在Java中创建独立程序包并从配置文件中进行一些设置

Question

Anyone plz let us know what to do when we have some configuration file which is basically xml.I want to for example give the path to save the image(for my java program) in a folder from some config file (xml in my case).In that case where should the config file be kept.Rt now every thing is converted to jar file when i create a java standalone package.But i want to give some setting from xml file.What to do in that case.How is it possible.This article only provides to create a single jar file for java project but talks nothing about the configuration settings that u can provide from some external source.

Regards Sagar

Answer 1

I'm not sure I fully understand your question, but if it is where to put the XML file with configuration information, you can place your xml file in the same directory as your jar file, and then pass the XML file name and path into the Jar on the command line when calling the Jar. If you're running it in Windows, this is often done using a shortcut. Then you can get the full path string for the Jar from the main method's String[] arg array that accepts the command parameters.

Answer 2

Sagar,

The fact your java program is a standalone package (.jar file) has no bearing on where your configuration file is stored. Your java package is a program and that program can read any file from the file system that it so desires; it does not have to be part of the code inside the IDE ie you don't have to write it when you write the program. What you do need is some way, when you start the program, to find and read said configuration file.

Depending on how you expect the program to be configured, you might put that file in a number of locations. For example, /etc/yourimageprogram/config.xml or c:\\program files\\yourimageprogram\\config.xml or perhaps c:\\users\\Sagar\\Application Settings\\yourimageprogram\\config.xml. Which you choose of those options really depends on what the use case is and that I can't help with.

However, there are some main points to reading any file:

• Does it exist?
• Are we allowed to open it for reading?
• Are we allowed to open it for writing? Might want to know if we want to update the config?

In Java, typically, you would test this with:

File configfile = new File("C:\test.xml");
if ( configfile.exists() && configfile.canRead() )
{
    // read the file
}
else
{
    // decide what to do if no config exists.
    // might be first run of app.
}

The next stage is to parse the file. There are a number of parsers available for XML including sax and org.w3c.dom. What you need to do is to use these to extract the information you require and store that in a class. Probably a singleton class as you're unlikely to have multiple configuration instances per instance of the program.

I suggest you read about XML Parsers and File Handling under Java . Also look at the File object. See all your options for file io in java. These should give you some indication of how to proceed.