关于Structs+Spring+Hibernate项目的构建(1)
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1. 准备活动:
MySQL 5.0安装好,mysql-connector-java-5.1.5-bin.jar的jar包
MyEclipse5.5M2 ,Tomcat5.5.23(感觉还是这个稳定性好)
2. 创建User表
create table user
(
id int(16) primary key auto_increment,
username varchar(64) not null,
password varchar(64) not null
);
(
id int(16) primary key auto_increment,
username varchar(64) not null,
password varchar(64) not null
);
新建项目,然后用MyEclipse加载框架按 Structs,Spring,Hibernate的顺序加载
注意:中间选择用Spring 管理Hibernate的Bean
3. 配置web.xml
在web.xml最上面加入
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:applicationContext.xml</param-value>
</context-param>
<servlet>
<servlet-name>springInitServlet</servlet-name>
<servlet-class>
org.springframework.web.context.ContextLoaderServlet
</servlet-class>
<load-on-startup>2</load-on-startup>
</servlet>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:applicationContext.xml</param-value>
</context-param>
<servlet>
<servlet-name>springInitServlet</servlet-name>
<servlet-class>
org.springframework.web.context.ContextLoaderServlet
</servlet-class>
<load-on-startup>2</load-on-startup>
</servlet>
好了 现在一个基本的Structs+Spring+Hibernate的框架搭好了
4 .验证
现在把这个项目发布到Tomcat下,运行下服务器看看有缺点jar包,缺的加上
MyEclipse用起来很方便有时缺jar包很麻烦,一定要验证下
5. 建立表的映射生成映射文件
Userhbm.xml
<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
<class name="com.yourcompany.vo.User" table="user" catalog="test">
<id name="id" type="java.lang.Integer">
<column name="id" />
<generator class="native" />
</id>
<property name="username" type="java.lang.String">
<column name="username" length="64" not-null="true" />
</property>
<property name="password" type="java.lang.String">
<column name="password" length="64" not-null="true" />
</property>
</class>
</hibernate-mapping>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
<class name="com.yourcompany.vo.User" table="user" catalog="test">
<id name="id" type="java.lang.Integer">
<column name="id" />
<generator class="native" />
</id>
<property name="username" type="java.lang.String">
<column name="username" length="64" not-null="true" />
</property>
<property name="password" type="java.lang.String">
<column name="password" length="64" not-null="true" />
</property>
</class>
</hibernate-mapping>
6.写个DAO类 Service 再写个Action
RegditUserAction.java execute方法
public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response) ...{
RegditUserForm regditUserForm = (RegditUserForm) form;
ApplicationContext context = WebApplicationContextUtils
.getWebApplicationContext(this.getServlet().getServletContext());
IUserService userService = (IUserService) context
.getBean("userService");
User user = new User();
user.setUsername(regditUserForm.getUsername());
user.setPassword(regditUserForm.getPassword());
if (!(regditUserForm == null && "".equals(regditUserForm))) ...{
userService.regedit(user);
return mapping.findForward("success");
} else ...{
return mapping.findForward("error");
}
HttpServletRequest request, HttpServletResponse response) ...{
RegditUserForm regditUserForm = (RegditUserForm) form;
ApplicationContext context = WebApplicationContextUtils
.getWebApplicationContext(this.getServlet().getServletContext());
IUserService userService = (IUserService) context
.getBean("userService");
User user = new User();
user.setUsername(regditUserForm.getUsername());
user.setPassword(regditUserForm.getPassword());
if (!(regditUserForm == null && "".equals(regditUserForm))) ...{
userService.regedit(user);
return mapping.findForward("success");
} else ...{
return mapping.findForward("error");
}
这样一个所谓的SSH就搭建好了
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