Java 捕获并显示 SqlExceptionHelper

Question

由于其复杂性，使用 Hibernate JPA 在 Oracle DB 中执行本机查询， 我想捕捉从SqlExceptionHelper 类抛出的" ORA-01722: Nombre non valide " 之类的异常，但捕捉到的是：

类 javax.persistence.PersistenceException: 无法提取 结果集

记录器错误跟踪我但未捕获：

jdbc.spi.SqlExceptionHelper   : ORA-01722: Nombre non valide


BigDecimal customerId = null;
try {
    Query q = entityManager.createNativeQuery(
            "select acc.account_id as customerId from Account ...");
    customerId = (BigDecimal) q.getSingleResult();

} catch (Exception e) {

    logger.info("CLASS : " + e.getClass());
    if (e instanceof PersistenceException) {  // should display ORA-01722: Nombre non valide ?
        logger.info("ERRROR : " + e.getMessage());
        throw new SQLException(e.getMessage());
    }else
    if (e instanceof SQLException) {
        logger.info("ERRROR : " + e.getMessage());
        throw new SQLException(e.getMessage());
    }
    logger.info("NOOOOOOOOOOOOO : " + e.getMessage());
    throw new Exception(e.getMessage());
}

Answer 1

@Entity
@Table(uniqueConstraints = { @UniqueConstraint(columnNames = { "continentName" }) })
public class Continent implements Serializable 

    @NotEmpty(message = "continentName is a required field")
    private String continentName;

在你的控制器中，你可以用这个来处理@NotEmpty：

@ExceptionHandler({ ConstraintViolationException.class })
    public ResponseEntity<String> handleError1(HttpServletRequest req, Exception ex) {

        String msg = null;
        if (ex.getCause().getCause() instanceof ConstraintViolationException) {
            ConstraintViolationException e = (ConstraintViolationException) ex.getCause().getCause();
            Optional<ConstraintViolation<?>> optional = e.getConstraintViolations().stream().findFirst();
            msg = optional.isPresent() ? optional.get().getMessageTemplate() : ex.getMessage();
        }

        return new ResponseEntity<>(msg, HttpStatus.CONFLICT);
    }

这将返回“continentName 是必填字段”

再次在控制器中，此方法将处理唯一约束违规。假设您已经输入了一个值“Brucrumus”并尝试再次输入：

@ExceptionHandler({ DataIntegrityViolationException.class })
    public ResponseEntity<String> handleError2(HttpServletRequest req, Exception ex) {

        String msg = ex.getMessage();
        if (ex.getCause().getCause() instanceof SQLException) {
            SQLException e = (SQLException) ex.getCause().getCause();

            if (e.getMessage().contains("Key")) {
                msg = e.getMessage().substring(e.getMessage().indexOf("Key"));
            }
        }

Answer 2

SQLException 作为参数传递给SQLGrammarException 构造函数。

您可以捕获它，提取根并显示本机 sql 错误或简单地重新抛出它：

} catch (SQLGrammarException e) {
    logger.info("CLASS : " + e.getClass());
    throw e.getSQLException();
} catch (Exception e){
   ...
}