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解决mybatis竟然报Invalid value for getInt()的问题

作者:It''''''''s my code life.

使用mybatis遇到一个非常奇葩的问题,总是报Invalid value for getInt()的问题,怎么解决呢?下面小编通过场景分析给大家代来了mybatis报Invalid value for getInt()的解决方法,感兴趣的朋友参考下吧

带你来看看mybatis为什么报"Invalid value for getInt()"这个错误

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() - 'wo'

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() - 'lh'
……

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=lh, 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=lh)

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=lh, id=2)

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