Question

我是Spring的新手，从简单的教程开始。我定义了简单的jsp和Controller，然后将它映射到xml文档并运行它，但只有我看到的是我的wev页面没有我在控制器中传递的值：

@Controller
public class HomeController {

@Autowired
private ExampleService exampleService;

@RequestMapping(value = "/", method = RequestMethod.GET)
public String home(Model model) {
    model.addAttribute("serverTime", exampleService.getSystemTime());
    model.addAttribute("appVersion", exampleService.getAppVersion());
    return "home";
}
}

@Component
public class ExampleService {

@Value("#{appProperties.appVersion}")
private String appVersion;

/**
 * Returns formatted system time.
 * 
 * @return
 */
public String getSystemTime() {
DateFormat dateFormat = DateFormat.getDateTimeInstance(DateFormat.LONG, DateFormat.LONG);
return dateFormat.format(new Date());
}

public String getAppVersion() {
return appVersion;
}
}

<?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" 
xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" 
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>Spring</display-name>

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/spring-config.xml</param-value>
<!--        <param-value>/WEB-INF/jdbc-config.xml</param-value> -->
</context-param>

<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<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/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>

<?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:util="http://www.springframework.org/schema/util"
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">

<annotation-driven />

<resources mapping="/resources/**" location="/resources/" />

<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
    <beans:property name="prefix" value="/views/" />
    <beans:property name="suffix" value=".jsp" />
</beans:bean>

<context:component-scan base-package="com.home.spring" />

enter image description here

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:util="http://www.springframework.org/schema/util"
xsi:schemaLocation="http://www.springframework.org/schema/beans 
    http://www.springframework.org/schema/beans/spring-beans-3.0.xsd 
    http://www.springframework.org/schema/util 
    http://www.springframework.org/schema/util/spring-util-3.0.xsd">

<!-- Root Context: defines shared resources visible to all other web components -->

<util:properties id="appProperties" location="properties.properties"/>

我已经学习了一些教程，并阅读了与我的工作相关的官方文档的Web部分，但我还不了解代码中的问题。

我坚持了下来。如果您有任何想法，请与我分享。谢谢。

Answer 1

它对我来说很好。相同的代码。你也可以发布jsp。

我有以下jsp：如果你的jsp看起来很相似，请告诉我们。

<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
${serverTime}
</body>
</html>

编辑：从web.xml中删除所有内容并放在下面：

<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/appServlet-servlet.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
</servlet>

<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>

Spring：如何调用简单的控制器？

1 个答案: