Spring JdbcTemplate - 插入blob并返回生成的密钥

Question

从Spring JDBC文档中,我知道如何使用JdbcTemplate插入blob

final File blobIn = new File("spring2004.jpg");
final InputStream blobIs = new FileInputStream(blobIn);
jdbcTemplate.execute(
  "INSERT INTO lob_table (id, a_blob) VALUES (?, ?)",
  new AbstractLobCreatingPreparedStatementCallback(lobhandler) {                         
      protected void setValues(PreparedStatement ps, LobCreator lobCreator) 
          throws SQLException {
        ps.setLong(1, 1L);
        lobCreator.setBlobAsBinaryStream(ps, 2, blobIs, (int)blobIn.length());           
      }
  }
);
blobIs.close();

以及如何检索新插入行的生成密钥:

KeyHolder keyHolder = new GeneratedKeyHolder();
jdbcTemplate.update(
    new PreparedStatementCreator() {
        public PreparedStatement createPreparedStatement(Connection connection) throws SQLException {
            PreparedStatement ps =
                connection.prepareStatement(INSERT_SQL, new String[] {"id"});
            ps.setString(1, name);
            return ps;
        }
    },
    keyHolder);

// keyHolder.getKey() now contains the generated key

有没有办法把两者结合起来？

Answer 1

我来到这里寻找相同的答案,但对接受的内容并不满意.所以我做了一些挖掘,并提出了我在Oracle 10g和Spring 3.0中测试过的解决方案

public Long save(final byte[] blob) {
  KeyHolder keyHolder = new GeneratedKeyHolder();
  String sql = "insert into blobtest (myblob) values (?)"; //requires auto increment column based on triggers
  getSimpleJdbcTemplate().getJdbcOperations().update(new AbstractLobPreparedStatementCreator(lobHandler, sql, "ID") {
    @Override
    protected void setValues(PreparedStatement ps, LobCreator lobCreator) throws SQLException, DataAccessException {
      lobCreator.setBlobAsBytes(ps, 1, blob);
    }
  }, keyHolder);

  Long newId = keyHolder.getKey().longValue();
  return newId;
}

这还需要以下抽象类,部分基于Spring的AbstractLobCreatingPreparedStatementCallback

public abstract class AbstractLobPreparedStatementCreator implements PreparedStatementCreator {
  private final LobHandler lobHandler;
  private final String sql;
  private final String keyColumn;
  public AbstractLobPreparedStatementCreator(LobHandler lobHandler, String sql, String keyColumn) {
    this.lobHandler = lobHandler;
    this.sql = sql;
    this.keyColumn = keyColumn;
  }
  public PreparedStatement createPreparedStatement(Connection con) throws SQLException {
    PreparedStatement ps = con.prepareStatement(sql, new String[] { keyColumn });
    LobCreator lobCreator = this.lobHandler.getLobCreator();
    setValues(ps, lobCreator);
    return ps;
  }
  protected abstract void setValues(PreparedStatement ps, LobCreator lobCreator) throws SQLException, DataAccessException;
}

此外,您在Oracle中创建的表应使用序列和触发器为id创建一个自动递增的列.触发器是必要的,因为否则你将不得不使用Spring的NamedParameterJdbcOperations(做你的SQL的sequence.nextval),这似乎并没有对持有者的支持(我用它来检索自动根ID).有关详细信息,请参阅此博客文章(而非我的博客):http://www.lifeaftercoffee.com/2006/02/17/how-to-create-auto-increment-columns-in-oracle/

create table blobtest (
id number primary key,
myblob blob);

create sequence blobseq start with 1 increment by 1;

CREATE OR REPLACE TRIGGER blob_trigger
BEFORE INSERT
ON blobtest
REFERENCING NEW AS NEW
FOR EACH ROW
BEGIN
SELECT blobseq.nextval INTO :NEW.ID FROM dual;
end;
/

Answer 2

所有这些对我来说似乎太复杂了.这很有效,也很简单.它用org.springframework.jdbc.core.namedparam.NamedParameterJdbcTemplate

import org.springframework.jdbc.core.namedparam.MapSqlParameterSource;
import org.springframework.jdbc.core.namedparam.NamedParameterJdbcTemplate;
import org.springframework.jdbc.core.support.SqlLobValue;
import org.springframework.jdbc.support.lob.DefaultLobHandler;


    public void setBlob(Long id, byte[] bytes) {
        try {
            jdbcTemplate = new NamedParameterJdbcTemplate(dataSource);
            MapSqlParameterSource parameters = new MapSqlParameterSource();
            parameters.addValue("id", id);
            parameters.addValue("blob_field", new SqlLobValue(new ByteArrayInputStream(bytes), bytes.length, new DefaultLobHandler()), OracleTypes.BLOB);
            jdbcTemplate.update("update blob_table set blob_field=:blob_field where id=:id", parameters);
        } catch(Exception e) {
            e.printStackTrace();
        }
    }