use*_*862 4 java spring jsp spring-mvc request
我有一个非常简单的问题.在我的.jsp文件中,我有**/registration的链接,方法viewRegistration被执行,一切正常.如果我有**/registration/getTags的链接?找不到tagName = blahblah页面.我不知道为什么,因为我认为我的requestMapping注释看起来是正确的......我将非常感谢你的帮助!
控制器:
@Controller
@RequestMapping(value = "/registration")
public class HelloController {
final static Logger logger = Logger.getLogger(HelloController.class);
@RequestMapping(method = RequestMethod.GET)
public String viewRegistration(Map<String, Object> model, HttpSession session) {
...
}
@RequestMapping(value = "/getTags", method = RequestMethod.GET)
public @ResponseBody
List<Tag> getTags(@RequestParam String tagName) {
....
}
}
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WEB.XML:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" version="3.0">
<display-name>aa</display-name>
<servlet>
<servlet-name>xxx</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>xxx</servlet-name>
<url-pattern>/registration/*</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/xxx-servlet.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
</web-app>
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xxx-servlet.xml:
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="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
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd">
<context:component-scan base-package="movies.controller" />
<context:property-placeholder location="/WEB-INF/properties/website.properties" />
<context:component-scan base-package="com" />
<context:annotation-config />
<mvc:annotation-driven />
<bean class="org.springframework.web.servlet.view.tiles3.TilesConfigurer"
id="tilesConfigurer">
<property name="definitions">
<list>
<value>/WEB-INF/tile/tilesJsp.xml</value>
</list>
</property>
</bean>
<mvc:resources mapping="/resources/**" location="/resources/" />
<bean class="org.springframework.web.servlet.view.UrlBasedViewResolver"
id="viewResolver">
<property name="viewClass"
value="org.springframework.web.servlet.view.tiles3.TilesView" />
</bean>
<bean id="messageSource"
class="org.springframework.context.support.ResourceBundleMessageSource">
<property name="basename" value="website" />
</bean>
</beans>
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编辑编辑:我甚至尝试过更简单:
@RequestMapping(value = "/getTags")
@ResponseBody
public List<Tag> getTags() {
String tagName="";
return simulateSearchResult(tagName);
}
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但仍然/ registration/getTags不起作用...,找不到页面.
的servlet
URL映射如下:
<servlet-mapping>
<servlet-name>xxx</servlet-name>
<url-pattern>/registration/*</url-pattern>
</servlet-mapping>
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这意味着所有urls
开头/registration/*
的servlet
东西都会被它所处理,它将由它来处理controller
.因此,如果您使用控制器配置控制器,@RequestMapping(value="/otherURL")
它将提供以下服务Url
:
http://localhost:xxxx/<appname>/registration/otherURL
在这种情况下,为了访问正确的方法,您应该:
@RequestMapping("/registration
)来自控制器,因为它已经映射到servlet级别,并调用:
http://localhost:xxxx/<appname>/registration/getTags?tagName=blahblah
会正常工作.
或者:
请拨打以下网址: http://localhost:xxxx/<appname>/registration/registration/getTags?tagName=blahblah
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