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>>
我实现的应用程序如下
员工.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;
}
}
员工道.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;
}
});
}
}
测试.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);
}
}
应用程序上下文.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>
创建的表
创建表员工(ID号(10),姓名瓦尔查尔2(100),工资号(10));
错误消息清楚地表明了这一点:
ResultSetExtractor 不是泛型的。
该接口在较新版本的 Spring 中是通用的。但无论如何,您仍然可以使用它的原始形式,尽管在使用extractData()
方法的结果时,您会被迫进行强制转换。
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;
}
});
}
附言:我假设您使用的是旧版本的 Spring 框架,因为在较新版本中,ResultSetExtractor
实际上是通用的。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),或者坚持使用上述代码片段中使用的方法。