Question

我正在使用struts2和Hibernate集成应用程序在数据库中插入值。但是从表单字段中插入值后，只有空值保存在数据库中; 这是我的表单jsp文件

employee.jsp

<%@ taglib  uri ="/struts-tags" prefix="s"%>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<s:form action="employee" method="post">
<s:textfield name ="name" label="ENTER your name"/>
<s:textfield name="address" label="Enter Address"/>
<s:submit label="submit">submit</s:submit>

</s:form>
</body>
</html>

实体类 Empmodel.java

package model;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;


import javax.persistence.Id;
@Entity

public class Empmodel {
@Id @GeneratedValue
private int serialno;
private String name;
private String address;
public String getName() {
    return name;
}
public void setName(String name) {
    this.name = name;
}
public String getAddress() {
    return address;
}
public void setAddress(String address) {
    this.address = address;
}   
}

Connectionfactory.java

import org.hibernate.SessionFactory;
import org.hibernate.cfg.Configuration;
import org.hibernate.service.ServiceRegistry;
import org.hibernate.service.ServiceRegistryBuilder;

public class ConnectionFactory {


private static  SessionFactory  sessionfactory = buildSessionFactory();
private static SessionFactory buildSessionFactory() {


    Configuration config =null;
     SessionFactory sf  =null;
    try{
        config= new Configuration().configure();
        ServiceRegistry servicereg = (ServiceRegistry) new ServiceRegistryBuilder().applySettings(config.getProperties()).buildServiceRegistry();

                sf= config.buildSessionFactory(servicereg); 


    }
    catch(Throwable e){
        System.out.println("Initial session factory creation failed"+ e);
        throw new ExceptionInInitializerError(e);
    }
    return sf;
}
 public static SessionFactory getSessionFactory(){
    return sessionfactory;
    }}

Empdao

import model.Empmodel;
import org.hibernate.Session;

public class Empdao {

public Empmodel add(){
    Empmodel model = new Empmodel();
    Session session=ConnectionFactory.getSessionFactory().getCurrentSession();

    session.beginTransaction();

    session.save(model);
    session.getTransaction().commit();
    return model;
}
}

行动类

import model.Empmodel;

import com.opensymphony.xwork2.ActionSupport;
import com.opensymphony.xwork2.ModelDriven;

import dao.Empdao;

public class Empaction extends ActionSupport implements ModelDriven<Empmodel> {
    private static final long serialVersionUID = 1L;
    private Empmodel model;


        public String execute() throws Exception{
            Empdao empdao = new Empdao();
            Empmodel queryresult =  empdao.add();

            if(queryresult != null) {
                return SUCCESS; 
            }
            else
                return ERROR;   
        }
        @Override
        public Empmodel getModel() {
            // TODO Auto-generated method stub
            return model;
        }
    }

Answer 1

如果你想先得到一些你需要放东西的东西。这条规则无处不在。在您的代码中，您需要将模型对象放到DAO中。然后DAO将使用模型中通过拦截器填充的值将其保存到数据库。例如

Empmodel queryresult = empdao.add(model);

为此，您需要更改方法签名以添加model的参数。

这件事似乎微不足道，但你需要删除在方法实现中重新创建model的语句，它不能正常工作。在开始交易之前，还要确保模型不为空。

DAO中的save方法应检查是否创建了null id值为if (model.getId() == null) session.save(model); else session.update(model); }的新对象。

{{1}}

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使用Struts2和hibernate在数据库中插入空值

1 个答案: