Using NamedParameterJdbcTemplate to send data to DataBase

混江龙づ霸主 提交于 2019-12-23 23:18:07

问题


package com.techm.template;
import java.sql.Types;
import java.util.Date;
import java.util.HashMap;
import java.util.Map;

import javax.sql.DataSource;
import org.springframework.jdbc.core.JdbcTemplate;
import org.springframework.jdbc.core.namedparam.MapSqlParameterSource;
import org.springframework.jdbc.core.namedparam.NamedParameterJdbcTemplate;
import org.springframework.jdbc.core.namedparam.SqlParameterSource;
import org.springframework.jdbc.datasource.DriverManagerDataSource;

public class InsertRecordInDatabaseWithJdbcTemplate {

    private static final String driverClassName = "oracle.jdbc.driver.OracleDriver";

    private static final String url = "jdbc:oracle:thin:@localhost:1521:xe";

    private static final String dbUsername = "SYSTEM";

    private static final String dbPassword = "9848451415";

    private static final String sql = "INSERT INTO employee(firstname,lastname,username,password) VALUES(?,?,?,?)";


    private static DataSource dataSource;



    public static void main(String[] args) throws Exception {

        dataSource = getDataSource();

        Employee e = new Employee();
        e.setFirstname("Surya Teja");
        e.setLastname("Tammana");
        e.setUsername("surya_kvm");
        e.setPassword("9848451415");

        int result = saveRecord(e);
        if(result != 0){
            System.out.println("Success");
        }
        else{
            System.out.println("Failure");
        }

    }



   public static int saveRecord(Employee e) {

       NamedParameterJdbcTemplate template = new NamedParameterJdbcTemplate(dataSource);

       SqlParameterSource params = new MapSqlParameterSource();

       Map<String, Object> parameters = new HashMap<String, Object>();
        parameters.put("firstnames", e.getFirstname());
        parameters.put("lastname", e.getLastname());
        parameters.put("username", e.getUsername());
        parameters.put("password", e.getPassword());

        int result = template.update(sql, parameters);
        return result;
   }



 public static DriverManagerDataSource getDataSource() {

 DriverManagerDataSource dataSource = new DriverManagerDataSource();
 dataSource.setDriverClassName(driverClassName);
 dataSource.setUrl(url);
 dataSource.setUsername(dbUsername);
 dataSource.setPassword(dbPassword);
 return dataSource;

    }
}   

Employee.java

package com.techm.template;

public class Employee {

private String firstname;
private String lastname;
private String username;
private String password;
public String getFirstname() {
    return firstname;
}
public void setFirstname(String firstname) {
    this.firstname = firstname;
}
public String getLastname() {
    return lastname;
}
public void setLastname(String lastname) {
    this.lastname = lastname;
}
public String getUsername() {
    return username;
}
public void setUsername(String username) {
    this.username = username;
}
public String getPassword() {
    return password;
}
public void setPassword(String password) {
    this.password = password;
}




}

Console

Jan 22, 2017 12:25:24 AM org.springframework.jdbc.datasource.DriverManagerDataSource setDriverClassName INFO: Loaded JDBC driver: oracle.jdbc.driver.OracleDriver Exception in thread "main" org.springframework.dao.InvalidDataAccessApiUsageException: SQL [INSERT INTO employee(firstname,lastname,username,password) VALUES(?,?,?,?)]: given 4 parameters but expected 0 at org.springframework.jdbc.core.PreparedStatementCreatorFactory$PreparedStatementCreatorImpl.(PreparedStatementCreatorFactory.java:221) at org.springframework.jdbc.core.PreparedStatementCreatorFactory$PreparedStatementCreatorImpl.(PreparedStatementCreatorFactory.java:199) at org.springframework.jdbc.core.PreparedStatementCreatorFactory.newPreparedStatementCreator(PreparedStatementCreatorFactory.java:173) at org.springframework.jdbc.core.namedparam.NamedParameterJdbcTemplate.getPreparedStatementCreator(NamedParameterJdbcTemplate.java:351) at org.springframework.jdbc.core.namedparam.NamedParameterJdbcTemplate.update(NamedParameterJdbcTemplate.java:287) at org.springframework.jdbc.core.namedparam.NamedParameterJdbcTemplate.update(NamedParameterJdbcTemplate.java:292) at com.techm.template.InsertRecordInDatabaseWithJdbcTemplate.saveRecord(InsertRecordInDatabaseWithJdbcTemplate.java:64) at com.techm.template.InsertRecordInDatabaseWithJdbcTemplate.main(InsertRecordInDatabaseWithJdbcTemplate.java:40)


回答1:


You are using positional parameters in your SQL query. This doesn't work because you are using the NamedParameterJdbcTemplate which requires named parameters. Just modify your query to use named parameters like that:

"INSERT INTO employee(firstname, lastname, username, password) VALUES(:firstname, :lastname, :username, :password)"

Also you could use the Employee object as a parameter source:

public static int saveRecord(Employee e) {
        NamedParameterJdbcTemplate template = new NamedParameterJdbcTemplate(dataSource);

        SqlParameterSource params = new BeanPropertySqlParameterSource(e);

        int result = template.update(sql, params);
        return result;
   }

The BeanPropertySqlParameterSourcewill use Employees properties to set the parameters.



来源:https://stackoverflow.com/questions/41783375/using-namedparameterjdbctemplate-to-send-data-to-database

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!