Spring: Basit kontrolcüyü nasıl çağırırsınız?

Question

Spring’de yeniyim ve basit eğiticilerle başladım. Ben jsp basit tanımlar ve Denetleyici, daha sonra bunu xml belgesini eşlenen ve çalıştırın, ama sadece ne var testere ben Kontrolcü içinde geçti değerler olmadan bir benim wev sayfası:

@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" /> 

<?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"/> 

Birkaç öğreticiyi öğrendim ve çalışmamla ilgili resmi belgelerin bir web bölümünü okudum ama henüz anlamadım, kodumda nerede sorun var.

Buna uyuyorum. Herhangi bir fikriniz varsa, lütfen benimle paylaşın. Teşekkürler.

Answer 1

Benim için iyi çalışıyor. Aynı kod. Lütfen jsp dosyasını da gönderir misiniz?

Aşağıda jsp var: Jsp'niz benzer görünüyorsa lütfen bize bildirin.

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

DÜZENLEME: web.xml''deki her şeyi kaldırmak ve aşağıda koyun:

<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>