用-classpath方式启动springboot

.github/workflows/build-latest-windows.yml

@@ -17,7 +17,9 @@ jobs:
       - name: Build with Maven
         run: |
           mvn -B package --file pom.xml
-          cp target/app.jar platform/windows/
+          unzip target/app.jar
+          cp target/BOOT-INF/classes platform/windows/
+          cp target/BOOT-INF/lib platform/windows/
           zip -r windows.zip platform/windows/ 
 
       # 获取 Maven 项目版本

platform/windows/start.bat

@@ -1,9 +1,24 @@
+@echo off
+setlocal enabledelayedexpansion
+
 set PWD=%~dp0
 set JAVA=%PWD%jre8/bin
 set DATA=%PWD%data
 set "PATH=%JAVA%;%PWD%;%PATH%"
-java -version
+set "directory=%PWD%\lib"
+set "classpath=%PWD%;classes"
+
+rem 遍历指定目录下的所有文件
+for /r "%directory%" %%I in (*) do (
+    set "classpath=!classpath!;%%~fI"
+)
+
+rem 去掉开头的分号
+if defined classpath (
+    set "classpath=!classpath:~1!"
+)
 start /b %PWD%aria2c.exe --conf-path=%PWD%aria2.conf > aria2.log
 echo "aria2RPC start"
 chcp 65001
-java -jar -Dfile.encoding=utf-8 app.jar --podcast2.data.data-path=%DATA% 
+java -Dfile.encoding=utf-8 -classpath "%classpath%" -Dpodcast2.data.data-path=%DATA% io.github.yajuhua.podcast2.Podcast2Application
+endlocal