Original code:
This would work on Unix platform, the returned path is "/home/..../xyz.xml"
However on Windows, it would return "invalid" path "/C:/.../xyz.xml", which is causing Exception from the get().
Solution in bugtracker suggested
1. Use toURI() method to convert getResource() result to URI, then use Paths.get()
2. Use Legacy File class.
https://bugs.openjdk.java.net/browse/JDK-8197918
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