I got HTTP status 406 even with explicit Accept header like this: curl -x localhost:8088 -H 'Accept: application/jso... The root cause was that I forgot to include Jackson JSON lib in the dependencies. Solution: add this to pom.xml <dependency> <groupId>org.codehaus.jackson</gr...

It's normal practice to import types from an external xsd file in WSDL like this: <wsdl:types> <xsd:schema xmlns:xsd="htt... When you use <dynamic-wsdl> and have Commons XMLSchema on the class path, Spring-WS inlines the xsd in the wsdl. But that doesn't happen when you use <static-wsdl> . You can define a SimpleXsdSchema bean to expose the xsd: <?xml version="1.0" encoding="UTF-8"?> <beans x... where the bean id "hello" should match the schemaLocation attribute in the WSDL (without the .xsd suffix). But note that the SimpleXsdSchema does not inline the xsd. It only makes the xsd available via an HTTP URL. Alternatively, you can simply put the xsd file under the content directory of the webapp (just link any CSS or JavaScript). Anyway, that's a lot of manual...

This is a step-by-step guide to create a "contract-first" web service with Apache CXF. It's a lot easier than doing the same thing with Spring-WS. The project uses standard Maven directory layout. Define the data types ( src/main/resources/hello.xsd ): <xs:schema xmlns:xs="http://www.w3.org/200... Define the service ( src/main/resources/hello.wsdl ): <?xml version='1.1' encoding='UTF-8'?> <wsdl:de... Create pom.xml : <project xmlns="http://maven.apache.org/POM/4.... Generate jaxb bindings: $ mvn generate-sources Code the web service ( src/main/java/com/example/cxfdemo/HelloPortImpl.java ): package com.example.cxfdemo; import javax.j... Declare the CXF servlet in web.xml ( src/main/webapp/WEB-INF/web.xml ): <?xml version="1.0" encoding="UTF-8"?> <web-app... Wire up the web service implementation ( src/main/webapp/WEB-INF/cxf-servlet.xml ): <?xml version="1.0" encoding="UTF-8"?> <beans x... Build the WAR: $ mvn clean package After the webapp is deployed (Tomcat running on port 8080), the web service (WSDL) is available via...

Spring-WS documentation says you can use a Jaxb object as parameter or return type, provided that it is annotated with javax.xml.bind.annotation.XmlRootElement , or is an instance of javax.xml.bind.JAXBElement . But that's a lot easier said than done! For example, if sayHelloResponse is defined as: <xs:element name="sayHelloResponse" type="tns:sayH... then the JAXB generated class is not annotated with XmlRootElement , therefore, unusable for Spring-WS. You have to define the type as: <xs:element name="sayHelloResponse"> <xs:compl... in order to generate a type annotated with XmlRootElement . But that is not always possible. Alternatively, you can use the Maven plugin maven-jaxb2-plugin with the jaxb2-basics-annotate plugin (yes, plugin inside plugin) to inject the XmlRootElement annotation into the generated JAXB class. This is the pom: <project xmlns="http://maven.apache.org/POM/4.0.0"... and the binding file: <?xml version="1.0" encoding="UTF-8" standalone="y......

This is a jaxb binding file that generates java.util.Calendar for xs:dateTime : <?xml version="1.0" encoding="UTF-8" standalone="y...

After a user resets a password, I want to force the user to change the password before she gets access to secured content. This is usually done with a servlet filter. But with Spring MVC, you can also use a HandlerInterceptor . According to Spring JavaDoc: HandlerInterceptor is basically similar to a Servlet 2.3 Filter, but in contrast to the latter it just allows custom pre-processing with the option of prohibiting the execution of the handler itself, and custom post-processing. Filters are more powerful, for example they allow for exchanging the request and response objects that are handed down the chain. Note that a filter gets configured in web.xml, a HandlerInterceptor in the application context. As a basic guideline, fine-grained handler-related preprocessing tasks are candidates...

Create a jax-ws web service with Spring, Apache CXF and Maven. Create the pom.xml : <project xmlns="http://maven.apache.org/POM/4.0.0"... Create the web service interface ( src/main/java/jaxws/JaxwsHello.java ): package jaxws; import javax.jws.WebService;... Implement the web service ( src/main/java/jaxws/JaxwsHelloImpl.java ): package jaxws; import javax.jws.WebService;... Create src/main/webapp/WEB-INF/cxf-servlet.xml : <?xml version="1.0" encoding="UTF-8"?> <beans x... Register the CXF servlet in src/main/webapp/WEB-INF/web.xml : <?xml version="1.0" encoding="UTF-8"?> <web-app... Build: mvn package The resulting WAR file can be deployed to any servlet container (for example, Tomcat).

This is probably the easiest way to create a web service in JAX-WS. There are no external dependencies other than Java EE. Assuming that you build the web service as a webapp (say jaxws-example.war), the pom.xml can be as simple as: <project xmlns="http://maven.apache.org/POM/4.0.0"... You can implement and deploy the web service in 3 easy steps: Code the service as a POJO (annotate class to expose it as a web service) package jaxws; import javax.jws.WebMethod; ... Declare the POJO as a servlet in WEB-INF/web.xml : <?xml version="1.0" encoding="UTF-8"?> <web-app... Build the webapp, and deploy the resulting war: mvn package The only catch is, this only works for a Java EE 5+ compliant container such as WebLogic or JBoss. It does not work for a simple servlet...

Use this link to lookup information on an IP address: http://bgp.he.net/ip/204.193.144.69 Whois lookup is also supported. Clean page, no clutter.

To find out the DNS server ip address your computer is using, load this page in your browser: http://myresolver.info/