CODEWITHSUNDEEP
javafile-io
Habib Ksentini
Habib Ksentini

Reputation: 402

FileNotFoundException with getResources

I used Class.getResource (String) to retrieve the url of a file, and it works very well, but when I try to 'insatncié a FileReader with the url returned, an exception:java.io.FileNotFoundException is triggered

URL bpmnResourceUrl = ConvertXmlToJson.class.getClassLoader().getResource("file.txt");
Reader reader = new FileReader(bpmnResourceUrl.toString());

Upvotes: 0

Views: 3390

Answers (2)

Roberto
Roberto

Reputation: 9080

Don't open it as a file, use it as InputStream, in your case, if you want a Reader to get the data then you can use:

InputStream is = ConvertXmlToJson.class.getClassLoader().getResourceAsStream("file.txt");

Reader reader = new InputStreamReader(is);

When you load a resource from Classpath, such resource can be located inside a jar file, so It's not accesible like a regular file in filesystem, but you can open a Stream to read it, as the code shows.

Upvotes: 0

Harmlezz
Harmlezz

Reputation: 8068

A Resource in Java is not a File. If the Resource is inside a JAR, for example, you can't access it like a File. You have to explode the JAR first. You may try:

Class.getResourceAsStream()

to read the content. Here is a short example:

public class Example {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(Example.class.getResourceAsStream("/META-INF/MANIFEST.MF")));
        String line;
        do {
            line = br.readLine();
            if (line != null) System.out.println(line);
        } while (line != null);
    }
}

Upvotes: 4

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