CODEWITHSUNDEEP
javajaryaml
dev
dev

Reputation: 1507

Java Class Read a Yaml Inside Jar

I am trying to read a file inside my jar from another class inside the jar. However I am continually getting the same error: Caught: class java.io.FileNotFoundException while attempting to read metrics: metrics.yml

At first I had my code do something like this, assuming it was from the path of the class:

String yamlPath = ".." + File.separator + ".." + File.separator + ".." + File.separator + ".." + File.separator + "myYaml.yml";

InputStream in = new FileInputStream(new File(yamlPath));
InputStreamReader isr = new InputStreamReader(in);
BufferedReader input = new BufferedReader(isr);
yamlObj = (HashMap) javaYAML.load(input);

I also did this assuming it might take the path from the base of the jar:

String yamlPath = "myYaml.yml";

InputStream in = new FileInputStream(new File(yamlPath));
InputStreamReader isr = new InputStreamReader(in);
BufferedReader input = new BufferedReader(isr);
yamlObj = (HashMap) javaYAML.load(input);

I then noticed this thread How to read a file from jar in Java? and found out that I needed a "/" before my path. I tried both methods above using the slash as well.

String yamlPath = File.seperator + ".." + File.separator + ".." + File.separator + ".." + File.separator + ".." + File.separator + "myYaml.yml";

OR

String yamlPath = File.seperator + "myYaml.yml";

I am completely lost on what to do about this error now. Is there something I am not getting about the jar structure? Why can my file not be found. Thanks in advance for any help / information.

Sorry, I forgot to mention where it is in the JAR: The class is in the following path: com/a/b/c/myclass.class The yaml is in the following path: myYaml.yml

Upvotes: 1

Views: 3677

Answers (3)

sendon1982
sendon1982

Reputation: 11234

Somehow for me, i have to add '/' in front of the file name.

InputStream resourceAsStream = Xx.class.getResourceAsStream("/test.yml");

From java doc:

Before delegation, an absolute resource name is constructed from the given resource name using this algorithm:

If the name begins with a '/' ('\u002f'), then the absolute name of the resource is the portion of the name following the '/'. Otherwise, the absolute name is of the following form: modified_package_name/name Where the modified_package_name is the package name of this object with '/' substituted for '.' ('\u002e').

And after i checked more detailed information, my solution is same as

InputStream in = Xx.class.getClassLoader().getResourceAsStream("test.yml");

Upvotes: 0

Jigar Joshi
Jigar Joshi

Reputation: 240898

File inside Jar is no longer a File, Change inputStream creation with this

InputStream in = YourClass.class.getResourceAsStream("myYaml.yml");

Assuming your .yml file is at root of jar

Upvotes: 2

Leos Literak
Leos Literak

Reputation: 9474

InputStream is = this.getClass().getClassLoader().getResourceAsStream("file");

Upvotes: 0

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