Question

如何将这部分代码由struts1改为struts2实现

由于我对struts1不熟 不知道如何修改。。。。。。。。部分代码如下：

struts.xml部分
<struts-config>
<action path="/control/product/type/list">
<forward name="list" path="test.jsp"/>
</action>
<controller>
<set-property property="processorClass" value="org.springframework.web.struts.DelegatingRequestProcessor"/>
</controller>
</struts-config>

action类部分：

@Controller("/contol/product/type/list")
public class ProductTypeAction extends Action {
@Resource(name="productTypeServiceBean")
private ProductTypeService productTypeService;

public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response) throws Exception {
ProductType productType = productTypeService.find(ProductType.class, 3);
request.setAttribute("productType", productType);
return mapping.findForward("list");
}
}

test.jsp页面中只有个取值 ${productType.name}

主要就是这部分了 ， 单独struts2测试成功，hibernate与spring整合成功，实体类 包等都修改好了， 如何将以上struts1修改为struts2，主要功能就是通过struts转到test.jsp页面，通过hibernate取到值，如何修改啊。。。。。感谢。。。。。

Answer 1

public class ProductTypeAction extends ActionSupport{
private ProductType productType ;
@Resource(name="productTypeServiceBean")
private ProductTypeService productTypeService;

public String getProductTypeById() throws Exception {
productType = productTypeService.find(ProductType.class, 3);
return "SUCCESS";
}
}
struts.xml
<package name="aaa" extends="struts-default">
<action method="getProductTypeById" name="getProductTypeById" class="xxx">
<result name="SUCCESS">/text.jsp</result>
</action>
</package>

Answer 2

<action class="control.product.type.list" method="execute">
<result name="list">/WEB-INF/test.jsp</result>
</action>