Reputation: 904
In/src/main/java/com/application/myGoogleAppEngine/controller/AddressController.java
, I want to use this instructions for using the beans by after :
ClassPathXmlApplicationContext ctx = new ClassPathXmlApplicationContext("/WEB-INF/spring-servlet.xml");
My spring-servlet.xml
file sits in /src/main/webapp/WEB-INF/
directory.
When I execute my spring application, I've the following error :
java.io.FileNotFoundException: class path resource [WEB-INF/spring-servlet.xml] cannot be opened because it does not exist
In my memories, I know that we can use a function to obtain the current directory of the project but I don't remember.
Otherwise, do you have any solutions ?
Thank you
Upvotes: 2
Views: 8629
Reputation: 1
you can check the web.xml location. If this is not correct it will throw an error.
you can check in properties of the project: please see in the attached image
[1]: https://i.sstatic.net/dLwsi.png
Upvotes: 0
Reputation: 545
use this way
new ClassPathXmlApplicationContext( "classpath*:**/servlet.xml")
This will search whole WEB-INF folder for given name
Upvotes: 1
Reputation: 279890
The WEB-INF
directory is not typically put on the classpath. Therefore you won't be able to get anything within it as a resource, ie. what ClassPathXmlApplicationContext
is trying to do.
Put your spring-servlet.xml
in a directory that is put on the classpath, ie. WEB-INF/classes
.
Upvotes: 6