How to get files from resources folder. Spring Framework

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 :(

try `getResource("classpath:xmlToParse/companies.xml")` – sidgate Apr 04 '16 at 16:10 — sidgate, Apr 04 '16 at 16:10
Wont you need another "/" before xmlToParse – LearningPhase Apr 04 '16 at 16:11 — LearningPhase, Apr 04 '16 at 16:11
Code was update. Check it out, please. – Tom Wally Apr 04 '16 at 16:33 — Tom Wally, Apr 04 '16 at 16:33
any resolve? I face the same problem – Mateusz Niedbal Nov 29 '22 at 05:39 — Mateusz Niedbal, Nov 29 '22 at 05:39

score 47 · Answer 1 · edited May 23 '17 at 10:31

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 Roth 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 :) E.g. for web services there are very good solutions that marshall your XMLs into Java Objects and back...

I tried this, works locally but cannot be resolve when deploy to dev. https://stackoverflow.com/q/57466273/5777189 — BabyishTank, Mar 11 '22 at 03:37

score 13 · Answer 2 · answered Apr 07 '17 at 02:58

13

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

answered Apr 07 '17 at 02:58

bearwithbeard

321
3
10

2

This break as soon as you deploy this spring project into a jar. See https://stackoverflow.com/q/20389255/5777189 – BabyishTank Jun 29 '22 at 01:31

score -2 · Answer 3 · answered Apr 04 '16 at 16:15

-2

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();
        }
    }
}

answered Apr 04 '16 at 16:15

Raphael Roth

26,751
15
88
145

Check whether the resources-folder is included in the deployment-assembly – Raphael Roth Apr 04 '16 at 17:15
I think, I have to bind this folder... Is there any easy way to do it? – Tom Wally Apr 04 '16 at 17:26

How to get files from resources folder. Spring Framework

3 Answers3

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