sysLoader.getResource() problem in java

Question

I am having following lines of code.

sysLoader = (URLClassLoader)Thread.currentThread().getContextClassLoader();
url = sysLoader.getResource(\"tempFile.txt\"         


        

Answer 1

This is known by Sun/Oracle, their advice is to use URI objects which will remove the %20 characters:

Instead of doing this:

FileInputStream fis = new FileInputStream(url.getFile());

you can force any %-escaped characters to be decoded by first converting the URL to a URI, and then use the path component of the URI as the filename:

URI uri = new URI(url.toString());
FileInputStream fis = new FileInputStream(uri.getPath());

Answer 2

Use URLDecoder.decode() method to replace %20 characters by spaces.

String path = URLDecoder.decode(url.getPath(), "UTF-8");

Please also keep in mind that when resource is located in jar file you have to handle it different way. See it e.g. here: How to access resources in jar where it can be present in multiple jar

Answer 3

To get the URL of the file from string, when the path contains spaces, this is what worked for me:

File file = new File("/Users/work space/tempFile.txt");
URL url = file.toURI().toURL();

According to Javadocs, file.toURL() is deprecated:

This method does not automatically escape characters that are illegal in URLs. It is recommended that new code convert an abstract pathname into a URL by first converting it into a URI, via the toURI method, and then converting the URI into a URL via the URI.toURL method.

Hence used file.toURI().toURL().

For Java 7+, this is approach can be taken instead:

URL url = Paths.get("/Users/work space/tempFile.txt").toURI().toURL());

Note: If the path begins with a / it is considered absolute else taken as a relative path.