Spring JdbcTemplate - Insert blob and return generated key

Question

From the Spring JDBC documentation, I know how to insert a blob using JdbcTemplate

final File blobIn = new File(\"spring2004.jpg\");
final InputStream blobIs         


        

Answer 1

All of this seemed way too complicated to me. This works and is simple. It uses 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();
        }
    }

Answer 2

Please use:

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);

}

Answer 3

Another solution with lambda (which is not required):

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

NB. Sorry this does not include KeyGenerator.

Answer 4

I got the same issue to update blob data -- need to update image into database. than i find some solution as below. for more details 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});

Answer 5

In case your underlying database is mysql, you can autogenerate your primary key. Then to insert a record into your db, you can use the following syntax for insertion:

INSERT INTO lob_table (a_blob) VALUES (?)

Answer 6

Maybe some like this:

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
}