I need to change the default location of log4j2 configuration file. I followed the documentation here
https://logging.apache.org/log4j/2.x/manual/webapp.html
But the only file log4j2 can see is log4j2.xml in the classpath. otherwise I get "no log4j2 configuration file found"
I tried:
-1 setting context parameters
-2 setting system property Log4jContextSelector to "org.apache.logging.log4j.core.selector.JndiContextSelector". and using the JNDI selector
as described here
https://logging.apache.org/log4j/2.x/manual/webapp.html#ContextParams
-3 lookups: web, env, sys, ctx and bundle. the first 4 failed only bundle worked but you can only lookup inside the classpath.
-4 set isLog4jAutoInitializationDisabled to true, and I am not sure how to configure the filter in this case. If I include them in the web.xml the app will not deploy.
jar in the project
./WEB-INF/lib/log4j-jcl-2.4.1.jar
./WEB-INF/lib/log4j-core-2.4.1.jar
./WEB-INF/lib/log4j-slf4j-impl-2.4.1.jar
./WEB-INF/lib/log4j-api-2.4.1.jar
In my situation with .propeties file I use code shown below
#Plugin(name = "LogsConfigurationFactory", category = ConfigurationFactory.CATEGORY)
public class CustomLogsConfigurationFactory extends PropertiesConfigurationFactory {
#Override
public Configuration getConfiguration(String name, URI configLocation) {
File propFile = new File("/path_to/log4j2.properties");
return super.getConfiguration(name, propFile.toURI());
}
#Override
protected String[] getSupportedTypes() {
return new String[] {".properties", "*"};
}
}
I think you can change CustomLogsConfigurationFactory on XmlConfigurationFactory, and change return typse in getSupportedTypes method. I hope this will help you.
Related
Below showing the project structure
Core Project
|-config project
|
|-Service project
After building the core project we get Service.jar file.
While running the service.jar am passing spring.config.additional.location as command line argument.
java -jar Service-1.0.jar --spring.config.additional-location=C:/Users/Administrator/Desktop/Springboot/
above spring.config.additional.location path having application.property file and some xml files.
I can able to read application property file in service project ,following logic
Application.propertes
external.config=C:/Users/Administrator/Desktop/Springboot/config/
Mian Class
#ImportResource(locations = {
"${external.config}"+"/spring/service-config.xml",
"${external.config}"+"/spring/datasource-config.xml"
})
public class ServiceMain {
public static void main(String[] args) {
ConfigurableApplicationContext applicationContext = new SpringApplicationBuilder(ServiceMain.class)
.build()
.run(args);
for (String name : applicationContext.getBeanDefinitionNames()) {
}
}
}
Similar kind of logic applied in config project is given below,its not working
#Configuration
public class ConfigurationFactory
{
#Value("${external.config}")
public String extConfPath;
public String REQ_CONF = extConfPath+"/Configuration.xml";
public static final String FILTER_XML_CONF = extConfPath+"/DocFilter.xml";
}
Is there any better way to do this? How can i read external application.properties in config project
Do we have any better way to do this in spring boot
As you are cleary developing a distributed web system the best practice is to used externalised configuration used by your different services allowing you to update settings without redeployment. Take a look at Spring Cloud Config
I am developing an application using Spring Boot. I have an externalized properties file on file system. Its location is stored in an environment variable as below--
export props=file://Path-to-file-on-filesystem/file.properties
Properties from this file are loaded on classpath and are made available to application like below--
List<String> argList = new ArrayList<>();
String properties = System.getenv().get("props");
try (InputStream is = BinaryFileReaderImpl.getInstance().getResourceAsStream(properties)) {
if (is != null) {
Properties props = new Properties();
props.load(is);
for(String prop : props.stringPropertyNames()) {
argList.add("--" + prop + "=" + props.getProperty(prop));
}
}
} catch(Exception e) {
//exception handling
}
This argList is passed to SpringBootApplication when it starts like below--
SpringApplication.run(MainApplication.class, argList);
I can access all the properties using ${prop.name}
However, I do not have access to these properties when I run JUnit Integration Tests. All my DB properties are in this externalized properties file. I do not want to keep this file anywhere in the application eg. src/main/resources
Is there any way I can load these properties in spring's test context?
I could finally read the externalized Properties file using #PropertySource on linux machine. Could not get it working on Windows instance though.
Changes done are as below-
export props=/path-to-file
Note that file:// and actual file name has been removed from environment variable.
#Configuration
#PropertySource({"file:${props}/file-test.properties", "classpath:some_other_file.properties"})
public class TestConfiguration {
}
Keep this configuration class in individual projects' src/test/java folder. Thank you Joe Chiavaroli for your comment above.
I want to create multiple application contexts in my Tomcat application.
Some of these application contexts have the same package and class names, but they all refer to different jars.
For example:
application0 use service.jar, model.jar
application1 use service-a.jar, model-a.jar
application2 use service-b.jar, model-b.jar
application0 context is OK because is in orign project.
I reference some web page to custom application1, I use my custom classloader to start applicationContext.
File file0 = new File("D://git/project1/service-a.jar");
File file1 = new File("D://git/project1/modele-a.jar");
// convert the file to URL format
URL url0 = file0.toURI().toURL();
URL url1 = file1.toURI().toURL();
List<URL> urls = new LinkedList<>();
List<File> libs = listFilesForFolder(new File("D://protal//apache-tomcat-8.0.39//lib"));
for(File lib : libs) {
urls.add(lib.toURI().toURL());
}
urls.add(url1);
urls.add(url0);
final URLClassLoader customClassLoader = new URLClassLoader(urls.toArray(new URL[urls.size()]));
ClassPathXmlApplicationContext context1 = new ClassPathXmlApplicationContext("applicationContext.xml") {
protected void initBeanDefinitionReader(XmlBeanDefinitionReader reader)
{
super.initBeanDefinitionReader(reader);
reader.setValidationMode(XmlBeanDefinitionReader.VALIDATION_NONE);
reader.setBeanClassLoader(customClassLoader);
}
};
allApplicationContexts.add(context1);
The Spring contexts start OK, but they fail to create the component-scan bean, and PropertyPlaceholderConfigurer isn't working. Everything else seems correct.
I sure my config is correct because it works without the custom classloader. Libs contains all spring lib.
Is it possible to get this working with multiple Spring contexts?
As mentioned in the title I have two applications with two different logging configurations. As soon as I use springs logging.file setting I can not seperate the configurations of both apps.
The problem worsens because one app is using logback.xml and one app is using log4j.properties.
I tried to introduce a new configuration parameter in one application where I can set the path to the logback.xml but I am unable to make the new setting work for all logging in the application.
public static void main(String[] args) {
reconfigureLogging();
SpringApplication.run(IndexerApplication.class, args);
}
private static void reconfigureLogging() {
if (System.getProperty("IndexerLogging") != null && !System.getProperty("IndexerLogging").isEmpty()) {
try {
JoranConfigurator configurator = new JoranConfigurator();
configurator.setContext(context);
// Call context.reset() to clear any previous configuration, e.g. default
// configuration. For multi-step configuration, omit calling context.reset().
System.out.println("SETTING: " + System.getProperty("IndexerLogging"));
System.out.println("SETTING: " + System.getProperty("INDEXER_LOG_FILE"));
context.reset();
configurator.doConfigure(System.getProperty("IndexerLogging"));
} catch (JoranException je) {
System.out.println("FEHLER IN CONFIG");
}
logger.info("Entering application.");
}
}
#Override
protected SpringApplicationBuilder configure(SpringApplicationBuilder application) {
reconfigureLogging();
return application.sources(applicationClass);
}
The above code works somehow. But the only log entry which is written to the logfile specified in the configuration, which ${IndexerLogging} points to, is the entry from logger.info("Entering application."); :(
I don't really like to attach that code to every class which does some logging in the application.
The application has to be runnable as tomcat deployment but also as spring boot application with integrated tomcat use.
Any idea how I can set the path from ${IndexerLogging} as the path to read the configuration file when first configuring logging in that application?
Take a look at https://github.com/qos-ch/logback-extensions/wiki/Spring you can configure the logback config file to use.
Is there a way we can lookup file resources using relative path in application.properties file in Spring boot application as specified below
spring.datasource.url=jdbc:hsqldb:file:${project.basedir}/db/init
I'm using spring boot to build a upload sample, and meet the same problem, I only want to get the project root path. (e.g. /sring-boot-upload)
I find out that below code works:
upload.dir.location=${user.dir}\\uploadFolder
#membersound answer is just breaking up the hardcoded path in 2 parts, not dynamically resolving the property. I can tell you how to achieve what you're looking for, but you need to understand is that there is NO project.basedir when you're running the application as a jar or war. Outside the local workspace, the source code structure doesn't exist.
If you still want to do this for testing, that's feasible and what you need is to manipulate the PropertySources. Your simplest option is as follows:
Define an ApplicationContextInitializer, and set the property there. Something like the following:
public class MyApplicationContextInitializer implements ApplicationContextInitializer<ConfigurableApplicationContext> {
#Override
public void initialize(ConfigurableApplicationContext appCtx) {
try {
// should be /<path-to-projectBasedir>/build/classes/main/
File pwd = new File(getClass().getResource("/").toURI());
String projectDir = pwd.getParentFile().getParentFile().getParent();
String conf = new File(projectDir, "db/init").getAbsolutePath();
Map<String, Object> props = new HashMap<>();
props.put("spring.datasource.url", conf);
MapPropertySource mapPropertySource = new MapPropertySource("db-props", props);
appCtx.getEnvironment().getPropertySources().addFirst(mapPropertySource);
} catch (URISyntaxException e) {
throw new RuntimeException(e);
}
}}
Looks like you're using Boot, so you can just declare context.initializer.classes=com.example.MyApplicationContextInitializer in your application.properties and Boot will run this class at startup.
Words of caution again:
This will not work outside the local workspace as it depends on the source code structure.
I've assumed a Gradle project structure here /build/classes/main. If necessary, adjust according to your build tool.
If MyApplicationContextInitializer is in the src/test/java, pwd will be <projectBasedir>/build/classes/test/, not <projectBasedir>/build/classes/main/.
your.basedir=${project.basedir}/db/init
spring.datasource.url=jdbc:hsqldb:file:${your.basedir}
#Value("${your.basedir}")
private String file;
new ClassPathResource(file).getURI().toString()