导致：java.io.FileNotFoundException：无法打开ServletContext资源[/applicationContext.xml] [英] Caused by: java.io.FileNotFoundException: Could not open ServletContext resource [/applicationContext.xml]
问题描述
我尝试通过使用spring hibernatetemplate将我的hibernate示例移植到Spring，但是我得到此错误。导致：java.io.FileNotFoundException：无法打开ServletContext资源[/applicationContext.xml]。请建议我运行我的项目。我更新鲜我的公司
I m trying to port my hibernate example to spring by using spring hibernatetemplate but i m getting this error Caused by: java.io.FileNotFoundException: Could not open ServletContext resource [/applicationContext.xml]. Please suggest me run my project. I m fresher in my company
我的web.xml文件
my web.xml file
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>spring</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>applicationContext.xml</param-value>
</context-param>
</web-app>
我的applicationContext.xml
my applicationContext.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans
xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:p="http://www.springframework.org/schema/p"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd">
<bean id="dataSource"
class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close">
<property name="driverClassName" value="com.mysql.jdbc.Driver"></property>
<property name="url" value="jdbc:mysql://my server/emp_app"></property>
<property name="username" value="root"></property>
<property name="password" value="root"></property>
</bean>
<bean id="sessionFactory" class="org.springframework.orm.hibernate3.LocalSessionFactoryBean">
<property name="dataSource"> <ref bean="dataSource" /> </property>
<property name="hibernateProperties">
<props>
<prop key="hibernate.dialect">
org.hibernate.dialect.MySQLDialect
</prop>
</props>
</property>
</bean>
<bean id="d" class="login.HrDao">
<property name="sessionFactory" ref="mysessionFactory"></property>
</bean>
</beans>
我的HrDAO.java文件 - >
my HrDAO.java file ->
package login;
import org.hibernate.SessionFactory;
import org.springframework.orm.hibernate3.HibernateTemplate;
public class HrDao {
HibernateTemplate template;
public void setSessionFactory(SessionFactory factory) {
template = new HibernateTemplate(factory);
}
public void saveStudent(HrModel e) {
template.save(e);
}
}
我的HrMain。 java文件 - >
MY HrMain.java File ->
package login;
import org.springframework.beans.factory.BeanFactory;
import org.springframework.beans.factory.xml.XmlBeanFactory;
import org.springframework.core.io.ClassPathResource;
import org.springframework.core.io.Resource;
public class HrMain {
public static void main(String[] args) {
Resource resource = new ClassPathResource("applicationContext.xml");
BeanFactory factory = new XmlBeanFactory(resource);
HrDao dao = (HrDao) factory.getBean("d");
HrModel student = new HrModel();
student.setEmployee_Name("Akash");
student.setPassword("jaisiaram");
dao.saveStudent(student);
}
}
页面 - >
<html>
<head>
<title></title>
</head>
<body>
<form action="HrMain" method="post">
<input type="text" name="Employee_Name"/>
<input type="text" name="password">
<input type="submit">
</form>
</body>
</html>
推荐答案
您的 applicationContext在哪里？ XML ？您可以将 web.xml 中的 applicationContext.xml 引用为：
Where is your applicationContext.xml? You can refer your applicationContext.xml in web.xml as:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/applicationContext.xml</param-value>
</context-param>
这篇关于导致：java.io.FileNotFoundException：无法打开ServletContext资源[/applicationContext.xml]的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
