2012-07-09 31 views
2

我有一个名为X项目,该项目具有wireup.xml奠定如下:Spring:如何从jar classpath加载@ContextConfiguration?我得到FileNotFoundException异常

X/ 
Module/ 
     src/ 
      main/ 
       resources/ 
         com.here/ 
           wireup.xml 

我进口的project XModuleproject Y作为

<dependency> 
     <groupId>com.org.X</groupId> 
     <artifactId>Module</artifactId> 
     <version>master-SNAPSHOT</version> 
    </dependency> 

现在,在测试中,我想豆是在wireup.xml,所以我做了以下操作:

@RunWith(SpringJUnit4ClassRunner.class) 
@ContextConfiguration(locations = {"classpath:com/org/X/Module/src/main/resources/com/here/wireup.xml"}) 
public class MongoSaverTest extends Case { 
    @Autowired 
    private SomeBeanInWireup variable; 
} 

但我得到上运行的测试说

Caused by: java.io.FileNotFoundException: class path resource [com/org/X/Module/src/main/resources/com/here/wireup.xml] cannot be opened because it does not exist 

如何解决这个问题?我如何知道正确的路径?

回答

3

这应该只是@ContextConfiguration(locations = {"classpath:com/here/wireup.xml"})

0

您的资源的路径是classpath:com/here/wireup.xml

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