运行可执行maven jar文件时Java加载资源失败：无法解析“file:/...”

Question

My issue is that when I build an executable jar with maven (using the shade plugin) and attempt to run the jar, I receive an error:

java.io.IOException: Unable to resolve "file:/C:/Users/name/Desktop/nameofjar-SNAPSHOT.jar!/path/to/file.gz

When I run the main program within eclipse, the GUI loads the resource file correctly. I printed the output of the directory of the file:

Loading parser from serialized file /C:/Users/name/git/guis/target/classes/path/to/file.gz

The IOException occurs at this line of code:

URL url = SentenceParser.class.getResource("file.gz");
props.setProperty("file", url.getPath());

The file.gz is located in src/main/resources/path/to/file.gz and the class is defined in a java src directory of src/main/java/path/to.

After research, my problem is almost exactly like this person's question: java, loading of resources fails: unable to resolve file:/my-jar.jar!/folder/my-file . They state that they must have the path of the file in order to put into a properties file. I, also, need the path of the file and cannot retrieve the resource as a stream. I need the path in order to add that path to a properties file which is then read by an external library.

Answer 1

Maybe it's best to place your file external to your jar file if it needs to be read in and manipulated by other jars in your application. Like so:

file.gz
myjar.jar
myjar2.jar

That way you can obtain an absolute path to the common resource file by using code like:

static final String WORKINGDIR = System.getProperty("user.dir");
static final String SEPARATOR = System.getProperty("file.separator");
String path =  WORKINGDIR + SEPARATOR + "file.gz";
URL url = SentenceParser.class.getResource(path);

Try this and see if it works.

The thing is that when you run your application directly from within Eclipse, the application itself is not inside a jar. You're just running the class files, unpackaged. Therefore, it can find the resource file just fine. But when you jar your application and run it, the resource file resides inside the jar and the application has difficulty locating it. If you keep your resource files external to the jar, they are easier to locate using the code I provided above. I find this works best for me, anyway.

Answer 2

Oh boy!

I ran into the same/similar issue yesterday; and scratched my (bald) head for better part of a day as to why the file in resources folder is accessible in Eclipse - but not by the executable JAR.

Reading project properties....
Jul 14, 2021 2:03:12 PM main.helper.CALotteryLogger logAbort
SEVERE: *******************************************************
Jul 14, 2021 2:03:12 PM main.helper.CALotteryLogger logAbort
SEVERE: * Alert:          InputStream for main/Resources/Common config With IDs.properties is NULL.
Jul 14, 2021 2:03:12 PM main.helper.CALotteryLogger logAbort
SEVERE: *******************************************************

Today, I found out that the path of the resource file is case-sensitive - even on Windows systems - when it comes to accessing it from the executable JAR.

My original path for the resource was:

private static final String CONFIG_PROPERTIES_FILEPATH = "main/Resources/Common config With IDs.properties";

which resulted in failure. But when I changed it to

private static final String CONFIG_PROPERTIES_FILEPATH = "main/resources/Common config With IDs.properties";

the problem was resolved.

Note that "Resources" in the file name was changed to "resources" to resolve the issue.

Usually, Windows is case-agnostic, but not when it comes to executable JAR files. Lesson learned!!

PS: Following syntax was used to open the resource file:

InputStream is = ReadConfigData.class.getClassLoader().getResourceAsStream(CONFIG_PROPERTIES_FILEPATH);