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Spring 4.3.3 - ParameterizableViewController POST method not more supported

After upgrading to Spring 4.3.3.RELEASE i get the error:
Request method 'POST' not supported
My application is a basic template and the home view is rendered via
<mvc:view-controller path="/" view-name="home.view"/>
It works fine on Spring 4.2.8.
Any hint to solve the problem?

We ran into the same problem. It turns out that, at some point, the ParameterizableViewController was changed to only support GET and HEAD requests.
We resolved this by replacing the definition with something like this:
<bean id="homeController" class="org.springframework.web.servlet.mvc.ParameterizableViewController">
<property name="supportedMethods" value="GET,POST,PUT,DELETE" />
<property name="viewName" value="home.view" />
</bean>
<bean class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<map>
<entry key="/" value-ref="homeController"/>
</map>
</property>
</bean>
Essentially, this allows you to create a ParameterizableViewController with whatever supported HTTP methods you wish. The second bean creates the mapping so that the path "/" resolves to the defined controller.

ParameterizableViewController default supported methods are GET,HEAD we are check it with the following code snippet.
ParameterizableViewController pvc=new ParameterizableViewController();
String[] str=pvc.getSupportedMethods();
for(String x:str) {
System.out.println(x);
}
in order to add POST or any HTTP method, we need to add this XML tag in our bean tag.
<bean id="testUrl"
class="org.springframework.web.servlet.mvc.ParameterizableViewController">
<property name="supportedMethods" value="GET,POST,PUT,DELETE" />
<property name="viewName" value="success" />
</bean>

Switch LDAP connection at runtime in Spring

I am new to spring. Admins of my spring based web app want to configure settings from the web interface, so users can authenticate against LDAP server with their company username and password.
Change in LDAP settings should be possible without restarting the application. This might happen during a 'migration' or whatever reason. I have a couple beans, which need to be refreshed after the admin saves new settings for the LDAP server:
<bean id="ldapServer" class="org.springframework.security.ldap.DefaultSpringSecurityContextSource">
<constructor-arg>
<list>
<value>${ldap.url1}</value>
...
</list>
</constructor-arg>
<constructor-arg value="${ldap.basedn}"</constructor-arg>
<property name="referral" value="${ldap.referral}" />
<property name="baseEnvironmentProperties">...</property>
<property name="userDn" value="${ldap.username}" />
<property name="password" value="${ldap.password}" />
</bean>
I am using Springframework 3.1.2. The problem is, there are constructor arguments, which I want to change and not affect other running jobs. I tried playing with Scoped proxy, but not to much success yet:
<bean id="ldapServer" scope="prototype" ...>
<aop:scoped-proxy/>
I was successful though to get ldapServer to reinstantiate, when using prototype scope by running this piece of code:
#Controller
public class LDAPSettingsController implements ApplicationContextAware {
public ModelAndView handleRequest(...) {
DefaultSpringSecurityContextSource ldap;
ldap = context.getParentBeanFactor().getBean("ldapServer");
System.out.println(ldap.hashCode());
return new ModelAndView(new RedirectView('login.jsp'));
}
...
}
Are scopes and proxies here the way to go, or is the another mechanism in Spring to reflect configuration changes into a running program instance?
UPDATE: Clear up the question.
UPDATE: The root problem with the AOP proxies was following root exception:
java.lang.IllegalArgumentException: Superclass has no null constructors but no arguments were given

What worked was adding proxy-target-class="false" attribute to the <aop:scoped-proxy/> tag. I created a new scope, which works better than prototype - It destroys beans on settings update. Now I have this in my beans.xml:
<bean class="org.springframework.beans.factory.config.CustomScopeConfigurer">
<property name="scopes">
<map>
<entry key="ldap">
<ref bean="ldapScope" />
</entry>
</map>
</property>
</bean>
<bean id="ldapScope" class="com.myapp.SettingsScope" />
<bean id="ldapServer" scope="ldap" ...>
<aop:scoped-proxy proxy-target-class="false"/>
<constructor-args>
<list><value>${ldap.url1}</value> .. </list>
</constructor-args>
...
</bean>
I also have a controller for LDAP settings into which I inject ldapScope and I call a method which destroys current life-cycle objects and starts a new life-cycle every time, user presses the apply button.
PS: Not sure if I handle the life-cycle "re-start" in the right way - people my way to look for auto-start beans and start them after such event happens (i.e.: Setting -> Apply)

Conditional initialization of classes in spring

I have a service which refers to a single source.
<bean id="XYZService" class="com.services.impl.DataService1">
<constructor-arg ref="DataSource1" />
</bean>
<bean id="DataSource1" class="com.source.impl.DataSource1">
<constructor-arg ref="DBDataSource"/>
<constructor-arg value="xyz"/>
</bean>
<bean id="DataSource2" class="com.source.impl.DataSource2">
<constructor-arg ref="MsgDataSource"/>
<constructor-arg value="xyz"/>
</bean>
Now if i want to perform a conditional check and my service should be able listen to particular source based on a input variable something like below.
<bean id="XYZService" class="com.services.impl.DataService1">
<constructor-arg ref=" $VARIABLE == true ? DataSource1 : DataSource2" />
</bean>
I did tried SPEL however no luck. I am beginner in spring. Any help will be appreciated.
Thanks.

There are many solutions. Here are two: You can use profiles for this. Define two profiles, define the DataSource beans with the same name but different profiles. (docs)
Alternatively, you can use a single bean and a static factory method (docs).
<bean id="DataSource" class="com.source.impl.DataSourceFactory"
factory-method="createInstance"/>
Inside of DataSourceFactory.createInstance(), you can check the flag and then create the correct data source in plain Java.
The latter is a bit easier to understand, IMO. Using profiles allows you to keep everything in XML (but you should really consider switching to the Java Configuration). The drawback with profiles is that you must not forget to activate at least one of the bean won't be defined.
A third option is to use three XML files and then modify the list of XML files that should be parsed when you pass it to the ApplicationContext. But that only works if you have control over this part of the code.

Assuming you are using Spring 3.1 or later, Spring Profiles may be the best solution.
Using the example of Production and Dev/QA environments, common bean declarations go in a shared file
<beans>
<bean id="XYZService" class="com.services.impl.DataService1">
<constructor-arg ref="DataSource" />
</bean>
</beans>
A separate configuration contains production references
<beans profiles="prod">
<bean id="DataSource" class="com.source.impl.DataSource1">
<constructor-arg ref="DBDataSource"/>
<constructor-arg value="xyz"/>
</bean>
</beans>
Another contains dev references
<beans profile="dev">
<bean id="DataSource" class="com.source.impl.DataSource2">
<constructor-arg ref="MsgDataSource"/>
<constructor-arg value="xyz"/>
</bean>
</beans>
To activate the given profile add -Dspring.profiles.active=prod to your JVM arguments
You can find more info here
Another approach uses factory methods.
<bean id="DataSource" class="com.source.impl.DataSourceFactory" factory-method="getInstance">
<constructor-arg value="#{VARIABLE}" />
</bean>
The above fragment assumes that you want your factory method to explcitly invoke the constructor of each of your services. If you dead set on using Spring to create the instances you can pass each datasource implementation as constructor arguments and use the constructor method as a simple dispatcher.

You need something like this:
<constructor-arg
ref="#{systemProperties.variable == 'true' ? 'DataSource1' : 'DataSource2'}" />
where "variable" is set like -Dvariable=true.

Is there a built-in Spring environment variable for the web context root?

I'm using Spring Web Services to expose a service as a web service. In my spring configuration xml file, I have a bean which is an instance of DefaultWsdl11Definition. One of the properties that needs to be set is the locationUri. This needs to be a fully qualified Uri, but I don't want to have to change this value when the application gets promoted from dev to uat and to production. Spring knows what the web application context root is, so is there a variable that I can specify in my configuration file to access it?
Something like:
<bean id="myWebServices"
class="org.springframework.ws.wsdl.wsdl11.DefaultWsdl11Definition">
<property name="schemaCollection">
<bean
class="org.springframework.xml.xsd.commons.CommonsXsdSchemaCollection">
<property name="xsds" ref="xsdList"/>
<property name="inline" value="true" />
</bean>
</property>
<property name="portTypeName" value="myWebServices" />
<property name="locationUri" value="${webContextRoot}/webServices" />
</bean>

With Spring 3.0, you should be able to access the servlet context through the servletContext bean in the web application context:
<property name="locationUri" value="#{servletContext.contextPath}/webServices" />
If you're using pre-Spring-EL (before 3.0), you should be able to do
<bean name="servletContextBean" class="org.springframework.web.context.support.ServletContextFactoryBean" />
<bean name="contextPath" factory-bean="servletContextBean" factory-method="getContextPath" />
<bean name="locationUri" factory-bean="contextPath" factory-method="concat">
<constructor-arg value="/webServices" />
</bean>
and inside your myWebservices bean
<property name="locationUri" ref="locationUri" />
EDIT:
I don't think getting the server name and port from the ServletContext, as depending on the setup the web container may not know the hostname (i.e. a HTTP server may be in front of the web container, e.g. tomcat may be behind an Apache web server or depending on the Websphere configuration).
However, the following may be part of a solution to get the hostname. With Spring 3.0, you could do the following:
<property name="host"
class="java.net.InetAddress"
factory-method="getLocalHost"/>
<property name="locationUri"
value="http://#{host.canonicalHostName}:8080/#{servletContext.contextPath}/webServices" />

I had a similar problem that you described, I use property files to do this
ws_dev.properties
ws_prod.properties
I configured my property file like this, The deployment property is java vm argument like
-Ddeployment=dev
<context:property-placeholder location="ws_${deployment}.properties"/>

May be late, but may some other need a solution too:
set property in servlet:
web.xml
<servlet>
<servlet-name>spring-ws</servlet-name>
<servlet-class>org.springframework.ws.transport.http.MessageDispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:/spring-ws-context.xml</param-value>
</init-param>
<init-param>
<param-name>transformWsdlLocations</param-name>
<param-value>true</param-value>
</init-param>
</servlet>
The bean declaration in spring-ws-context.xml:
<bean id="WebService"
class="org.springframework.ws.wsdl.wsdl11.DefaultWsdl11Definition"
p:portTypeName="App" p:locationUri="/WebServices" p:requestSuffix="Request"
p:responseSuffix="Response">
<property name="schema">
<bean class="org.springframework.xml.xsd.SimpleXsdSchema" p:xsd="classpath:/requestTypes.xsd" />
</property>
</bean>

You can add ApplicationContextAware interface to your bean, cast it to WebApplicationContext and then get ServletContext. Also see class org.springframework.web.context.ContextLoader

Spring MVC - Form Mapping

Probably missing something completely obvious here, but here goes. I'm starting out with Spring MVC. I have a form controller to process inbound requests to /share/edit.html. When I hit this url from my browser, I get the following error:
The requested resource (/inbox/share/share/edit) is not available.
Here is my applicationContext-mvc.xml:
<bean id="publicUrlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping" >
<property name="mappings" >
<value>
/share/edit.html=shareFormController
/share/list.html=shareController
/share/view.html=shareController
/folders.json=foldersController
/studies.json=studiesController
</value>
</property>
</bean>
<bean id="internalPathMethodNameResolver" class="org.springframework.web.servlet.mvc.multiaction.InternalPathMethodNameResolver" />
<bean id="shareFormController" class="com.lifeimage.lila.controller.ShareFormController" />
<bean id="shareController" class="com.lifeimage.lila.controller.ShareController" >
<property name="methodNameResolver" ref="internalPathMethodNameResolver" />
</bean>
and my form Controller:
public class ShareFormController extends SimpleFormController {
public ShareFormController() {
setCommandClass( Share.class );
}
#Override
protected ModelAndView onSubmit(HttpServletRequest request, HttpServletResponse response, Object command, BindException errors)
throws Exception {
//controller impl...
}
}

You should look at your view resolver. Make sure that it is resolving the logical name in your controller as you think it should. Looks like the name it is resolving it to does not exist currently

I think I've resolved this issue. There were two problems:
1) Implementations of SimpleFormController require a form and success view; which I had not configured here. As this is a server method for an AJAX client, I added a Spring-JSON view as follows:
<?xml version="1.0" encoding="UTF-8"?>
http://www.springframework.org/schema/beans/spring-beans-2.5.xsd"
default-lazy-init="false" default-autowire="no"
default-dependency-check="none">
<bean name="jsonView" class="org.springframework.web.servlet.view.json.JsonView">
<property name="jsonErrors">
<list>
<ref bean="statusError" />
<ref bean="modelflagError" />
</list>
</property>
</bean>
<bean name="statusError"
class="org.springframework.web.servlet.view.json.error.HttpStatusError">
<property name="errorCode"><value>311</value></property>
</bean>
<bean name="modelflagError"
class="org.springframework.web.servlet.view.json.error.ModelFlagError">
<property name="name"><value>failure</value></property>
<property name="value"><value>true</value></property>
</bean>
which can be used for all controllers that return JSON.
2) I switched from a SimpleURLHandlerMapping to ControllerClassNameHandlerMapping and relied on Spring naming conventions ( controllerClassName/method.html ), which fixed the routing issue. Might not be a long term solution, but got me through the task.

Did you check your log output? Spring MVC is generally pretty verbose in what it outputs.
Also, the URL you've posted (/inbox/share/share/edit) does not seem to match what you are configuring (/share/edit.html).

#jordan002 when I see all the hoops you had to jump to accomplish your task, I feel obliged to share a very powerful Java MVC framework that requires much less configuration. The framework is called Induction, check out the article Induction vs. Spring MVC, http://www.inductionframework.org/induction-vs-spring-mvc.html