Tengo un problema con mi declaración preparada pero no puedo averiguar dónde está el error. Estoy tratando de insertar un enlace URI en la base de datos.
@Repository public interface LoggerDao extends CrudRepository<Logger, Long> { @Query("select t from Logger t where t.user.id=?#{principal.id}") List<Logger> findAll(); @Modifying @Query(value = "insert into Logger t (t.redirect, t.user.id) VALUES (:insertLink,?#{principal.id})", nativeQuery = true) @Transactional void logURI(@Param("insertLink") String insertLink);Error
2017-03-11 19:52:59.157 WARN 65154 --- [nio-8080-exec-8] ohengine.jdbc.spi.SqlExceptionHelper : SQL Error: 42001, SQLState: 42001 2017-03-11 19:52:59.157 ERROR 65154 --- [nio-8080-exec-8] ohengine.jdbc.spi.SqlExceptionHelper : Syntax error in SQL statement "INSERT INTO LOGGER T[*] (T.REDIRECT, T.USER.ID) VALUES (?,?) "; expected "., (, DIRECT, SORTED, DEFAULT, VALUES, SET, (, SELECT, FROM"; SQL statement: insert into Logger t (t.redirect, t.user.id) VALUES (?,?) [42001-190] 2017-03-11 19:52:59.181 ERROR 65154 --- [nio-8080-exec-8] oaccC[.[.[/].[dispatcherServlet] : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.dao.InvalidDataAccessResourceUsageException: could not prepare statement; SQL [insert into Logger t (t.redirect, t.user.id) VALUES (?,?)]; nested exception is org.hibernate.exception.SQLGrammarException: could not prepare statement] with root cause org.h2.jdbc.JdbcSQLException: Syntax error in SQL statement "INSERT INTO LOGGER T[*] (T.REDIRECT, T.USER.ID) VALUES (?,?) "; expected "., (, DIRECT, SORTED, DEFAULT, VALUES, SET, (, SELECT, FROM"; SQL statement: insert into Logger t (t.redirect, t.user.id) VALUES (?,?) [42001-190] at org.h2.engine.SessionRemote.done(SessionRemote.java:624) ~[h2-1.4.190.jar:1.4.190] at org.h2.command.CommandRemote.prepare(CommandRemote.java:68) ~[h2-1.4.190.jar:1.4.190] at org.h2.command.CommandRemote.<init>(CommandRemote.java:45) ~[h2-1.4.190.jar:1.4.190] at org.h2.engine.SessionRemote.prepareCommand(SessionRemote.java:494) ~[h2-1.4.190.jar:1.4.190] at org.h2.jdbc.JdbcConnection.prepareCommand(JdbcConnection.java:1188) ~[h2-1.4.190.jar:1.4.190] at org.h2.jdbc.JdbcPreparedStatement.<init>(JdbcPreparedStatement.java:72) ~[h2-1.4.190.jar:1.4.190] at org.h2.jdbc.JdbcConnection.prepareStatement(JdbcConnection.java:276) ~[h2-1.4.190.jar:1.4.190] at org.apache.tomcatLogré resolver el problema. Agregué una identificación a los parámetros para que pueda pasar la identificación del usuario, usando Principal en el controlador.
@Repository public interface LoggerDao extends CrudRepository<Logger, Long> { @Query("select t from Logger t where t.user.id=?#{principal.id}") List<Logger> findAll(); @Modifying @Query(value = "insert into Logger (redirect,user_id) VALUES (:insertLink,:id)", nativeQuery = true) @Transactional void logURI(@Param("insertLink") String insertLink, @Param("id") Long id);Hay una manera de hacer inserciones usando consultas obj (no nativas) (usando @Query & @Modifying) pero depende de la base de datos que esté usando. A continuación funcionó para mí en Oracle (usando la tabla Dual):
@Repository public interface DualRepository extends JpaRepository<Dual,Long> { @Modifying @Query("insert into Person (id,name,age) select :id,:name,:age from Dual") public int modifyingQueryInsertPerson(@Param("id")Long id, @Param("name")String name, @Param("age")Integer age); }Aquí hay un enlace que muestra en la parte inferior qué base de datos admite stmts seleccionados sin una cláusula from: http://modern-sql.com/use-case/select-without-from