spring-3

Is it possible to run Tiles 3.0 with Spring 3.0? What is the needed configuration? The current version of Spring only officially supports Tiles 2. In order to get Tiles 3 working, you'd need to implement your own View Class, Tiles Configurer (if that is needed), etc. This is something the Spring community would appreciate very much, but no easy feat I'm sure. Hope this helps. 来源： https://stackoverflow.com/questions/11739777/tiles-3-0-spring-mvc-settings

Spring 3.0 MVC First of all, I haven't found any documentation regarding messages.properties @ springsource Everything I've found about overriding error messages has been on various forums. If anyone has a reference to where messages.properties is documented that'd be fantastic. Maybe messages.properties comes not from spring but a java spec? I have tried following the advice on JSR-303 Type Checking Before Binding My goal is to replace some type mismatch error messages with my own user friendly error messages My situation is as follows: Model public class Test { private int numberbomb; public

I am using Spring 3 ,java based configuration, with BootStrap. I have downloaded the bootstrap and put the css and js under resources directory. The issue that I cann't use these .css from within the freemarker page. Howeve that I imported them. As I am using the java based configuration,I have added the "addResourceHandler" as follows: WebAppConfig: @Configuration @EnableWebMvc @ComponentScan("com.springway") public class WebConfig implements WebApplicationInitializer { @Override public void onStartup(final ServletContext servletContext) throws ServletException { final

Hi I am running a maven test using maven 3.0.3 with hibernate 4.0.0 Final release and spring 3.1 on jdk7 update 2. I get the following error. Caused by: java.lang.VerifyError: Expecting a stackmap frame at branch target 63 in method ${myDomainClass}.equals(Ljava/lang/Object;)Z at offset 24 at java.lang.Class.getDeclaredMethods0(Native Method) at java.lang.Class.privateGetDeclaredMethods(Class.java:2442) at java.lang.Class.getDeclaredMethods(Class.java:1808) at org.hibernate.property.BasicPropertyAccessor.getterMethod(BasicPropertyAccessor.java:352) at org.hibernate.property

I have a java configuration class providing fooBean directly and barBean by component scan. @Configuration @ComponentScan(basePackages = { "com.blah" }) public class Config { @Bean public FooBean fooBean { return new FooBean(); } } and i want to reuse it in the test cases and i need to replace the beans with mocks: @Configuration @Import(Config.class) public class TestConfig { @Bean public FooBean fooBean { return new FooBeanMock(); } @Bean public BarBean barBean { return new BarBeanMock(); } } (here it does not make much sense to reuse Config, but in real life i have 1000 Beans and i need to

i've set up a roo application. The default view resolver built-in in roo is for jspx files. Is it possible to support also jsp files?. I tried configuring two viewResolvers but it seems that no urlBasedViewResolvers can coexist, its either one or the other. Changing the order does not affect the behaviour. If I set order =1 to the jspx then if i search for any .jsp file it gives me 404. The same if I search for jspx but jsp viewResolver is set with order =1. Is there anyway to do this? Thanks! here is my webmvc-config.xml <bean id="viewResolver" class="org.springframework.web.servlet.view

I'm new to Spring MVC and trying to learn from the internet tutorial such as these good ones from journaldev , springsource , and codejava All of those tell me use STS as the IDE and to go for New Project - Spring Template Project - Spring MVC Project . Though in my case, using the version 3.6.1.RELEASE of STS, I can NOT find any Spring MVC Project options there. Do you have the same issue using STS? p.s. I found the solution here for previous version of STS, though it not works for version 3.6.1. Seeing no MVC template snapshot. I've just solved the problem. These are my way: - Go to New

What's the proper way to programmatically log a web visitor in under a particular username in Spring and Spring Security 3.1? It seems the way I was doing it under 2.5 has changed a little. I'm sure there's a much better way of doing this now. Basically, when I create a new user, I need to also log them in at the same time. Create an Authentication (usually a UsernamePasswordAuthenticationToken ) and then call SecurityContextHolder.getContext().setAuthentication(authentication) This three lines of code do the work for me: Authentication request = new UsernamePasswordAuthenticationToken(

In my web application when I tries to logout it goes to j_spring_security_logout instead of the given page. In my spring-security.xml page i have added <logout logout-success-url="/login" delete-cookies="JSESSIONID" /> The problem is this worked earlier when I used spring security 3.1.4.RELEASE version. Now I'm using 3.2.2.RELEASE I've tried the following also. Didn't work <logout logout-url="/logout" delete-cookies="JSESSIONID" /> spring-security.xml <beans:beans xmlns="http://www.springframework.org/schema/security" xmlns:beans="http://www.springframework.org/schema/beans" xmlns:xsi="http:/