How to configure port for a Spring Boot application

There are two main ways to change the port in the Embedded Tomcat in a Spring Boot Application.\n\nModify application.properties\n\nFirst you can try the application.properties file in the /resources folder:\n\nserver.port = 8090\n\n\n\n\nModify a VM option\n\nThe second way, if you want to avoid modifying any files and checking in something that you only need on your local, you can use a vm arg:\n\nGo to Run -> Edit Configurations -> VM options\n\n-Dserver.port=8090\n\n\n\n\nAdditionally, if you need more information you can view the following blog post here: Changing the port on a Spring Boot Application",

Since Spring Boot provides various configuration externalization mechanism (through various PropertySource implementations and/or processors wired into Environment object in order), you can set any property outside of your jar archive through following methods:\n\n\nPass property through command line argument as application argument\n\njava -jar <path/to/my/jar> --server.port=7788\n\nFrom property in SPRING_APPLICATION_JSON (Spring Boot 1.3.0+)\n\n\nDefine environment variable in U*IX shell:\n\nSPRING_APPLICATION_JSON='{\"server.port\":7788}' java -jar <path/to/my/jar>\n\nBy using Java system property:\n\njava -Dspring.application.json='{\"server.port\":7788}' -jar <path/to/my/jar>\n\nPass through command line argument:\n\njava -jar <path/to/my/jar> --spring.application.json='{\"server.port\":7788}'\n\n\nDefine JVM system property\n\njava -Dserver.port=7788 -jar <path/to/my/jar>\n\nDefine OS environment variable\n\n\nU*IX Shell\n\nSERVER_PORT=7788 java -jar <path/to/my/jar>\n\nWindows\n\nSET SERVER_PORT=7788\njava -jar <path/to/my/jar>\n\n\nPlace property in ./config/application.properties configuration file\n\nserver.port=7788\n\n\nand run:\n\n java -jar <path/to/my/jar>\n\nPlace property in ./config/application.yaml\n\nserver:\n port: 7788\n\n\nand run:\n\n java -jar <path/to/my/jar>\n\nPlace property in ./application.properties\n\nserver.port=7788\n\n\nand run:\n\n java -jar <path/to/my/jar>\n\nPlace property in ./application.yaml\n\nserver:\n port: 7788\n\n\nand run:\n\n java -jar <path/to/my/jar>\n\n\n\n\n\nYou can combine above methods all together, and the former configuration in the list take precedence over the latter one.\n\nFor example:\n\nSERVER_PORT=2266 java -Dserver.port=5566 -jar <path/to/my/jar> --server.port=7788\n\n\nThe server will start and listen on port 7788.\n\nThis is very useful providing default properties in PropertySources with lower precedence (and usually packaged in the archive or coded in the source), and then override it in the runtime environment. And it is the design philosophy of Spring Boot:\n\n\n Be opinionated out of the box, but get out of the way quickly as requirements start to diverge from the defaults.\n\n\n\n\nSERVER_NAME to server.name conversion was done by Relaxed Binding.",

Also, you can configure the port programmatically.\nFor Spring Boot 2.x.x:\n@Configuration\npublic class CustomContainer implements WebServerFactoryCustomizer<ConfigurableServletWebServerFactory> {\n public void customize(ConfigurableServletWebServerFactory factory){\n factory.setPort(8042);\n }\n}\n\nFor older versions:\n@Configuration\npublic class ServletConfig {\n @Bean\n public EmbeddedServletContainerCustomizer containerCustomizer() {\n return (container -> {\n container.setPort(8012);\n });\n }\n}\n",

If you would like to run it locally, use this -\n\nmvn spring-boot:run -Drun.jvmArguments='-Dserver.port=8085'\n\nAs of Spring Boot 2.0, here's the command that works (clues were here):\n\nmvn spring-boot:run -Dspring-boot.run.arguments=--server.port=8085\n",

As said in docs either set server.port as system property using command line option to jvm -Dserver.port=8090 or add application.properties in /src/main/resources/ with\nserver.port=8090\n\nFor a random port use:\nserver.port=0\n\nSimilarly add application.yml in /src/main/resources/ with:\nserver:\n port: 8090\n",