I have migrated an application from a standard Java WEB project to Spring Boot. In the process I had to migrate also a Servlet that its main scope was to initialize some components (reading property files, etc).
The legacy code:
<servlet>
<servlet-name>startupservlet</servlet-name>
<servlet-class>my.package.name.StartupServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
The new code:
#Bean
public ServletRegistrationBean<StartupServlet> startupServletBean() {
ServletRegistrationBean<StartupServlet> bean = new ServletRegistrationBean<StartupServlet>(
new StartupServlet());
bean.setLoadOnStartup(1);
return bean;
}
Doing so, the StartupServlet gets invoked when accessing any URL, messing with the app behavior.
As a workaround, I have added a dummy path in order the Servlet init code to be called, but never to get through the get and post methods.
I believe I am missing something from how Spring Boot is managing servlet declaration but I still did not find what.
[Edit] It is not a real issue, because, as per #jbx's comment, I should transform it into a #Service. But I just wanted to find out why Spring enables the servlet on every path by default or how I can inhibit this.
I was trying with latest spring 5.0.0.Final with my EAR project which has a parent context defined in web.xml using context-param
with param names locatorFactorySelector and parentContextKey but spring could not able to load the parent context. When i checked the ContextLoaderListener source code it seems like there is no logic applied to pick parent context. Here my question is does spring 5 provide any default implementation of ContextLoader which caters loading of parent context or spring 5 dropped, if not what is the approach to support this, do i have to implement our own ?
The loading of the parent context based on locatorFactorySelector were handled at ContextLoader#loadParentContext(). But they changed it to return null in this commit.
As said by the javadoc , I think you can create a new ContextLoaderListener and override this method to return the parent context:
public class FooContextLoaderListener extends ContextLoaderListener{
#Override
protected ApplicationContext loadParentContext(ServletContext servletContext) {
//load and return the parent context ......
}
}
Then use this ContextLoaderListener to start up Spring :
<listener>
<listener-class>org.foo.bar.FooContextLoaderListener</listener-class>
</listener>
For me this below piece of code worked fine.
public class BeanFactoryContextLoaderListener extends ContextLoaderListener {
private static Logger log = Logger.getLogger(BeanFactoryContextLoaderListener.class);
#Override
protected ApplicationContext loadParentContext(ServletContext servletContext) {
ApplicationContext ctx = new ClassPathXmlApplicationContext("beanRefFactory.xml");
return ctx;
}
}
Obviously I added a listener too in web.xml.
My team recently bumped into the same problem. We wanted to start using Webflux and it requires Spring 5.
Here is what I did:
Manually reintroduce BeanFactoryLocator mechanism. Take following classes from Spring 4, put it into your code and fix packages:
AbstractUrlMethodNameResolver
AnnotationMethodHandlerAdapter
BeanFactoryLocator
BeanFactoryReference
BootstrapException
ContextSingletonBeanFactoryLocator
DefaultAnnotationHandlerMapping
HandlerMethodInvocationException
HandlerMethodInvoker
HandlerMethodResolver
InternalPathMethodNameResolver
MethodNameResolver
NoSuchRequestHandlingMethodException
ServletAnnotationMappingUtils
SingletonBeanFactoryLocator
SourceHttpMessageConverter
WebUtils
XmlAwareFormHttpMessageConverter
Following Subhranil's advice from this thread, use custom ContextLoaderListener which loads parent context same as in Spring 4. Then use it in web.xml.
In each WAR's spring-servlet.xml add DefaultAnnotationHandlerMapping so it scans for controllers. Accompanying beans like AnnotationMethodHandlerAdapter are also needed.
It worked for us.
If all you need is your context-param in any of your spring managed class you are looking for ServletContextAware.
Just implement that class and override its method to get the ServletContext object. Later you can also get the context-params using the ServletContext object.
Check out a very similar question.
Apparently, the mechanism for locating the parent context was removed with SPR-15154 (see also the corresponding Github issue spring-framework#19720).
One workaround is to extend org.springframework.web.context.ContextLoaderListener and re-implement the loadParentContext method is described in this stackoverflow answer.
There is probably a better way to solve parent context loading in Spring 5.x, which I still have to figure out.
In most of the spring tutorials that I see the spring bean instantiation is like below in the class with the main method. But in real world scenario this is not how it works right? Say in case of a web application, how can we do the below and were do we need to do it?
ApplcationContext context = new ClassPathXmlApplicationContext("applicaitonContext.xml");
context.getBean("blah");
Web applications with spring usually have defined a little something called listener in web.xml like this:
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
This listener is responsible for initialization XmlWebApplicationContext (it's implementation of ApplicationContext just like ClassPathXmlApplicationContext) with /WEB-INF/applicationContext.xml as a default configuration file location if not redefined.
So
ApplicationContext context = new
ClassPathXmlApplicationContext("applicaitonContext.xml");
is done automatically.
As for the
context.getBean("blah");
well it is possible, you just need to inject ApplicationContext into any bean like:
#Component
class AnyBean {
#Autowired
private ApplicationContext context;
void doSomethingWithBlah() {
context.getBean("blah").doSomething();
}
}
But that is not what spring was meant for. You supposed to inject "blah" bean with #Autowired or #Inject instead of interacting with application context directly.
Please refer to spring documentation for details.
I'm writing a Spring MVC application deployed on Tomcat. See the following minimal, complete, and verifiable example
public class Application extends AbstractAnnotationConfigDispatcherServletInitializer {
protected Class<?>[] getRootConfigClasses() {
return new Class<?>[] { };
}
protected Class<?>[] getServletConfigClasses() {
return new Class<?>[] { SpringServletConfig.class };
}
protected String[] getServletMappings() {
return new String[] { "/*" };
}
}
Where SpringServletConfig is
#Configuration
#ComponentScan("com.example.controllers")
#EnableWebMvc
public class SpringServletConfig {
#Bean
public InternalResourceViewResolver resolver() {
InternalResourceViewResolver vr = new InternalResourceViewResolver();
vr.setPrefix("/WEB-INF/jsps/");
vr.setSuffix(".jsp");
return vr;
}
}
Finally, I have a #Controller in the package com.example.controllers
#Controller
public class ExampleController {
#RequestMapping(path = "/home", method = RequestMethod.GET)
public String example() {
return "index";
}
}
My application's context name is Example. When I send a request to
http://localhost:8080/Example/home
the application responds with an HTTP Status 404 and logs the following
WARN o.s.web.servlet.PageNotFound - No mapping found for HTTP request with URI `[/Example/WEB-INF/jsps/index.jsp]` in `DispatcherServlet` with name 'dispatcher'
I have a JSP resource at /WEB-INF/jsps/index.jsp I expected Spring MVC to use my controller to handle the request and forward to the JSP, so why is it responding with a 404?
This is meant to be a canonical post for questions about this warning message.
Your standard Spring MVC application will serve all requests through a DispatcherServlet that you've registered with your Servlet container.
The DispatcherServlet looks at its ApplicationContext and, if available, the ApplicationContext registered with a ContextLoaderListener for special beans it needs to setup its request serving logic. These beans are described in the documentation.
Arguably the most important, beans of type HandlerMapping map
incoming requests to handlers and a list of pre- and post-processors
(handler interceptors) based on some criteria the details of which
vary by HandlerMapping implementation. The most popular implementation
supports annotated controllers but other implementations exists as
well.
The javadoc of HandlerMapping further describes how implementations must behave.
The DispatcherServlet finds all beans of this type and registers them in some order (can be customized). While serving a request, the DispatcherServlet loops through these HandlerMapping objects and tests each of them with getHandler to find one that can handle the incoming request, represented as the standard HttpServletRequest. As of 4.3.x, if it doesn't find any, it logs the warning that you see
No mapping found for HTTP request with URI [/some/path] in DispatcherServlet with name SomeName
and either throws a NoHandlerFoundException or immediately commits the response with a 404 Not Found status code.
Why didn't the DispatcherServlet find a HandlerMapping that could handle my request?
The most common HandlerMapping implementation is RequestMappingHandlerMapping, which handles registering #Controller beans as handlers (really their #RequestMapping annotated methods). You can either declare a bean of this type yourself (with #Bean or <bean> or other mechanism) or you can use the built-in options. These are:
Annotate your #Configuration class with #EnableWebMvc.
Declare a <mvc:annotation-driven /> member in your XML configuration.
As the link above describes, both of these will register a RequestMappingHandlerMapping bean (and a bunch of other stuff). However, a HandlerMapping isn't very useful without a handler. RequestMappingHandlerMapping expects some #Controller beans so you need to declare those too, through #Bean methods in a Java configuration or <bean> declarations in an XML configuration or through component scanning of #Controller annotated classes in either. Make sure these beans are present.
If you're getting the warning message and a 404 and you've configured all of the above correctly, then you're sending your request to the wrong URI, one that isn't handled by a detected #RequestMapping annotated handler method.
The spring-webmvc library offers other built-in HandlerMapping implementations. For example, BeanNameUrlHandlerMapping maps
from URLs to beans with names that start with a slash ("/")
and you can always write your own. Obviously, you'll have to make sure the request you're sending matches at least one of the registered HandlerMapping object's handlers.
If you don't implicitly or explicitly register any HandlerMapping beans (or if detectAllHandlerMappings is true), the DispatcherServlet registers some defaults. These are defined in DispatcherServlet.properties in the same package as the DispatcherServlet class. They are BeanNameUrlHandlerMapping and DefaultAnnotationHandlerMapping (which is similar to RequestMappingHandlerMapping but deprecated).
Debugging
Spring MVC will log handlers registered through RequestMappingHandlerMapping. For example, a #Controller like
#Controller
public class ExampleController {
#RequestMapping(path = "/example", method = RequestMethod.GET, headers = "X-Custom")
public String example() {
return "example-view-name";
}
}
will log the following at INFO level
Mapped "{[/example],methods=[GET],headers=[X-Custom]}" onto public java.lang.String com.spring.servlet.ExampleController.example()
This describes the mapping registered. When you see the warning that no handler was found, compare the URI in the message to the mapping listed here. All the restrictions specified in the #RequestMapping must match for Spring MVC to select the handler.
Other HandlerMapping implementations log their own statements that should hint to their mappings and their corresponding handlers.
Similarly, enable Spring logging at DEBUG level to see which beans Spring registers. It should report which annotated classes it finds, which packages it scans, and which beans it initializes. If the ones you expected aren't present, then review your ApplicationContext configuration.
Other common mistakes
A DispatcherServlet is just a typical Java EE Servlet. You register it with your typical <web.xml> <servlet-class> and <servlet-mapping> declaration, or directly through ServletContext#addServlet in a WebApplicationInitializer, or with whatever mechanism Spring boot uses. As such, you must rely on the url mapping logic specified in the Servlet specification, see Chapter 12. See also
How are Servlet url mappings in web.xml used?
With that in mind, a common mistake is to register the DispatcherServlet with a url mapping of /*, returning a view name from a #RequestMapping handler method, and expecting a JSP to be rendered. For example, consider a handler method like
#RequestMapping(path = "/example", method = RequestMethod.GET)
public String example() {
return "example-view-name";
}
with an InternalResourceViewResolver
#Bean
public InternalResourceViewResolver resolver() {
InternalResourceViewResolver vr = new InternalResourceViewResolver();
vr.setPrefix("/WEB-INF/jsps/");
vr.setSuffix(".jsp");
return vr;
}
you might expect the request to be forwarded to a JSP resource at the path /WEB-INF/jsps/example-view-name.jsp. This won't happen. Instead, assuming a context name of Example, the DisaptcherServlet will report
No mapping found for HTTP request with URI [/Example/WEB-INF/jsps/example-view-name.jsp] in DispatcherServlet with name 'dispatcher'
Because the DispatcherServlet is mapped to /* and /* matches everything (except exact matches, which have higher priority), the DispatcherServlet would be chosen to handle the forward from the JstlView (returned by the InternalResourceViewResolver). In almost every case, the DispatcherServlet will not be configured to handle such a request.
Instead, in this simplistic case, you should register the DispatcherServlet to /, marking it as the default servlet. The default servlet is the last match for a request. This will allow your typical servlet container to chose an internal Servlet implementation, mapped to *.jsp, to handle the JSP resource (for example, Tomcat has JspServlet), before trying with the default servlet.
That's what you're seeing in your example.
I resolved my issue when in addition to described before:`
#Bean
public InternalResourceViewResolver resolver() {
InternalResourceViewResolver vr = new InternalResourceViewResolver();
vr.setPrefix("/WEB-INF/jsps/");
vr.setSuffix(".jsp");
return vr;
}
added tomcat-embed-jasper:
<dependency>
<groupId>org.apache.tomcat.embed</groupId>
<artifactId>tomcat-embed-jasper</artifactId>
<scope>provided</scope>
</dependency>
`
from: JSP file not rendering in Spring Boot web application
In my case, I was following the Interceptors Spring documentation for version 5.1.2 (while using Spring Boot v2.0.4.RELEASE) and the WebConfig class had the annotation #EnableWebMvc, which seemed to be conflicting with something else in my application that was preventing my static assets from being resolved correctly (i.e. no CSS or JS files were being returned to the client).
After trying a lot of different things, I tried removing the #EnableWebMvc and it worked!
Edit: Here's the reference documentation that says you should remove the #EnableWebMvc annotation
Apparently in my case at least, I'm already configuring my Spring application (although not by using web.xml or any other static file, it's definitely programmatically), so it was a conflict there.
Try to amend your code with the following change on your config file. Java config is used instead of application.properties.
Do not forget to enable configuration in configureDefaultServletHandling method.
WebMvcConfigurerAdapter class is deprecated, so we use WebMvcConfigurer interface.
#Configuration
#EnableWebMvc
#ComponentScan
public class WebConfig implements WebMvcConfigurer {
#Override
public void configureViewResolvers(ViewResolverRegistry registry) {
registry.jsp("/WEB-INF/views/", ".jsp");
}
#Override
public void configureDefaultServletHandling(DefaultServletHandlerConfigurer configurer) {
configurer.enable();
}
}
I use gradle, your should have the following dependencies in pom.xml:
dependencies {
compile group: 'org.springframework.boot', name: 'spring-boot-starter-web', version: '2.3.0.RELEASE'
compile group: 'org.apache.tomcat.embed', name: 'tomcat-embed-jasper', version: '9.0.35'
}
I came across another reason for the same error. This could also be due to the class files not generated for your controller.java file. As a result of which the the dispatcher servlet mentioned in web.xml is unable to map it to the appropriate method in the controller class.
#Controller
Class Controller{
#RequestMapping(value="/abc.html")//abc is the requesting page
public void method()
{.....}
}
In eclipse under Project->select clean ->Build Project.Do give a check if the class file has been generated for the controller file under builds in your workspace.
Clean your server. Maybe delete the server and add the project once again and Run.
Stop the Tomcat server
Right click the server and select "Clean"
Right click server again and select "Clean Tomcat Work Directory"
In my case using a tutorial for SpringBoot(2.7.3) RestController, startup failed with java.lang.ClassNotFoundException: javax.validation.ParameterNameProvider
I thought that Spring REST does not use WebMvc so I removed 'spring-boot-starter-web' and that resolved the startup problem. However, POST requests failed with the '404 Not Found' issue described here and spent several hours experimenting with
server.servlet.context-path
#RestController vs #Controller etc, and
SecurityConfig options
I finally resolved the 404 issue by
restoring dependency 'spring-boot-starter-web'
adding dependency
javax.validation
validation-api
and after undoing my 101 debug hacks it worked.
Very painful because even with root logger at DEBUG, there were no server-side logs to help.
For me, I found that my target classes were generated in a folder pattern not same as source. This is possibly in eclipse I add folders for containing my controllers and not add them as packages. So I ended up defining incorrect path in spring config.
My target class was generating classes under app and I was referring to com.happy.app
<context:annotation-config />
<context:component-scan
base-package="com.happy.app"></context:component-scan>
I added packages (not folders) for com.happy.app and moved the files from folders to packages in eclipse and it resolved the issue.
In my case, I was playing around with import of secondary java config files into a main java config file. While making secondary config files, I had changed the name of the main config class, but I had failed to update the name in web.xml. So, every time that I had restarted my tomcat server, I was not seeing mapping handlers noted in the Eclipse IDE console, and when I tried to navigate to my home page I was seeing this error:
Nov 1, 2019 11:00:01 PM org.springframework.web.servlet.PageNotFound
noHandlerFound WARNING: No mapping found for HTTP request with URI
[/webapp/home/index] in DispatcherServlet with name 'dispatcher'
The fix was to update the web.xml file so that the old name "WebConfig" would be instead "MainConfig", simply renaming it to reflect the latest name of the main java config file (where "MainConfig" is arbitrary and the words "Web" and "Main" used here are not a syntax requirement). MainConfig was important, because it was the file that did the component scan for "WebController", my spring mvc controller class that handles my web requests.
#ComponentScan(basePackageClasses={WebController.class})
web.xml had this:
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
com.lionheart.fourthed.config.WebConfig
</param-value>
</init-param>
web.xml file now has:
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
com.lionheart.fourthed.config.MainConfig
</param-value>
</init-param>
Now I am seeing the mapping in the console window:
INFO: Mapped "{[/home/index],methods=[GET]}" onto public
org.springframework.web.servlet.ModelAndView
com.lionheart.fourthed.controller.WebController.gotoIndex()
And my web page is loading again.
In my case, I had created Config.java (class) and also config.xml and mapping was done partially in both of them. And since config.java uses #Configuration annotation , it was considered priority. And was not considering config.xml.
If anyone gets in trouble like this , just delete config.java with annotation and try to keep config.xml , it works fine.
For me, the issue was hidden in the web.xml file.
Inside the servlet tag, you'll find:
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/todo-servlet.xml</param-value>
</init-param>
Make sure that in the <param-value> you have kept the correct location of the dispatcher servlet (aka Front Controller).
I had kept an incorrect location, hence I was able to view the homepage but all other pages were giving HTTP 404 error.
I had same problem as **No mapping found for HTTP request with URI [/some/path] in DispatcherServlet with name SomeName**
After I analyzed for 2 to 4 days I found out the root cause. Class files was not generated after I run the project. I clicked the project tab.
Project-->CloseProject-->OpenProject-->Clean-->Build project
Class files for source code have been generated. It solved my problem. To check whether class files have been generated or not, Please check the Build folder in your project folder.
So the problem can be as simple as an additional space in the path of the project. Make sure that there is no space in the path which took me quite some time to solve.
I have a web application which accesses an external web service. I'm writing an automated acceptance test suite for the web application. I do not want to invoke the external web service as it has serious overhead, I want to mock this web service. How can acheive this without changing the application context of the web application? We recently migrated to Spring 3.1 so I'm tempted to use the new environment features. Would these new feature help me in overriding this single web service and leave application context as it is?
I would use the Spring #Profile feature, which I assume is the "environment features" you were referring to.
For example:
#Service #Profile("dev")
public class FakeWebService implements WebService {
}
#Service #Profile("production")
public class ExternalWebService implements WebService {
}
EDIT
And to specify which profile to use in your test:
#RunWith(SpringJUnit4ClassRunner.class)
#ContextConfiguration("/app-config.xml")
#ActiveProfiles("dev")
public class MyAcceptanceTest {
}
See this section of the Spring docs for more details.
There are several ways to set the active profile in production, but the method I've used before is in the web.xml:
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>spring.profiles.active</param-name>
<param-value>production</param-value>
</init-param>
</servlet>
I would use a BeanFactoryPostProcessor to do this, which is only registered in the test scenarios that you want this mocked.
The BeanFactoryPostProcessor allows you to modify the Application Context immediately after it is created and populated. You can look for the name of your specific bean, and register a different bean for it.
public class SystemTestBeanFactoryPostProcessor implements BeanFactoryPostProcessor
{
#Override
public void postProcessBeanFactory(final ConfigurableListableBeanFactory factory) throws BeansException
{
final MyInterface myInterface = new MyInterfaceStub();
factory.registerSingleton("myInterfaceBeanName", myInterface);
}
}
This will allow you to overwrite only the beans that you want to stub/mock.
I am not sure that this is the 'newest 3.x' way to do this sort of thing. But it is very straightforward and easy to implement.