Question

我正在使用Spring和Oracle DB，并且由于需要在包含相同列的单独表中存储不同的XML文件，我想使用单个DAO来对这些表进行操作。我是Spring的新手，所以我问这种方法是否可行，如果可行，那么它是如何完成的。

Answer 1

创建一个抽象的实体基类

public abstract class YourBaseEntityClass {
    private String fooProperty;
    private Long barProperty;
    // + getters and setters
}

创建通用DAO接口

public interface Dao<T> {
    T get(Long id);
    // you'll probably want more methods here :-)
}

创建通用抽象RowMapper

public abstract class GenericRowMapper<T extends YourBaseEntityClass>
implements RowMapper<T> {

    public T mapRow(final ResultSet rs, final int rowNum)
    throws SQLException {
        final T entity = instantiateEntityClass();
        entity.setFooProperty(rs.getString("foo"));
        entity.setBarProperty(rs.getLong("bar"));
        return entity;
    }

    protected abstract T instantiateEntityClass();

}

创建通用DAO实现

public class GenericJdbcDao<T> implements Dao<T> {

    private String tableName;
    public void setTableName(final String tableName) {
        this.tableName = tableName;
    }

    private JdbcTemplate jdbcTemplate;
    public void setJdbcTemplate(final JdbcTemplate jdbcTemplate) {
        this.jdbcTemplate = jdbcTemplate;
    }

    private RowMapper<T> rowMapper;
    public void setRowMapper(final RowMapper<T> rowMapper) {
        this.rowMapper = rowMapper;
    }

    public T get(final Long id) {
        return jdbcTemplate.queryForObject(
            // please don't do it like this, this is just a quick example
            "select * from " + tableName + " where id=" + id, rowMapper);
    }
}

现在，对于每种特定的实体类型，您需要：

子类YourBaseEntityClass
子类GenericRowMapper，以便创建新的实体类型
使用Spring，使用正确的表名和rowmapper

就是这样！

每个多个表具有相同结构的一个DAO

1 个答案: