java:30)

My guess is it is not able to find the file.
But how do I modify the above code to read config file from a different folder?
Thanks

Edit: Code from the link:

import java.util.*;
import java.util.Properties;

public class Config 
{
   Properties configFile;
   public Config()
   {
    configFile = new java.util.Properties();
    try {           
      configFile.load(this.getClass().getClassLoader().
      getResourceAsStream("myapp/config.cfg"));         
    }catch(Exception eta){
        eta.printStackTrace();
    }
   }

   public String getProperty(String key)
   {
    String value = this.configFile.getProperty(key);        
    return value;
   }
}

Answer 1

The code that you posted expects the configuration-file to be on the classpath (that is, in the same sorts of places that Java looks for your .class files). So, you can either include the directory containing the configuration-file in the classpath:

java   -classpath .:/path/to   packagename.conf.Config

Or else you can modify the code to expect the configuration-file to be a regular filesystem file:

final InputStream cfg = new FileInputStream("/path/to/config.cfg");
try
    { configFile.load(cfg); }
finally
    { cfg.close(); }