如何从资源文件夹中获取文件。 Spring框架

Question

I'm trying to unmarshal my xml file:

public Object convertFromXMLToObject(String xmlfile) throws IOException {
    FileInputStream is = null;
    File file = new File(String.valueOf(this.getClass().getResource("xmlToParse/companies.xml")));
    try {
        is = new FileInputStream(file);
        return getUnmarshaller().unmarshal(new StreamSource(is));
    } finally {
        if (is != null) {
            is.close();
        }
    }
}

But I get this errors: java.io.FileNotFoundException: null (No such file or directory)

Here is my structure:

Why I can't get files from resources folder? Thanks.

Update.

After refactoring,

URL url = this.getClass().getResource("/xmlToParse/companies.xml"); File file = new File(url.getPath());

I can see an error more clearly:

java.io.FileNotFoundException: /content/ROOT.war/WEB-INF/classes/xmlToParse/companies.xml (No such file or directory)

It tries to find WEB-INF/classes/ I have added folder there, but still get this error :(

Answer 1

I had the same problem trying to load some XML files into my test classes. If you use Spring, as one can suggest from your question, the easiest way is to use org.springframework.core.io.Resource - the one Raphael Rot h already mentioned.

The code is really straight forward. Just declare a field of the type org.springframework.core.io.Resource and annotate it with org.springframework.beans.factory.annotation.Value - like that:

@Value(value = "classpath:xmlToParse/companies.xml")
private Resource companiesXml;

To obtain the needed InputStream, just call

companiesXml.getInputStream()

and you should be okay :)

But forgive me, I have to ask one thing: Why do you want to implement a XML parser with the help of Spring? There are plenty build in :) Eg for web services there are very good solutions that marshall your XMLs into Java Objects and back...

Answer 2

ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());

Answer 3

you are suppose to give an absolute path (so add a loading ´/´, where resource-folder is the root-folder):

public Object convertFromXMLToObject(String xmlfile) throws IOException {
    FileInputStream is = null;
    File file = new File(String.valueOf(this.getClass().getResource("/xmlToParse/companies.xml")));
    try {
        is = new FileInputStream(file);
        return getUnmarshaller().unmarshal(new StreamSource(is));
    } finally {
        if (is != null) {
            is.close();
        }
    }
}