• mybatis竟然报"Invalid value for getInt()"


    背景

    使用mybatis遇到一个非常奇葩的问题,错误如下:

    Cause: org.apache.ibatis.executor.result.ResultMapException: Error attempting to get column 'name' from result set.  Cause: java.sql.SQLException: Invalid value for getInt() - 'hhhh'
    

    场景

    还原一下当时的情况:

    public interface UserMapper {
        @Results(value = {
                @Result(property = "id", column = "id", javaType = Long.class, jdbcType = JdbcType.BIGINT),
                @Result(property = "age", column = "age", javaType = Integer.class, jdbcType = JdbcType.INTEGER),
                @Result(property = "name", column = "name", javaType = String.class, jdbcType = JdbcType.VARCHAR)
        })
        @Select("SELECT id, name, age FROM user WHERE id = #{id}")
        User selectUser(Long id);
    }
    
    @Data
    @Builder
    public class User {
        private Long id;
        private Integer age;
        private String name;
    }
    
    public class MapperMain {
        public static void main(String[] args) throws Exception {
            MysqlConnectionPoolDataSource dataSource = new MysqlConnectionPoolDataSource();
            dataSource.setUser("root");
            dataSource.setPassword("root");
            dataSource.setUrl("jdbc:mysql://localhost:3306/test?useUnicode=true&characterEncoding=utf-8");
    
            TransactionFactory transactionFactory = new JdbcTransactionFactory();
            Environment environment = new Environment("development", transactionFactory, dataSource);
            Configuration configuration = new Configuration(environment);
            configuration.addMapper(UserMapper.class);
            SqlSessionFactory sqlSessionFactory = new SqlSessionFactoryBuilder().build(configuration);
    
            try (SqlSession session = sqlSessionFactory.openSession()) {
                UserMapper userMapper = session.getMapper(UserMapper.class);
                System.out.println(userMapper.selectUser(1L));
            }
        }
    }
    

    数据库如下:

    上面是一个很简单的例子,就是根据id选出用户的信息,运行结果如下:

    User(id=1, age=2, name=3)
    

    没有任何问题,但是我再往数据库里插入一条数据,如下:

    MapperMain类中增加一行代码,如下:

    System.out.println(userMapper.selectUser(2L));
    

    运行结果如下:

    User(id=1, age=2, name=3)
    ### Error querying database.  Cause: org.apache.ibatis.executor.result.ResultMapException: Error attempting to get column 'name' from result set.  Cause: java.sql.SQLException: Invalid value for getInt() - 'hhhh'
    ……
    

    可以看出第一条查询没有问题,第二条查询就报错了

    初探

    其实我的直觉告诉我,是不是因为User类里字段顺序和SQL语句里select字段的顺序不一致导致的,那就来试一下吧

    改一下User类里字段的顺序:

    @Data
    @Builder
    public class User {
        private Long id;
        private String name;
        private Integer age;
    }
    

    结果如下:

    User(id=1, name=3, age=2)
    User(id=2, name=hhhh, age=3)
    

    果不其然,直觉还是很6的

    或者改一下SQL语句里select字段的顺序:

    @Data
    @Builder
    public class User {
        private Long id;
        private Integer age;
        private String name;
    }
    
    public interface UserMapper {
        @Results(value = {
                @Result(property = "id", column = "id", javaType = Long.class, jdbcType = JdbcType.BIGINT),
                @Result(property = "age", column = "age", javaType = Integer.class, jdbcType = JdbcType.INTEGER),
                @Result(property = "name", column = "name", javaType = String.class, jdbcType = JdbcType.VARCHAR)
        })
        @Select("SELECT id, age, name FROM user WHERE id = #{id}")
        User selectUser(Long id);
    }
    

    以我们的直觉,结果肯定也没问题,果不其然,如下:

    User(id=1, age=2, name=3)
    User(id=2, age=3, name=hhhh)
    

    再探

    其实到上一步,问题已经解决了,可以继续干活了,但是搞不懂为什么,心里总觉得不踏实。

    bugdebug开始,从下面的入口开始:

    追踪到如下:

    可以看出User这个类是有构造函数的,而且是包含所有字段的构造函数
    利用这个构造函数创建实例的时候,参数的顺序就是SQL语句选择字段的顺序,不会根据映射关系去选择
    所以就出现了类型不匹配

    那我们再来看一下问什么会有一个这样的构造函数产生,直觉告诉我是@Builder这个注解

    一起来看一下User编译后的结果:

    public class User {
        private Long id;
        private String name;
        private Integer age;
    
        User(final Long id, final String name, final Integer age) {
            this.id = id;
            this.name = name;
            this.age = age;
        }
    
        public static User.UserBuilder builder() {
            return new User.UserBuilder();
        }
    
        public static class UserBuilder {
            private Long id;
            private String name;
            private Integer age;
    
            UserBuilder() {
            }
    
            public User.UserBuilder id(final Long id) {
                this.id = id;
                return this;
            }
    
            public User.UserBuilder name(final String name) {
                this.name = name;
                return this;
            }
    
            public User.UserBuilder age(final Integer age) {
                this.age = age;
                return this;
            }
    
            public User build() {
                return new User(this.id, this.name, this.age);
            }
        }
    }
    

    果然如此,UserBuilder.build()方法就是利用这个构造函数来生成的。

    结局

    最终解决方案就是给User类加上无参的构造函数就OK了,如下:

    @Builder
    @AllArgsConstructor
    @NoArgsConstructor
    public class User {
        private Integer age;
        private String name;
        private Long id;
    }
    

    字段顺序随便放,最后再执行一下:

    User(age=2, name=3, id=1)
    User(age=3, name=hhhh, id=2)
    
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  • 原文地址:https://www.cnblogs.com/eaglelihh/p/15459073.html
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