Java Properties, getting file path

Question

logpath = LoggerUtils.getProperties().getProperty("log.path");
System.out.println("logpath: " + logpath);

The above code returns:

logpath: C:UsersMauriceDesktopLogs

In the properties file is:

log.path    C:\Users\Maurice\Desktop\Logs

How do I retain the file separators? I want this to work on Linux as well and not just Windows.

Answer 1

Actually, you need to put this in the property file:

log.path    C:\\Users\\Maurice\\Desktop\\Logs

See this:

• http://docs.oracle.com/javase/6/docs/api/java/util/Properties.html

more precisely the load method:

• http://docs.oracle.com/javase/6/docs/api/java/util/Properties.html#load(java.io.Reader)

Scroll down a bit and you will see this among other things:

The method does not treat a backslash character, \, before a non-valid escape character as an error; the backslash is silently dropped. For example, in a Java string the sequence "\z" would cause a compile time error. In contrast, this method silently drops the backslash. Therefore, this method treats the two character sequence "\b" as equivalent to the single character 'b'.

Backslash \ is an escape character that is silently dropped otherwise.