Created a jar file for a spring boot multimodule application and ran the jar file using java -jar command. While starting the application, it gives ResourceFinderException. When I analyzed it, the issue is happening because in my ResourceConfig file, i have used the package for my api end points. If I use register(service.class), the application starts fine. Any suggestion how can I provide the package instead of using register? The reason I want to use package is because I have lots of services inside multiple packages and the code looks very ugly if i use register for all the services. The ResourceConfig file looks like below.
public class AppResourceConfig extends ResourceConfig {
public AppResourceConfig {}{
super();
property("jersery.config.beanValidation.enableOutputValidationErrorEntity.server");
**packages("com.api");**
register(GsonProvider.class);
register(RequestContextFilter.class);
register(NotFoundExceptionMapper.class);
register(DefaultExceptionMapper.class);
}
}
Here the issue is with highlighted line: packages("com.api")
If I comment out this code application will be up. Otherwise it is giving org.glassfish.jersey.server.internal.scanning.ResourceFinderException: java.io.FileNotFoundException: api-01.03.00.04-snapshot.jar (No such file or direcotry)
Note: api-01.03.00.04-snapshot.jar is the jar file for one of the module in a project
Related
Been searching for others that have run into this issue, and not finding much out there, so it can't be that common.
I have a spring-boot project that I want to convert into a jar project, running with embedded tomcat. It's using yml files (application.yml and then the profile versions - eg appplication-dev.yml.) It ran fine as war with the yml files, however, when I convert it to a jar, and kick off the jar, the embedded tomcat never starts UNLESSS I add an empty application.properties file as well. (No errors just no Tomcat startup unless the empty application.properties file is added.)
I believe it's somehow related to one of our internal jar dependencies (also spring), since if I remove that dependency from the pom (and any of the code referencing it) I can get the jar to startup the embedded tomcat just fine (without providing the empty application.properties file.)
I could also, of course, forgo using yml files and just use .properties files, but I'd like to use yml files if possible. Why adding an empty applcation.properties file causes things to work has me stumped.
If it helps, the config in the dependency project that causes the issue we're seeing is set up as:
#Configuration
#EnableConfigurationProperties(OracleDataSourceProperties.class)
#EnableTransactionManagement
#ComponentScan(basePackages = {"com.foo.data.services","com.foo.data.domain", "com.foo.utility", "com.foo.cipher.utility"})
#MapperScan(value = {"com.foo.data.services.mapper","com.foo.data.services.batchmapper"})
public class DataServicesPersistenceConfig { ... }
and the OracleDataSourceProperties class:
#ConfigurationProperties(prefix="oradb", ignoreUnknownFields = true)
public class OracleDataSourceProperties extends BaseVO implements InitializingBean{
I have a spring boot application jar
#SpringBootApplication
#EnableScheduling
#PropertySource({"classpath:1.properties","classpath:2.properties"})
public class MyClass {
//My class
}
1.properties is in src/main/resources
2.properties is in server location project location
| MyClass.jar
| config
| 2.properties
But I am getting file not found for 2.properties when starting the application.Please let me know what I could be missing here.
Like mentioned in the comment, your 2.properties file is not under your classpath. You can only use classpath if your file really exists in your jar or war.
To get the 2.properties you should use the command file: instead of classpath:.
#PropertySource("classpath:1.properties","file:${application_home}2.properties")
I am not quite sure if its still necessary to have an OS environment variable or a system property to set the path to your property file. In this case I named it application_home.
I can't find an answer to this question on stackoverflow hence im asking here so I could get some ideas.
I have a Spring Boot application that I have deployed as a war package on Tomcat 8. I followed this guide Create a deployable war file which seems to work just fine.
However the issue I am currently having is being able to externalize the configuration so I can manage the configuration as puppet templates.
In the project what I have is,
src/main/resources
-- config/application.yml
-- config/application.dev.yml
-- config/application.prod.yml
-- logback-spring.yml
So how can I possibly load config/application.dev.yml and config/application.prod.yml externally and still keep config/application.yml ? (contains default properties including spring.application.name)
I have read that the configuration is load in this order,
A /config subdirectory of the current directory.
The current directory
A classpath /config package
The classpath root
Hence I tried to load the configuration files from /opt/apache-tomcat/lib to no avail.
What worked so far
Loading via export CATALINA_OPTS="-Dspring.config.location=/opt/apache-tomcat/lib/application.dev.yml"
however what I would like to know is,
Find out why loading via /opt/apache-tomcat/lib classpath doesn't work.
And is there a better method to achieve this ?
You are correct about load order. According to Spring boot documentation
SpringApplication will load properties from application.properties files in the following locations and add them to the Spring Environment:
A /config subdirectory of the current directory.
The current directory
A classpath /config package
The classpath root
The list is ordered by precedence (properties defined in locations higher in the list override those defined in lower locations).
[Note]
You can also use YAML ('.yml') files as an alternative to '.properties'.
This means that if you place your application.yml file to /opt/apache-tomcat/lib or /opt/apache-tomcat/lib/config it will get loaded.
Find out why loading via /opt/apache-tomcat/lib classpath doesn't work.
However, if you place application.dev.yml to that path, it will not be loaded because application.dev.yml is not filename Spring is looking for. If you want Spring to read that file as well, you need to give it as option
--spring.config.name=application.dev or -Dspring.config.name=application.dev.
But I do not suggest this method.
And is there a better method to achieve this ?
Yes. Use Spring profile-specific properties. You can rename your files from application.dev.yml to application-dev.yml, and give -Dspring.profiles.active=dev option. Spring will read both application-dev.yml and application.yml files, and profile specific configuration will overwrite default configuration.
I would suggest adding -Dspring.profiles.active=dev (or prod) to CATALINA_OPTS on each corresponding server/tomcat instance.
I have finally simplified solution for reading custom properties from external location i.e outside of the spring boot project. Please refer to below steps.
Note: This Solution created and executed windows.Few commands and folders naming convention may vary if you are deploying application on other operating system like Linux..etc.
1. Create a folder in suitable drive.
eg: D:/boot-ext-config
2. Create a .properties file in above created folder with relevant property key/values and name it as you wish.I created dev.properties for testing purpose.
eg :D:/boot-ext-config/dev.properties
sample values:
dev.hostname=www.example.com
3. Create a java class in your application as below
------------------------------------------------------
import org.springframework.boot.context.properties.ConfigurationProperties;
import org.springframework.context.annotation.PropertySource;
#PropertySource("classpath:dev.properties")
#ConfigurationProperties("dev")
public class ConfigProperties {
private String hostname;
//setters and getters
}
--------------------------------------------
4. Add #EnableConfigurationProperties(ConfigProperties.class) to SpringBootApplication as below
--------------------------------------------
#SpringBootApplication
#EnableConfigurationProperties(ConfigProperties.class)
public class RestClientApplication {
public static void main(String[] args) {
SpringApplication.run(RestClientApplication.class, args);
}
}
---------------------------------------------------------
5. In Controller classes we can inject the instance using #Autowired and fetch properties
#Autowired
private ConfigProperties configProperties;
and access properties using getter method
System.out.println("**********hostName******+configProperties.getHostName());
Build your spring boot maven project and run the below command to start application.
-> set SPRING_CONFIG_LOCATION=<path to your properties file>
->java -jar app-name.jar
I'm new to Groovy, Spring Boot, and Gradle (using gradle as well) and am trying to build a small test program.
I have a main class EvalMain and a InputObj class in a com.eval package.
#Controller
class EvalMain {
#RequestMapping("/")
#ResponseBody
public static String textTest() {
def iO = new InputObj("dsa", "dasdsa", "U1dBRw==");
return iO.xorString();
}
}
when running "spring run EvalMain.groovy" I get the following error:
startup failed:
file:<filepath>EvalMain.groovy: 14: unable to resolve class InputObj
# line 14, column 18.
def iO = new InputObj("dsa", "dasdsa", "U1dBRw==");
I tried throwing an import at the top but then spring boot complained about this too? What am I missing here? Any help would be much appreciated, thanks!
after adding the import:
import com.eval.InputObj;
I get this error:
file:/<filePath>/EvalMain.groovy: 2: unable to resolve class com.eval.InputObj
# line 2, column 1.
import com.eval.InputObj
^
1 error
Not sure if this is a good way to do it, but I was able to run it with "spring run .groovy .groovy" I wonder if there's the equivalent of a makefile?
Try to run spring run *.groovy.
Spring Boot CLI is for quick prototyping a single Groovy script. It tries to autoload all Spring Boot dependencies, but its your responsibility to include your additional dependencies
If you have just multiple source files you have to pass it to the CLI.
If it is an external dependency, include it with a #Grab, see here.
However, you have Gradle and more than one source file. Therefor it might be the time to switch from the Spring Boot CLI to a normal Gradle project.
Just go to the guides and choose "Build with Gradle". Then you will see an example Gradle build file. Also the Gradle Spring Boot plugin documentation provides samples.
You can start your application with gradlew bootRun.
I don't know how to call a Job defined in Spring Batch using CommandLineJobRunner, documentation details are not enough for me.
I've followed the Spring Batch official guide to write Jobs in Spring Batch using Java annotations e.g. #EnableBatchProcessing because I wanted to avoid XML configuration files for the description of the job, the steps, etc.
So far I have:
a configuration class (com.package.bla.bla.ClassContainingTheBatchConfiguration see below) where I've put all the stuff defining ItemReader, ItemProcessor, ItemWriter, Job, and Step (with return jobs.get("nameOfTheJob") see below) using a #Bean annotaion.
a class with a main method with SpringApplication.run(...) and and annotation with #ImportResource("classpath:META-INF/spring/applicationContext.xml") to import some beans I need when processing the data in the Job.
On the Maven side I am currently using some plugins:
maven-jar-plugin specifying <addClasspath>true</addClasspath> and the class containing the main method in the tag <mainClass>
maven-assembly-plugin because I would like a unique executable jar containing all the stuff in the dependencies, I am using <phase>package</package> to be able to build the jar in the package phase, I am also using <goal>single</goal> to be able to properly build the jar using the assembly
maven-compiler-plugin specifying I am using Java 1.7
I think I've configured all the things I need to configure, however after having a Maven BUILD SUCCESS I am not able to run the job from the command line:
java -cp ./target/JAR_FILE_NAME.jar org.springframework.batch.core.launch.support.CommandLineJobRunner com.package.bla.bla.ClassContainingTheBatchConfiguration nameOfTheJob
Is throwing IOException due to the java.io.FileNotFoundException regarding com.package.bla.bla.ClassContainingTheBatchConfiguration. How should I specify the parameters in the command line in order to get the Job executed?
If you are already using SpringApplication from Spring Boot, why not finish the job and use #EnableAutoConfiguration as well, and also the Maven plugin (see for example this guide)? That way you will get something working pretty quickly and you can always add your own features later.
If the first argument to the CommandLineJobRunner is your #Configuration FQCN instead of a resource path, the ClassPathXmlApplicationContext constructor that's called from the CommandLineJobRunner's start() method will break.
int start(String jobPath, String jobIdentifier, String[] parameters, Set<String> opts) {
ConfigurableApplicationContext context = null;
try {
context = new ClassPathXmlApplicationContext(jobPath);
If you've already written a class with a main(), that replaces the CLJR, you shouldn't be passing CLJR as the class name in the command line. Pass that instead.
dont use spring.batch.job.enabled=false
then run using java -jar [jar-files] --spring.batch.job.names=[job-name]