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Home/ Questions/Q 7941261
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T23:36:58+00:00 2026-06-03T23:36:58+00:00

I have a java swing database application which needs to be run on Windows

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I have a java swing database application which needs to be run on Windows and Linux. My database connection details are stored in a XML file and I load them.

This application can load this properties on Linux properly but it is not working on Windows.

How do I load files on multiple platforms properly using Java?


This is the code:

PropertyHandler propertyWriter = new PropertyHandler();

List keys = new ArrayList();
keys.add("ip");
keys.add("database");
Map localProps = propertyWriter.read(keys, "conf" + File.separatorChar + "properties.xml", true);//if false load from the local properties

//get properties from the xml in the internal package
List seKeys = new ArrayList();
seKeys.add("driver");
seKeys.add("username");
seKeys.add("password");

Map seProps = propertyWriter.read(seKeys, "conf" + File.separatorChar + "properties.xml", true);

String dsn = "jdbc:mysql://" + (String) localProps.get("ip") + ":3306/" + (String) localProps.get("database");
jDBCConnectionPool = new JDBCConnectionPool((String) seProps.get("driver"), dsn, (String) seProps.get("username"), (String) seProps.get("password"));

File reader method:

public Map read(List properties, String path, boolean isConfFromClassPath)
{
    Properties prop = new Properties();
    Map props = new HashMap();
    try {

        if (isConfFromClassPath) {
            InputStream in = this.getClass().getClassLoader().getResourceAsStream(path);
            prop.loadFromXML(in);

            for (Iterator i = properties.iterator(); i.hasNext();) {
                String key = (String) i.next();
                props.put(key, prop.getProperty(key));
            }
            in.close();

        } else {
            FileInputStream in = new FileInputStream(path);
            prop.loadFromXML(in);

            for (Iterator i = properties.iterator(); i.hasNext();) {
                String key = (String) i.next();
                props.put(key, prop.getProperty(key));
            }
            in.close();
        }

    } catch (Exception ex) {
        ex.printStackTrace();
    }
    return props;
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-03T23:37:00+00:00Added an answer on June 3, 2026 at 11:37 pm

    If the file is in a jar file and accessed by the classpath then you should always use /.

    The JavaDocs for the ClassLoader.getResource say that “The name of a resource is a ‘/’-separated path name that identifies the resource.”

    http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/ClassLoader.html#getResource(java.lang.String)

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