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I want to have a form that creates one of several subclasses in a type hierarchy. Say it's AbstractPerson with the subclasses Employee and Visitor.

Can I do that with a single Action / REST-Controller Bean?

Usually I use the form-ids smart, so it assigns values directly to the setters of my Action. So if I have a member like

AbstractPerson member;

I would try to use a form with an input field called "member.name".

However, struts must create an instance of AbstractPerson first - and it can't because it's abstract! It would be very cool if I could give struts2 a hint that it should actually create a Empolyee or Visitor object (depending on the form content). Is that or sth similar possible?

Cheers!

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1 回答 1

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这类似于我最近通过一小组 crud 操作访问实体类所做的事情。那是一些 crud 操作允许您对某个包中的所有实体类进行操作。您应该能够将此策略应用于您的 Employee 和 Visitor 类。

简而言之，它是如何工作的：

1) 您需要在命名空间或操作名称中指定应创建的类的名称。

2）您使用struts2s 可准备接口来创建模型。（反射性地创建从步骤 1 确定的类。

3）使用模型驱动接口，返回步骤2中定义的对象。这样对象在栈顶，你可以简单地说“name”，知道它是步骤中确定的Class的name属性1. 你可以避免这一步，但它不是那么漂亮。

现在这样做有一个小故障，您会发现要执行上述三个步骤，您需要一个自定义堆栈，即“staticParams-prepare-params”堆栈。

首先是一个示例，然后是该堆栈的定义以使其工作，如果您有任何问题，请告诉我：

package com.quaternion.demo.action.crud;

import com.quaternion.demo.orm.ActionValidateable;
import com.quaternion.demo.service.CrudService;
import com.quaternion.demo.util.ActionUtils;
import com.opensymphony.xwork2.ActionSupport;
import com.opensymphony.xwork2.ModelDriven;
import com.opensymphony.xwork2.Preparable;
import java.util.HashMap;
import java.util.Map;
import java.util.logging.Level;
import java.util.logging.Logger;
import org.apache.struts2.convention.annotation.Namespace;
import org.apache.struts2.convention.annotation.ParentPackage;
import org.apache.struts2.convention.annotation.Result;
import org.springframework.beans.factory.annotation.Autowired;

// Adds a new record to the database
@ParentPackage("staticParams-prepare-parms")
@Namespace("/crud/{entityName}")
@Result(type = "kjson") //TODO: could rid of this line by setting the result as the default for the package
public class AddAction extends ActionSupport implements Preparable, ModelDriven {

    private static final Logger log = Logger.getLogger(AddAction.class.getName());
    @Autowired
    private CrudService crudService;
    private String entityName; 
    private Object entityModel; 
    private Map jsonModel = new HashMap(); //for output, return the newly created object
    private Class clazz;

    @Override
    public String execute() throws ClassNotFoundException, InstantiationException, IllegalAccessException {
        log.log(Level.INFO, "In execute entityName is set with {0}", entityName);
        //If an id is passed in it will merge the object with that id, null will be used for unset attributes
        String status = SUCCESS;
        boolean error = false;
        Object entity = null;
        try { 
            entity = crudService.create(clazz, entityModel);
        } catch (Exception e) {
            error = true;
            status = ERROR;
            jsonModel.put("message", e.getMessage());
        }
        if (error == false) {
            jsonModel.put("entity", entity);
        }
        jsonModel.put("status", status);
        return SUCCESS;
    }

    public Object getEntityModel() {
        return entityModel;
    }

    public void setEntityModel(Object entityModel) {
        this.entityModel = entityModel;
    }

    public Object getJsonModel() {
        return jsonModel;
    }

    @Override
    public Object getModel() {
        return this.entityModel;
    }

    @Override
    public void prepare() throws Exception {
        log.log(Level.INFO, "In prepare entityName is set with {0}", entityName);
        clazz = ActionUtils.initClazz(entityName);
        entityModel = clazz.newInstance();
    }

    public String getEntityName() {
        return entityName;
    }

    public void setEntityName(String entityName) {
        this.entityName = entityName;
    }

    //TODO: validation would be a good idea can't implement in this class need to delegate
    //if entity implements a validate method, this validate should
    //call that validate
    @Override
    public void validate(){
        if (entityModel instanceof ActionValidateable){
            ((ActionValidateable)entityModel).validate(this);
        }
    }
}

这是堆栈的定义：

<package name="staticParams-prepare-parms" extends="struts-default">
    <result-types>
        <result-type name="kjson" default="true" class="com.quaternion.demo.result.Kjson"/>
    </result-types>
    <interceptors>
        <interceptor-stack name="staticParamsPrepareParamsStack">
            <interceptor-ref name="exception"/>
            <interceptor-ref name="alias"/>
            <interceptor-ref name="i18n"/>
            <interceptor-ref name="checkbox"/>
            <interceptor-ref name="multiselect"/>
            <interceptor-ref name="staticParams"/>
            <interceptor-ref name="actionMappingParams"/>
            <interceptor-ref name="servletConfig"/>
            <interceptor-ref name="prepare"/>
            <interceptor-ref name="chain"/>
            <interceptor-ref name="modelDriven"/>
            <interceptor-ref name="fileUpload"/>
            <interceptor-ref name="staticParams"/>
            <interceptor-ref name="actionMappingParams"/>
            <interceptor-ref name="params">
                <param name="excludeParams">dojo\..*,^struts\..*,^session\..*,^request\..*,^application\..*,^servlet(Request|Response)\..*,parameters\...*</param>
            </interceptor-ref>
            <interceptor-ref name="conversionError"/>
            <interceptor-ref name="validation">
                <param name="excludeMethods">input,back,cancel,browse</param>
            </interceptor-ref>
            <interceptor-ref name="workflow">
                <param name="excludeMethods">input,back,cancel,browse</param>
            </interceptor-ref>
        </interceptor-stack>
    </interceptors>
    <default-interceptor-ref name="staticParamsPrepareParamsStack"/>
</package>

您可能想知道 kjson 结果类型是什么。struts2-json 插件在其他几个动作上让我感到挑战。我创建了一个通用的分页和读取操作，“flexjson”默认情况下不会序列化集合，这可以防止延迟加载问题（在没有加载集合的情况下，这些简单的服务总是会出现这种情况）所以 kjson 只是一个使用 flexjson 返回 json 的结果类型。

于 2013-03-30T05:28:36.437 回答