Develop Reference

Tomcat does not unpack spring boot application

thank you for your attention in reading my problem;
I'm trying to deploy a spring boot application but when I put it in the tomcat webapps folder it doesn't load my application, does anyone know what could be happening?
I already tried to run several versions of java changing the system variables, I used several versions of tomcat but none unzip the application
I'm using java version 17
and version apache-tomcat-9.0.71
Log Tomcat
Tomcat Past

I thought Spring Boot came with a pre-installed web server, so there is no need to use this method. Why don't you use the current method which supports auto deployment of your application?

An Spring Boot application is a ready to run jar file which contains tomcat/jetty and can be run with java -jar <your app>.jar.
When you want to run an application in tomcat you should build a .war file.
======================================
As found here: https://stackoverflow.com/a/27905557/2144466
Did you follow this guide: https://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#howto.traditional-deployment
and do you have a class which extends SpringBootServletInitializer andd overwrites the configure method:
`
#SpringBootApplication
public class MyApplication extends SpringBootServletInitializer {
#Override
protected SpringApplicationBuilder configure(SpringApplicationBuilder application) {
return application.sources(MyApplication.class);
}
public static void main(String[] args) {
SpringApplication.run(MyApplication.class, args);
}
`

Please follow below.
Rename your war file at location webapps from web-0.0.1-SNAPSHOT.war to web.war
restart tomcat.
open url (http://localhost:8080/web) in browser.

I managed to solve the problem.
Apparently it was version incompatibility between spring, tomcat and java, by adjusting the system variables I can deploy both java 11 and java 17 applications.
Even though I managed to find a solution, I thank the friends above for dedicating their time to trying to solve my problem.

Spring boot not serving static files from inside jar

Spring boot is not serving static files placed inside a jar.
I've had a backend app which I decided to add frontend to. Setup tasks to copy static frontend files to src/main/resources/static. Went through a bunch of answers here on SO, all of them advise the static content (index.html, .js and .css files) should reside under src/main/resources/static or src/main/resources/public, which i both tried. I open the built .jar and the static files are there, but starting application using java -jar myApp.jar and opening localhost:8080 gives me default whitelabel error page. My application works, since i can access the api i have running on it.
Application doesn't have #EnableWebMvc or any other custom configuration.
If i manually copy/paste same static resources to project's src/main/resources/static and ran the application using #SpringBootApplication class in IDE - the resources are loaded without problem and index.html opens upon visiting localhost:8080, so it's only a problem when files are in .jar.
Should the static files be somewhere different when they're in runnable spring boot .jar file?
Spring boot version 2.1.1

I am facing the same problem.
In case it is helpfull, I can manage to serve the static files properly by adding the following configuration :
import org.springframework.context.annotation.Configuration;
import org.springframework.web.servlet.config.annotation.EnableWebMvc;
import org.springframework.web.servlet.config.annotation.ResourceHandlerRegistry;
import org.springframework.web.servlet.config.annotation.WebMvcConfigurer;
#Configuration
#EnableWebMvc
public class StaticFilesConfig implements WebMvcConfigurer {
#Override
public void addResourceHandlers(ResourceHandlerRegistry registry) {
registry.addResourceHandler("/static/**").addResourceLocations("classpath:/static/");
}
}
... but it overrides the "implicit" auto configuration from spring boot and the rest of my filters etc. doesn't work anymore... => this is because #EnableWebMvc deactivates spring's default configuration.
[EDIT] In the end, I happened to understand that the jar containing the static files was not included in the built bootJar. You may want to check that.
HTH!

Run Spring's .class instead of .jar file

I have a Java Spring application which I built with maven, and a .jar file together with .class files were generated.
Instead of running the .jar file, is there a way for me to run the .class files instead?
I tried executing java -cp . hello.Application in the same directory as the Application.class (source code of Application.java below), but was given Error: Could not find or load main class hello.Application. Removing the hello. did not work either.
package hello;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
#SpringBootApplication
public class Application {
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}

I am doing the Spring boot application.I think you can also use maven to create a jar file and then use linux commend to run it.

you can run a class by executing follow command :
java Application
for only simple console application. to run Spring Boot application you need maven to build jar file and executing jar file to call main method to load spring boot context. Spring Boot need several requirements and dependencies to run. Spring load context and create a special structure to load. for more you can see following links:
https://spring.io/guides/gs/spring-boot/
http://www.logicbig.com/tutorials/spring-framework/spring-boot/boot-exploded-structure/

What is the difference between spring cloud config server exec.jar and .jar suffix

We are currently using spring cloud config server spring-cloud-config-server-1.1.0.M2-exec.jar and would like to upgrade to the latest version. However, i've noticed that beyond 1.1.0.M2 version there's only standard jars and no exec.jar in Maven repo http://maven.springframework.org/milestone/org/springframework/cloud/spring-cloud-config-server/
Could someone please explain what the difference is? Will I be able to just substitute the exec one with the standard non exec one?
Thanks

The exec jar contains an executable version of the config server (as a Spring Boot application). The non-exec jar contains only the config server classes. So you can't just replace the exec jar with the other one. What you basically have to do is to create a basic Spring Boot application with the config server dependencies and the appropriate annotations (like in the example):
#SpringBootApplication
#EnableDiscoveryClient
#EnableConfigServer
public class ConfigServerApplication {
public static void main(String[] args) {
SpringApplication.run(ConfigServerApplication.class, args);
}
}

Could we implement resful webservice via Jersey without manven and web.xml on Tomcat 8

I am trying to implement a very simple restful webservice via Jersey and I keep getting a 404 error page without any other error on the console.
Here is the tutorial that I diligently followed.
http://www.doublecloud.org/2013/08/rest-api-service-made-easy-with-jersey-framework-in-java/
Since I am not using maven, I did not document any dependencies.
I am using Eclipse Java EE luna / Tomcat 8 / Java 8 / Servlet 3.0 ( I don't have any web.xml)
I loaded the Jersey 2.22.1 libraries
Is there something that I am missing to make it work? Should I convert it to a maven project?
Thanks a lof for your help in advance.
*********Response***************
1. Download the jersey library
2. Upload them in the application in the web-inf/lib folder and add them in the Java build path
3. Create an HelloApp.java class
`package testwebservice.rest.demo;
import javax.ws.rs.ApplicationPath;
import org.glassfish.jersey.server.ResourceConfig;
#ApplicationPath("/rest")
public class HelloApp extends ResourceConfig {
public void AppConfig() {
packages("testwebservice.rest.demo");
}
}`
4. Create HelloWorldService.java class
`package testwebservice.rest.demo;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.core.Response;
#Path("/hi")
public class HelloWorldService
{
#GET
#Path("/{name}")
public Response getMessage(#PathParam("name") String name)
{
String outMsg = "Hello " + name + "!";
return Response.status(200).entity(outMsg).build();
}
}`
Project / Clean in eClips
Start the server
Test the link "localhost:8080/testwebservice/rest/hi/Test"
result= Hello Test!

If you have all the jars from the Jersey JAX-RS 2.0 RI bundle, you don't need a web.xml. Just just need an javax.ws.rs.core.Application sublcass, annotated with #ApplicationPath. So this is the most minimal required configuration
#ApplicationPath("/api")
public class MyApplication extends Application {}
Just having this empty class will trigger class-path scanning, which will scan for classes annotated with #Path and #Provider classes.
As mentioned here class-path scanning is not recommended though. Better to use Jersey's package scanning capabilities. First instead of extending Application, you should extend ResourceConfig (which is an Application subclass). Then just call packages("..") to list the package to scan for your resources
#ApplicationPath("/api")
public class AppConfig extends ResourceConfig {
public AppConfig() {
packages("your.packages.to.scan");
}
}
And that's it. You should be up and running, as long as your resource class is structured properly. And make sure all your jars are actually in your WEB-INF/lib when deployed to Tomcat.
See Also:
Servlet Based Deployment for other deployment options (information)

Im pretty sure the tomcat servlet requires a web.xml at the very least.
Maven/gradle makes dependency management and building/deploying applications extremely easy, i would recommend this.
On another note: If you just need a simple small web service, take a look at spring-boot. It has an embedded tomcat server. No setup needed what so ever, other than a simple maven / gradle setup.
Take a look here: https://spring.io/guides/gs/rest-service/