FileReader throws exception because files path has wrong syntax?

Question

I put a file to be read into my resources folder

src
|_main
   |_resources
     |_graphqls
       |_test.graphqls

The following snippet reads the file

final String pathToSchemaFile = this.getClass().getClassLoader().getResource("graphqls/test.graphqls").getFile();
final File = new File(pathToSchemaFile);

this is the result I get when I evaluate the File object returned by .getFile() from the preceding snippet.

file:\C:\maven_repository\com\...\app.jar!\graphqls\test.graphqls

enter image description here

When running the following code new FileReader(file) this exception is being thrown

Method threw 'java.io.FileNotFoundException' exception.
file:\C:\maven_repository\com\...\app.jar!\graphqls\test.graphqls (The filename, directory name, or volume label syntax is incorrect)
java.io.FileNotFoundException: file:\C:\maven_repository\com\...\app.jar.jar!\graphqls\test.graphqls (The filename, directory name, or volume label syntax is incorrect)

Why do you need the file, it looks like you should just use getResourceAsStream. — matt
– matt, Commented May 9, 2018 at 8:37
The GraphQL Framework expects a file that represents the graphQL schema file. Streaming it content would be my second option but I'd like to just put the file reference — xetra11
– xetra11, Commented May 9, 2018 at 8:38
Your example shows a FileReader which is an InputStreamReader, can you show the api where you need an actual file? — matt
– matt, Commented May 9, 2018 at 8:41
It is this one here: github.com/graphql-java/graphql-java/blob/master/src/main/java/… see line 40 — xetra11
– xetra11, Commented May 9, 2018 at 8:45
The second constructor uses a Reader so you can use the input stream. — matt
– matt, Commented May 9, 2018 at 8:47

Robin van Breukelen · Accepted Answer · 2018-05-09 08:38:20Z

2

You're accessing a file that is actually inside a JAR file (like a ZIP).

If your jar is on the classpath:

InputStream is = YourClass.class.getResourceAsStream("1.txt");

If it is not on the classpath, then you can access it via:

URL url = new URL("jar:file:/absolute/location/of/yourJar.jar!/1.txt");
InputStream is = url.openStream();

answered May 9, 2018 at 8:38

Robin van Breukelen

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3 Comments

xetra11 Over a year ago

I know I am accessing a file that is inside a jar. But what is wrong with this approach? Why is FileReader crying like a little girl?

Robin van Breukelen Over a year ago

I'm not 100% sure, but as far as I know the File was used to represent something on the filesystem. And the file within the jar is not handled like any ordinary file on the file system.

xetra11 Over a year ago

Hm okay that might explain things if this is a true statement

matt · Accepted Answer · 2018-05-09 09:15:34Z

0

You can avoid the file by creating an InputStreamReader.

InputStreamReader reader = new InputStreamReader(
    this.getClass().getResourceAsStream("/graphqls/test.graphqls"), 
    StandardCharsets.UTF_8
);

It is advised to use YourClassName.class.getResourceAsStream() unless you have some specific reason to use the .getClass && .getClassLoader.

edited May 9, 2018 at 9:15

answered May 9, 2018 at 8:50

matt

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4 Comments

xetra11 Over a year ago

I actually have to use .getClassLoader() because it won't find the file otherwise.

matt Over a year ago

You're using the relative pathname. So if your class is in a package, then then when it looks it up, it will look relative to the package you're in.

xetra11 Over a year ago

To be honest I do not have a lot of experience using Files and the specific filepath conventions. Is the way I do it in any way problematic? If yes - what would you recommend me to do?

matt Over a year ago

@xetra11 It's good that you're using a resource instead of just using a file. The rest are detail, that are good practice but not necessary. here is a pretty concise but thorough article.

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