sysLoader.getResource() problem in java

Question

I am having following lines of code.

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

It is giving an weird problem. If I run this from a path where there is no space in the path (Folder names) then it is running fine. But if the path contains any spaces (line "c:\New Foler...") then it is not working.

How to solve this?

EDIT: In more detail - I inspected the sysloader object.

sysloader -> UCP -> path

Is having a path with character %20 instead of space

And therefore all the URLs are null.

How to resolve this?

Define "not working". Does it return `null`? Or does the returned `URL` not work in some context? — Joachim Sauer, Jul 16 '10 at 09:36

score 50 · Accepted Answer · edited Aug 20 '16 at 08:24

50

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());

edited Aug 20 '16 at 08:24

Matthias Braun

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answered Nov 20 '12 at 09:57

snoopygee

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Here is the corresponding but report marked as "Won't fix": http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4466485 . This bug/feature still exists in Java 1.7 . – asmaier Nov 08 '13 at 16:53
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A shorter version without explicitly converting to string and back to URI would be: `URI uri = url.toURI();` `FileInputStream fis = new FileInputStream(uri.getPath());` – Thomas Dec 02 '14 at 11:03

score 12 · Answer 2 · edited May 23 '17 at 11:46

12

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

edited May 23 '17 at 11:46

Community

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answered Feb 19 '11 at 05:16

libor

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Once you get the URL from getResource that is used, while passing the path in the file, we can use the method URLDecoder#decode. Something like this: File file = new File(URLDecoder.decode(testFile.getFile(), "UTF-8")); Here, testFile is a URL object. I am adding a comment since even though I found the answer useful, I had to somehow think about how to use this. Thanks anyway. – Raj Kundalia May 07 '21 at 08:18

Nitin Labhishetty · Answer 3 · 2017-04-13T08:55:30.197

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.

sysLoader.getResource() problem in java

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