Question

I have a bean that is created by Spring. The actual class resides in a different JAR than Spring. This bean is passed a path as a constructor argument. However, I am having difficulty retrieving a handle to the file. The file is in WEB-INF/classes/. I’ve tried relative pathing based on WEB-INF, but obviously that didn’t work.

XML:

 <bean id="configurationManager" class="package.ConfigurationManager" 
      scope="singleton">           
      <property name="configurationMapping">
            <bean class="package.PropertiesFileConfigurationMapper">
                <constructor-arg type="java.lang.String">
                    <value>/path/to/file</value>
                </constructor-arg>
            </bean> 
      </property>                     
</bean> 

Bean:

public class ConfigurationMapper {

    public ConfigurationMapper(String resource) {
            _map = new HashMap<String, String>();
        String property = null;
        BufferedReader reader = null;
        try {
            FileReader file = new FileReader(resourcePath);
            reader = new BufferedReader(file);
            while ((property = reader.readLine()) != null) {
                if (property.matches("(.+)=(.+)")) {
                    String[] temp = property.split("(.+)=(.+)");
                    _map.put(temp[0], temp[1]);
                }
            }
        } catch (Exception ex){
            ex.printStackTrace();
        } finally {
            if (reader != null)
                reader.close();
        }
    }

    //other methods to manipulate settings
}

How can I get the proper path to the rm.properties file and pass it to the bean at runtime?

Edit: Added constructor code.

Edit: I got it. I changed the constructor argument to no longer take a path. It now takes a Resource, so Spring has found the resource that I wanted loaded.

Answer 1

java.io.File and FileReader only work for actual files. A resource packed inside a JAR file isn’t itself a file.

The easiest way to load it is as a classpath resource:

Replace this:

FileReader file = new FileReader(resourcePath);
reader = new BufferedReader(file);

with something like this:

InputStream inputStream = getClass().getResourceAsStream();
reader = new BufferedReader(new InputStreamReader(inputStream));

Better yet, use Spring’s Resource abstraction, by declaring the constructor parameter as org.springframework.core.io.Resource:

public ConfigurationMapper(Resource resource) {
   ...
   InputStream inputStream = resource.getInputStream();
   reader = new BufferedReader(new InputStreamReader(inputStream));

When you then supply the path:

<constructor-arg value="classpath:/path/to/file"/>

Spring will automatically create a ClasspathResource for that path (using a classpath) , and pass it to your constructor.