Question

我正在学习春天。在创建一个例子时我遇到了错误。

The type ResultSetExtractor is not generic; it cannot be parameterized with arguments <List<Employee>>

我按照以下方式实施应用

Employee.java

package com.develop;
public class Employee {
    private int id;  
    private String name;  
    private float salary;   
    public Employee(){} 
    public Employee(int id, String name, float salary){
        this.id = id;
        this.name = name;
        this.salary = salary;
    }

    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public float getSalary() {
        return salary;
    }
    public void setSalary(float salary) {
        this.salary = salary;
    } 

}

EmployeeDao.java

package com.develop;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.util.ArrayList;
import java.util.List;

import org.springframework.dao.DataAccessException;
import org.springframework.jdbc.core.JdbcTemplate;
import org.springframework.jdbc.core.ResultSetExtractor;

public class EmployeeDao {
private JdbcTemplate template;

public void setJdbcTemplate(JdbcTemplate template) {
    this.template = template;
}

public List<Employee> getAllEmployees(){  
     return template.query("select * from employee",new ResultSetExtractor<List<Employee>>(){  
        @Override  
         public List<Employee> extractData(ResultSet rs) throws SQLException,  
                DataAccessException {  

            List<Employee> list=new ArrayList<Employee>();  
            while(rs.next()){  
            Employee e=new Employee();  
            e.setId(rs.getInt(1));  
            e.setName(rs.getString(2));  
            e.setSalary(rs.getInt(3));  
            list.add(e);  
            }  
            return list;  
            }  
        });  
      }     
}

Test.java

 package com.develop;
    import java.util.List;
    import org.springframework.context.ApplicationContext;
    import org.springframework.context.support.ClassPathXmlApplicationContext;    
    public class Test {
        public static void main(String[] args) {
            ApplicationContext ctx=new ClassPathXmlApplicationContext("applicationContext.xml");              
             EmployeeDao dao=(EmployeeDao)ctx.getBean("edao");  
                List<Employee> list=dao.getAllEmployees();                        
                for(Employee e:list)  
                    System.out.println(e);                        
                } 
    }

的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="ds" class="org.springframework.jdbc.datasource.DriverManagerDataSource">  
<!-- <property name="driverClassName" value="oracle.jdbc.driver.OracleDriver" />  
<property name="url" value="jdbc:oracle:thin:@localhost:1521:xe" />  
<property name="username" value="system" />  
<property name="password" value="oracle" />  -->

<property name="driverClassName" value="com.mysql.jdbc.Driver" />  
<property name="url" value="jdbc:mysql://localhost:3306/springdatabase" />  
<property name="username" value="root" />  
<property name="password" value="admin123" />

</bean>  

<bean id="jdbcTemplate" class="org.springframework.jdbc.core.JdbcTemplate">  
<property name="dataSource" ref="ds"></property>  
</bean>  

<bean id="edao" class="com.develop.EmployeeDao">  
<property name="template" ref="jdbcTemplate"></property>  
</bean>  

</beans>

创建表

CREATE TABLE员工（身份证号码（10），NAME varchar2（100），工资号码（10））;

Answer 1

错误信息明确表示：

ResultSetExtractor不是通用的。

界面在更新版本的Spring中变得通用。但无论如何，你仍然可以使用它的原始形式，尽管你在使用extractData()方法的结果时被强迫你做一个演员。

public List<Employee> getAllEmployees(){  
    return template.query("select * from employee",new ResultSetExtractor(){  
        @Override  
         public Object extractData(ResultSet rs) throws SQLException,  
            DataAccessException {  

            List<Employee> list=new ArrayList<Employee>();  
            while(rs.next()) {  
                Employee e=new Employee();  
                e.setId(rs.getInt(1));  
                e.setName(rs.getString(2));  
                e.setSalary(rs.getInt(3));  
                list.add(e);  
            }  
            return list;  
        }  
    });  
  }

P.S .: 我假设您使用旧版本的Spring Framework，因为在较新的版本中，ResultSetExtractor is actually generic。因此，您可以更新Spring版本（但要小心，因为这可能会导致编译问题和其他问题），或者坚持使用上面代码片段中使用的方法。

ResultSetExtractor类型不是通用的;它不能使用参数<list <employee>＆gt; </list <employee>进行参数化

1 个答案: