CODEWITHSUNDEEP
javaspringclasspath
MaVVamaldo
MaVVamaldo

Reputation: 2535

load a (xml) file from the classpath in a spring web app

I'd like to take a xml file from my classpath to unmarshal it and use it for testing purposes. My problem is to get it as an InputStream. I wrote these lines but I always get a null result.

InputStream is = getClass().getResourceAsStream("WebContent/WEB-INF/classes/testing/"+ COMPLETE_DOCUMENT + ".xml");

of course the path you see in the method argument is the one to my file. I tried several combinations:

WebContent/WEB-INF/classes/testing/
classpath:testing/
classpath*:testing/

but I always get the InputStream = null.

I even tried to switch to

 ClassLoader.getResourceAsStream(...)

but nothing happens. I suppose the path to the resource is somehow wrong, but I can't figure out where. From my servlet.xml I use some resource in the classpath configuring PropertyPlaceholderConfigurer or Jaxb2Marshaller just with the syntax 

"classpath:folder/file.xsd"

and it works perfectly. The folder I want to load my xml from is a sibling of the one in the example above. What am I missing?

EDIT : I try to follow the spring ClassPathResource helper class approach and I get a strange behaviour: as I said before I already have some resources loaded from the classpath by some spring beans at the startup. If I use the path to such resources in the code suggested by dardo in as follows:

ClassPathResource cpr = new ClassPathResource("xmlschemas/lrinode.xsd");
InputStream is = cpr.getInputStream();

I Still get a FileNotFound Exception! Of course "xmlschemas/lrinode.xsd" is a xsd I load at the startup successfully. It doesn't work even if I use the full path to the resource, starting from the root of the application.

I'm starting to think I'm missing something trivial.

Upvotes: 11

Views: 35674

Answers (3)

dardo
dardo

Reputation: 4970

Spring provides a helper class named ClassPathResource

So something like:

ClassPathResource cpr = new ClassPathResource("folder/file.xsd");
InputStream is = cpr.getInputStream();

Should work, hope this helps!

Link to API Doc: http://static.springsource.org/spring/docs/3.0.x/api/org/springframework/core/io/ClassPathResource.html

Sidenote

Also, if you're using it for testing purposes, might want to wire a bean mapped to the xsd.

Might want to look into a JAXB marshaller

http://static.springsource.org/spring-ws/site/reference/html/oxm.html#oxm-jaxb2-xsd

Upvotes: 20

Eugene Kuleshov
Eugene Kuleshov

Reputation: 31795

When you call Class.getResourceAsStream() with resource name without leading slash, it assumes you are requesting resource relative to a current package (i.e. package of the caller class). To make it an absolute resource path you need to add leading slash to resource name, e.g. in your case "/testing/"+ COMPLETE_DOCUMENT + ".xml"

Upvotes: 0

skaffman
skaffman

Reputation: 403601

You need a combination you didn't try:

getClass().getResourceAsStream("/testing/"+ COMPLETE_DOCUMENT + ".xml");

The WebContent/WEB-INF/classes directory should already be on the classpath.

The classpath: syntax only works if you're using Spring's ResourceLoader abstraction, which you aren't. Your usage of classpath:folder/file.xsd in your servlet.xml woprks because Spring's passing it through a ServletContextResourceLoader, which resolves classpath: automatically.

Upvotes: 3

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