java Spring引导代码源返回为null
公平的警告我是一个新的春天靴子
任何人遇到试图获取codeSource/URL时抛出空指针异常的场景,我已经将其缩小到spring代码库中的这段代码
ProtectionDomain protectionDomain = getClass().getProtectionDomain();
CodeSource codeSource = protectionDomain.getCodeSource();
this is line 119: URI location = (codeSource != null) ? codeSource.getLocation().toURI() : null;
这是堆栈跟踪
org.springframework.boot.loader.JarLauncher.main exited with an exception.
java.lang.IllegalStateException: java.lang.NullPointerException
at org.springframework.boot.loader.ExecutableArchiveLauncher.<init>(ExecutableArchiveLauncher.java:41)
at org.springframework.boot.loader.JarLauncher.<init>(JarLauncher.java:35)
at org.springframework.boot.loader.JarLauncher.main(JarLauncher.java:51)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:483)
at com.kabira.platform.MainWrapper.invokeMain(MainWrapper.java:66)
Caused by: java.lang.NullPointerException
at org.springframework.boot.loader.Launcher.createArchive(Launcher.java:119)
at org.springframework.boot.loader.ExecutableArchiveLauncher.<init>(ExecutableArchiveLauncher.java:38)
... 7 more
共 (0) 个答案