Java、Spring、Apache Tiles 错误：无法解析名称为“spring”的 servlet 中名称为“contact”的视图

Java, Spring, Apache Tiles error : Could not resolve view with name 'contact' in servlet with name 'spring'

提问by Jaanus

controller:

控制器：

@Controller
@SessionAttributes
public class ContactController {

    @RequestMapping(value = "/addContact", method = RequestMethod.POST) 
    public String addContact(@ModelAttribute("contact") 
                            Contact contact, BindingResult result) {

         System.out.println("First Name:" + contact.getFirstName() +
                 "Last Name:" + contact.getLastName());

     return "redirect:contacts.html";
    }

    @RequestMapping("/contact")
    public ModelAndView showContacts() {

        return new ModelAndView("contact", "command", new Contact());
    }
}

this is my tiles.xml:

这是我的tiles.xml：

<tiles-definitions>
    <definition name="base.definition"
        template="/WEB-INF/jsp/layout.jsp">
        <put-attribute name="title" value="" />
        <put-attribute name="header" value="/WEB-INF/jsp/header.jsp" />
        <put-attribute name="menu" value="/WEB-INF/jsp/menu.jsp" />
        <put-attribute name="body" value="" />
        <put-attribute name="footer" value="/WEB-INF/jsp/footer.jsp" />
    </definition>

    <definition name="contact" extends="base.definition">
        <put-attribute name="title" value="Contact Manager" />
        <put-attribute name="body" value="/WEB-INF/jsp/contact.jsp" />
    </definition>

error is this:

错误是这样的：

org.apache.jasper.JasperException: javax.servlet.ServletException: Could not resolve view with name 'contact' in servlet with name 'spring'
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:584)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:456)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:389)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:333)
javax.servlet.http.HttpServlet.service(HttpServlet.java:722)

feel free to ask for any more source code

随意索取更多源代码

回答by Ralph

In your last questionyou showed that you used a UrlBasedView Resolver with an pre and suffix.

在上一个问题中，您表明您使用了带有前缀和后缀的 UrlBasedView 解析器。

Remove that pre and suffix.

删除该前缀和后缀。

That should be enogth:

那应该是：

<bean class="org.springframework.web.servlet.view.UrlBasedViewResolver"
    id="tilesViewResolver">
    <property name="viewClass"
        value="org.springframework.web.servlet.view.tiles2.TilesView" />
</bean>
<bean class="org.springframework.web.servlet.view.tiles2.TilesConfigurer"
    id="tilesConfigurer">
    <property name="definitions">
        <list>
            <value>/WEB-INF/layouts/layouts.xml</value>
            <!-- Scan views directory for Tiles configurations -->
            <value>/WEB-INF/views/**/views.xml</value>
        </list>
    </property>
</bean>

btw: this configruation allowes two tzps (only there sermatic differs) of tiles configuration files

顺便说一句：这个配置允许瓷砖配置文件的两个tzps（只有sermatic不同）

• the /WEB-INF/layouts/layouts.xmlcontains all the (lets call it) "base definitions"
• the /WEB-INF/views/**/views.xmlcontains the concrete definitions that extends the "base definitions" (for example the "contact" definition) -- You can have a seperate view.xmlfor each folder -- this is usefull if you group your views, for example all views (create, update, show and list) for each entity, in a separete folder

• 在/WEB-INF/layouts/layouts.xml包含了所有的（可以称之为）“基本定义”
• 在/WEB-INF/views/**/views.xml包含扩展“基本定义”（例如“联系人”的定义）的具体定义-你可以有一个单独的view.xml每个文件夹-这是有用的，如果你组的意见，例如，所有的意见（创建，更新, 显示和列出) 每个实体, 在一个单独的文件夹中

If you do not need this feature, then simply remove that line.

如果您不需要此功能，则只需删除该行即可。