Spring JdbcTemplate - 插入blob并返回生成的密钥

时间：2010-05-05 05:43:34

标签： java jdbctemplate spring-jdbc

从Spring JDBC文档中，我知道如何insert a blob using JdbcTemplate 

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retrieve the generated key of a newly inserted row：

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

有没有办法将两者结合起来？

11 个答案:

答案 0 :(得分：15)

我来到这里寻找相同的答案，但对接受的内容并不满意。所以我做了一点挖掘，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，因为否则你必须使用Spring的NamedParameterJdbcOperations（在SQL中执行sequence.nextval），它似乎不支持KeyHolder（我用它来检索自动生成id）。有关详细信息，请参阅此博客文章（不是我的博客）：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;
/

答案 1 :(得分：8)

所有这些对我来说似乎太复杂了。这很有效，也很简单。它使用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();
        }
    }

答案 2 :(得分：2)

我最后只执行了两个查询，一个用于创建行，另一个用于更新blob。

int id = insertRow();
updateBlob(id, blob);

查看Spring源代码并提取所需的部分，我想出了这个：

final KeyHolder generatedKeyHolder = new GeneratedKeyHolder();
getJdbcTemplate().execute(
    "INSERT INTO lob_table (blob) VALUES (?)",
    new PreparedStatementCallback() {
        public Object doInPreparedStatement(PreparedStatement ps) throws SQLException {
            LobCreator lobCreator = lobHandler.getLobCreator();
            lobCreator.setBlobAsBinaryStream(ps, 2, blobIs, (int)blobIn.length());

            int rows = ps.executeUpdate();
            List generatedKeys = generatedKeyHolder.getKeyList();
            generatedKeys.clear();
            ResultSet keys = ps.getGeneratedKeys();
            if (keys != null) {
                try {
                    RowMapper rowMapper = new ColumnMapRowMapper();
                    RowMapperResultSetExtractor rse = new RowMapperResultSetExtractor(rowMapper, 1);
                    generatedKeys.addAll((List) rse.extractData(keys));
                }
                finally {
                    JdbcUtils.closeResultSet(keys);
                }
            }
            if (logger.isDebugEnabled()) {
                logger.debug("SQL update affected " + rows + " rows and returned " + generatedKeys.size() + " keys");
            }
            return new Integer(rows);
        }
    }
);

我不能说我完全理解这里发生了什么。我不确定在这个简单的情况下是否需要提取生成密钥的复杂方法，而且我还不完全清楚在代码变得多毛时使用JdbcTemplate的好处。

无论如何，我测试了上面的代码并且它有效。对于我的情况，我认为它会使我的代码太复杂。

答案 3 :(得分：2)

package com.technicalkeeda.dao;

import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.InputStream;
import java.sql.Types;

import javax.sql.DataSource;

import org.springframework.dao.DataAccessException;
import org.springframework.jdbc.core.JdbcTemplate;
import org.springframework.jdbc.core.support.SqlLobValue;
import org.springframework.jdbc.support.lob.DefaultLobHandler;
import org.springframework.jdbc.support.lob.LobHandler;

public class ImageDaoImpl implements ImageDao {

    private DataSource dataSource;

    private JdbcTemplate jdbcTemplate;

    public void setDataSource(DataSource dataSource) {
        this.dataSource = dataSource;
        this.jdbcTemplate = new JdbcTemplate(this.dataSource);
    }

    @Override
    public void insertImage() {
        System.out.println("insertImage" + jdbcTemplate);

        try {
            final File image = new File("C:\\puppy.jpg");
            final InputStream imageIs = new FileInputStream(image);

            LobHandler lobHandler = new DefaultLobHandler(); 

            jdbcTemplate.update(
                     "INSERT INTO trn_imgs (img_title, img_data) VALUES (?, ?)",
                     new Object[] {
                       "Puppy",
                       new SqlLobValue(imageIs, (int)image.length(), lobHandler),
                     },
                     new int[] {Types.VARCHAR, Types.BLOB});


        } catch (DataAccessException e) {
            e.printStackTrace();
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }

    }
}

答案 4 :(得分：1)

2012年，SimpleJdbcTemplate已弃用。这就是我所做的：

KeyHolder keyHolder = new GeneratedKeyHolder();

List<SqlParameter> declaredParams = new ArrayList<>();

declaredParams.add(new SqlParameter(Types.VARCHAR));
declaredParams.add(new SqlParameter(Types.BLOB));
declaredParams.add(new SqlParameter(Types.VARCHAR));
declaredParams.add(new SqlParameter(Types.INTEGER));
declaredParams.add(new SqlParameter(Types.INTEGER));

PreparedStatementCreatorFactory pscFactory = 
    new PreparedStatementCreatorFactory(SQL_CREATE_IMAGE, declaredParams);

pscFactory.setReturnGeneratedKeys(true);

getJdbcTemplate().update(
    pscFactory.newPreparedStatementCreator(
        new Object[] {
            image.getName(), 
            image.getBytes(), 
            image.getMimeType(), 
            image.getHeight(),
            image.getWidth() 
        }), keyHolder);

image.setId(keyHolder.getKey().intValue());

SQL看起来像这样：

INSERT INTO image (name, image_bytes, mime_type, height, width) VALUES (?, ?, ?, ?, ?)

答案 5 :(得分：0)

如果底层数据库是mysql，您可以自动生成主键。然后，要将记录插入数据库，可以使用以下语法进行插入：

INSERT INTO lob_table (a_blob) VALUES (?)

答案 6 :(得分：0)

这仅在MySql上测试，我只粘贴相关部分。 运行我的测试类后，结果如下所示： “通过template.update（psc，kh）添加记录：1添加并获得密钥36”

final byte[] bytes = "My Binary Content".getBytes();
final ByteArrayInputStream bais = new ByteArrayInputStream(bytes);        
PreparedStatementCreator psc = new PreparedStatementCreator() {
        PreparedStatement ps = null;
        public PreparedStatement createPreparedStatement(
                Connection connection) throws SQLException {
            dummy.setStringCode("dummy_jdbc_spring_createPS_withKey_lob");
            ps = connection
                    .prepareStatement(
                            "INSERT INTO DUMMY (dummy_code, dummy_double, dummy_date, dummy_binary) VALUES (?, ?, ?,?)",
                            Statement.RETURN_GENERATED_KEYS);
            ps.setString(1, dummy.getStringCode());
            ps.setDouble(2, dummy.getDoubleNumber());
            ps.setDate(3, dummy.getDate());
            new DefaultLobHandler().getLobCreator().setBlobAsBinaryStream(
                    ps, 4, bais, bytes.length);

            return ps;
        }
    };
KeyHolder holder = new GeneratedKeyHolder();
System.out.println("record added via template.update(psc,kh): "
            + template.update(psc, holder)+" added and got key " + holder.getKey());

答案 7 :(得分：0)

使用lambda的另一种解决方案（不是必需的）：

jdbcTemplate.update(dbcon -> {
    PreparedStatement ps = dbcon.prepareStatement("INSERT INTO ...");
    ps.setString(1, yourfieldValue);
    ps.setBinaryStream(2, yourInputStream, yourInputStreamSizeAsInt));
    return ps;
});

NB。对不起，这不包括KeyGenerator。

答案 8 :(得分：0)

更新blob数据时遇到了同样的问题-需要将映像更新到数据库中。比我发现一些解决方案如下。有关更多详细信息，update image into database     

 LobHandler lobHandler = new DefaultLobHandler();
 statusRes = jdbcTemplate.update("update  USERS set FILE_CONTENT = ?, FILE_NAME = ? WHERE lower(USER_ID) = ?",
               new Object[] {new SqlLobValue(image, lobHandler),fileName,userIdLower},
               new int[] {Types.BLOB,Types.VARCHAR,Types.VARCHAR});

答案 9 :(得分：-1)

也许有人这样：

public class JdbcActorDao implements ActorDao {
private SimpleJdbcTemplate simpleJdbcTemplate;
private SimpleJdbcInsert insertActor;

public void setDataSource(DataSource dataSource) {
    this.simpleJdbcTemplate = new SimpleJdbcTemplate(dataSource);
    this.insertActor =
            new SimpleJdbcInsert(dataSource)
                    .withTableName("t_actor")
                    .usingGeneratedKeyColumns("id");
}

public void add(Actor actor) {
    Map<String, Object> parameters = new HashMap<String, Object>(2);
    parameters.put("first_name", actor.getFirstName());
    parameters.put("last_name", actor.getLastName());
    Number newId = insertActor.executeAndReturnKey(parameters);
    actor.setId(newId.longValue());
}

//  ... additional methods
}

答案 10 :(得分：-1)

请使用：

addValue("p_file", noDataDmrDTO.getFile_data(), Types.BINARY)

noDataDmrDTO.getFile_data() is byte array.


{
 simpleJdbcCall =
          new SimpleJdbcCall(jdbcTemplate).withProcedureName("insert_uploaded_files").withCatalogName("wct_mydeq_stg_upld_pkg")
              .withSchemaName("WCT_SCHEMA");

 SqlParameterSource sqlParms =
        new MapSqlParameterSource().addValue("p_upload_idno", Integer.parseInt("143"))
            .addValue("p_file_type_idno", Integer.parseInt(noDataDmrDTO.getFile_type_idno())).addValue("p_file_name", noDataDmrDTO.getFile_name())
            .addValue("p_file", noDataDmrDTO.getFile_data(), Types.BINARY).addValue("p_comments", noDataDmrDTO.getComments())
            .addValue("p_userid", noDataDmrDTO.getUserid());


    simpleJdbcCallResult = simpleJdbcCall.execute(sqlParms);

}
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