Question

我尝试通过jdbcTemplate对MySql数据库执行简单请求，但是当框架加载和解析xml文件时，我有一个错误来定义我的数据源：

<?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:util="http://www.springframework.org/schema/util"
xmlns:tx="http://www.springframework.org/schema/tx"
xsi:schemaLocation="http://www.springframework.org/schema/beans
    http://www.springframework.org/schema/beans/spring-beans-3.0.xsd

    http://www.springframework.org/schema/tx
    http://www.springframework.org/schema/tx/spring-tx-3.0.xsd">

    <bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
      <property name="driverClassName" value="com.mysql.jdbc.Driver"/>
      <property name="url" value="jdbc:mysql://localhost:3306/spring_training"/>
      <property name="username" value="root"/>
      <property name="password" value="pass"/>
    </bean>
</beans>

的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app 
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xmlns="http://java.sun.com/xml/ns/javaee" 
xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" 
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>SpringTrainingTemplate</display-name>

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/spring-config.xml /WEB-INF/jdbc-config.xml</param-value>
</context-param>

<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<servlet>
    <servlet-name>hello</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/servlet-context.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>hello</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>

和调用它的Controller：

@Controller
public class HomeController {

@Autowired
private ExampleService exampleService;

@RequestMapping(value = "/details", method = RequestMethod.GET)
public String details(Model model) {

    ApplicationContext context = new ClassPathXmlApplicationContext("jdbc-config.xml");
    ExampleDao dao = (ExampleDao) context.getBean("ExampleDao");
    List<Application> list = dao.getAllApplications();

    model.addAttribute("application", list.get(0).getName());
    model.addAttribute("descriptionOfApplication", list.get(0).getDescription());

    return "details";
}
}

public class ExampleDao {

private String request = "select * from application";

private JdbcTemplate jdbcTemplate;

@Autowired
private DataSource dataSource;

public ExampleDao(DataSource dataSource) {
    this.jdbcTemplate = new JdbcTemplate(dataSource);
}

public List<Application> getAllApplications() {
    List<Application> applications = this.jdbcTemplate.query(request, new RowMapper<Application>() {
        @Override
        public Application mapRow(ResultSet rs, int i) throws SQLException {
            Application application = new Application();
            application.setName(rs.getString("name"));
            application.setType(rs.getString("type"));
            application.setDescription(rs.getString("description"));
            application.setDownloads(rs.getInt("downloads"));
            return application;
        }
    });
    return applications;
}
}

enter image description here

我运行它并输入http://localhost:8080/details我有一个带有此消息的堆栈跟踪500异常：

root cause
org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from class path resource [jdbc-config.xml]; nested exception is java.io.FileNotFoundException: class path resource [jdbc-config.xml] cannot be opened because it does not exist

你能解释一下如何以严格的方式配置jdbc连接，或者我的方法是否正确，我应该寻找我的问题的解决方案？所有帮助将不胜感激。谢谢。

Answer 1

Spring无法找到您的jdbc-config.xml配置文件。

您可以将它放在类路径而不是WEB-INF文件夹中，并将其加载到您的web.xml中，如下所示：

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>classpath:spring-config.xml,classpath:jdbc-config.xml</param-value>
</context-param>

一个好的做法是在src文件夹中创建文件夹main和resources，并将它们添加到类路径中。然后，您可以将spring配置文件放在src / resources文件夹中。

Answer 2

类路径资源[jdbc-config.xml]无法打开，因为它确实如此不存在

文件名是否正确，它位于何处？指定数据库连接的文件不是你应该在的地方 - 在类路径上。

Spring：如何定义数据库配置？

2 个答案: