Аз съм абсолютно начинаещ с Java и Spring и искам да се уча от пример.
Използвам конфигурация/инсталация извън кутията
- Mac OSX
- Springsource Tool Suite като IDE
- Spring 2.8.1.RELEASE
- vfabric-tc-server-developer-2.6.1.RELEASE
Опитах се да генерирам нов проект, базиран на "Spring Template Project". След това избрах „Пролетен MVC проект“. Примерният проект е генериран. След това, без да променям нищо, се опитах да изпълня страницата "home.jsp" чрез "Run As". Уеб сървърът стартира и накрая получих грешката в раздела на конзолата.
Не е намерено съпоставяне за HTTP заявка с URI [/myproject/] в DispatcherServlet с име „appServlet“
И този друг резултат в тези уеб страници:
http://localhost:8080/myproject/WEB-INF/views/home.jsphttp://localhost:8080/myproject
Тук можете да видите изображение за това как е структуриран моят проект (генериран автоматично за STS):
Какво не е наред?
Тук можете да видите съдържанието на файла web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
Файлът root-context.xml има това съдържание.
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd">
<!-- Root Context: defines shared resources visible to all other web components -->
</beans>
И накрая servlet-context.xml има това съдържание.
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->
<!-- Enables the Spring MVC @Controller programming model -->
<annotation-driven />
<!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>
<context:component-scan base-package="com.mycompany.myapp" />
</beans:beans>
Някой има ли идея за решаването му?
