我一直在尝试将Tiles与基于Struts 2注释的操作集成,但它无法正常工作。
由于我没有struts-config.xml
,并且在网络上提供的每个教程中,他们都使用struts-config.xml
引用它。
首先,可以将基于注释的struts动作与切片集成。如果是,那么如何?
@Action(value="/login",results={@Result(name="success",location="/home",type=TilesResult.class),
@Result(name="login",location="/jsp/userLogin.jsp")})
public String execute() {
这是我的代码,但它始终在TilesResult.class
的MyEclipse中给我错误
Type mismatch: cannot convert from Class<TilesResult> to String
我对我的pom依赖:
<dependency>
<groupId>org.apache.struts</groupId>
<artifactId>struts2-tiles-plugin</artifactId>
<version>2.1.8</version>
</dependency>
任何人都可以帮助我如何在基于注释的操作中添加切片
我使用type="tiles"
代替type=TilesResult.class
,然后它给了我以下异常
Caused by: The Result type [tiles] which is defined in the Result annotation on the class [class com.actions.LoginAction] or determined by the file extension or is the default result type for the PackageConfig of the action, could not be found as a result-type defined for the Struts/XWork package [com.actions#convention-default#] - [unknown location]
at org.apache.struts2.convention.DefaultResultMapBuilder.createResultConfig(DefaultResultMapBuilder.java:422)
at org.apache.struts2.convention.DefaultResultMapBuilder.createFromAnnotations(DefaultResultMapBuilder.java:394)
at org.apache.struts2.convention.DefaultResultMapBuilder.build(DefaultResultMapBuilder.java:202)
at org.apache.struts2.convention.PackageBasedActionConfigBuilder.createActionConfig(PackageBasedActionConfigBuilder.java:800)
at org.apache.struts2.convention.PackageBasedActionConfigBuilder.buildConfiguration(PackageBasedActionConfigBuilder.java:586)
at org.apache.struts2.convention.PackageBasedActionConfigBuilder.buildActionConfigs(PackageBasedActionConfigBuilder.java:318)
at org.apache.struts2.convention.ClasspathPackageProvider.loadPackages(ClasspathPackageProvider.java:53)
at com.opensymphony.xwork2.config.impl.DefaultConfiguration.reloadContainer(DefaultConfiguration.java:204)
at com.opensymphony.xwork2.config.ConfigurationManager.getConfiguration(ConfigurationManager.java:55)
答案 0 :(得分:4)
试试这些:
使用type="tiles"
代替type="TilesResult.class"
在结果位置使用目标图块定义location="tiles-definition-name"
,而不是JSP页面location="/jsp/userLogin.jsp"
在web.xml
<context-param>
<param-name>org.apache.tiles.impl.BasicTilesContainer.DEFINITIONS_CONFIG</param-name>
<param-value>/WEB-INF/tiles.xml</param-value>
</context-param>
<listener>
<listener-class>org.apache.struts2.tiles.StrutsTilesListener</listener-class>
</listener>
在struts.xml
中有以下内容(如果您单独使用注释而没有struts.xml
,那么您必须为此创建一个最小的注释,因为没有可用于定义自定义结果的注释型)
<struts>
<constant name="struts.convention.default.parent.package" value="codeoftheday.blogspot.com"/>
<package name="codeoftheday.blogspot.com" extends="struts-default">
<result-types>
<result-type name="tiles" class="org.apache.struts2.views.tiles.TilesResult" />
</result-types>
</package>
</struts>
注意:我已就此问题撰写了详细的博客文章 - Maven, Struts2 Annotations and Tiles Integration Example via Convention / Codebehind / Zero Config plugin using Eclipse IDE
答案 1 :(得分:0)
请将您的web.xml与以下代码进行比较。
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<filter>
<filter-name>struts2</filter-name>
<filter-class>org.apache.struts2.dispatcher.filter.StrutsPrepareAndExecuteFilter</filter-class>
<init-param>
<param-name>actionPackages</param-name>
<param-value>com.demo.action</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>struts2</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<context-param>
<!-- <param-name>org.apache.tiles.impl.BasicTilesContainer.DEFINITIONS_CONFIG</param-name> -->
<param-name>org.apache.tiles.definition.DefinitionsFactory.DEFINITIONS_CONFIG</param-name>
<param-value>/WEB-INF/config/tiles.xml</param-value>
</context-param>
<listener>
<listener-class>org.apache.struts2.tiles.StrutsTilesListener</listener-class>
</listener>