如何从 Java 代码(JSP/Servlet)中停止另一个 tomcat
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【中文标题】如何从 Java 代码(JSP/Servlet)中停止另一个 tomcat【英文标题】:How to stop another tomcat from Java code (JSP/Servlet) 【发布时间】:2020-11-24 19:19:03 【问题描述】:我有两个tomcat安装如下:
-
Tomcat-A 路径:C:\TomcatA,端口:连接器端口 =“8080”
redirectPort="8443" 关闭端口为 -1
Tomcat-B 路径:C:\TomcatB,端口:连接器端口 =“8090”
redirectPort="8444" 关闭端口为 8015
我有一个在 Tomcat-A 上运行的 JSP/Servlet 应用程序。我的要求是关闭 Tomcat-B 并删除/取消部署托管在 Tomcat-B 上的应用程序。下面是我的 Java 代码。
private void StopTomcat() throws Exception
if(logger.isDebugEnabled()) logger.debug("StopTomcat");
try
if (logger.isTraceEnabled())logger.trace("Shutting down tomcat: ");
String Directory = "C:\TomcatB\bin\";
Process process = Runtime.getRuntime().exec(Directory + "shutdown.bat");
if (logger.isTraceEnabled())logger.trace("Executed shutdown.bat at " + Directory );
BufferedReader stdInput = new BufferedReader(new InputStreamReader(process.getInputStream()));
BufferedReader stdError = new BufferedReader(new InputStreamReader(process.getErrorStream()));
if (logger.isTraceEnabled())logger.trace("Printing shutdown process logs" );
String s = null;
while ((s = stdInput.readLine()) != null)
if (logger.isTraceEnabled())
logger.trace(s);
String Error = "";
while ((s = stdError.readLine()) != null)
Error = Error + s;
if (!Error.equals(""))
logger.error(Error);
throw new Exception("Tomcat Shutdown Failed with error: " + Error);
catch (Exception e)
e.printStackTrace();
logger.error( Arrays.toString( e.getStackTrace() ));
throw new Exception(e.getMessage());
在我得到的日志中:
2020-08-04 21:11:00 TRACE MigrateApp:55 - Executed shutdown.bat at C:\TomcatB\bin\
2020-08-04 21:11:00 TRACE MigrateApp:60 - Printing shutdown process logs
2020-08-04 21:11:00 TRACE MigrateApp:64 - Using CATALINA_BASE: "TomcatA"
2020-08-04 21:11:00 TRACE MigrateApp:64 - Using CATALINA_HOME: "TomcatA"
2020-08-04 21:11:00 TRACE MigrateApp:64 - Using CATALINA_TMPDIR: "TomcatA\temp"
2020-08-04 21:11:00 TRACE MigrateApp:64 - Using JRE_HOME: "C:\Program Files\Java\jre1.8.0_251"
2020-08-04 21:11:00 TRACE MigrateApp:64 - Using CLASSPATH: "TomcatA\bin\bootstrap.jar;TomcatA\bin\tomcat-juli.jar"
2020-08-04 21:11:00 ERROR MigrateApp:72 - 04-Aug-2020 21:11:00.668 SEVERE [main] org.apache.catalina.startup.Catalina.stopServer No shutdown port configured. Shut down server through OS signal. Server not shut down.
2020-08-04 21:11:00 ERROR MigrateApp:78 - [sailpoint.support.applicationdeployer.util.MigrateApp.StopTomcat(MigrateApp.java:73), sailpoint.support.applicationdeployer.util.MigrateApp.run(MigrateApp.java:36)]
2020-08-04 21:11:00 ERROR MigrateApp:79 - Tomcat Shutdown Failed with error: 04-Aug-2020 21:11:00.668 SEVERE [main] org.apache.catalina.startup.Catalina.stopServer No shutdown port configured. Shut down server through OS signal. Server not shut down.
2020-08-04 21:11:00 ERROR MigrateApp:39 - Tomcat Shutdown Failed with error: 04-Aug-2020 21:11:00.668 SEVERE [main] org.apache.catalina.startup.Catalina.stopServer No shutdown port configured. Shut down server through OS signal. Server not shut down.
我无法理解为什么我的代码试图关闭 Tomcat-A。任何人都可以帮助我吗?
【问题讨论】:
【参考方案1】:我可以通过如下修改 exec 参数来实现。它现在可以正常工作了。
String Directory = "C:\TomcatB\bin\";
File workingDirecotry = new File(Directory);
Process process = Runtime.getRuntime().exec(Directory + "shutdown.bat", null, workingDirecotry);
【讨论】:
【参考方案2】:如果您尚未编辑 Tomcat B shutdown.bat
默认脚本,您将看到 C:\TomcatB\bin\shutdown.bat
确定 CATALINA_HOME 并执行:
%CATALINA_HOME%\bin\catalina.bat
看起来您的 TomcatA 虚拟机正在将自己的 CATALINA_HOME 传递给 Tomcat B shutdown.bat
,这会覆盖 C:\TomcatB\bin 使用的默认值。您可以创建一个不假定 CATALINA_HOME 环境变量的新脚本 shutdown2.bat,或者切换到 ProcessBuilder
并在 ProcessBuilder
启动命令中设置 CATALINA_HOME,如下所示:
File exe = new File(Directory, "shutdown.bat");
String [] cmd = new String[] exe.getAbsolutePath() ;
ProcessBuilder pb = new ProcessBuilder(cmd);
pb.environment().put("CATALINA_HOME", exe.getParentFile().getParentFile().getAbsolutePath());
Process process = pb.start();
如果进程锁定,您可能需要在后台线程中使用输出/错误流或通过在 start() 之前调用将它们重定向到文件:
pb.redirectOutput(outfile);
pb.redirectError(errfile);
【讨论】:
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