Question

当列与数据库列不匹配时，我们将在日志中获取异常，如下所示。但它并没有抛出异常。我甚至保留了试试。我怎么能抓住这个例外？

WARN 2015-11-04 17：25：44,155 [http-apr-8080-exec-4] [SqlExceptionHelper.java:144]：SQL错误：904，SQLState：42000 错误2015-11-04 17：25：44,455 [http-apr-8080-exec-4] [SqlExceptionHelper.java:146]：ORA-00904：＆＃34; SH_RATE_DESCRIPTIO＆＃34;：无效识别费里

 import javax.persistence.EntityManager;
 import javax.persistence.PersistenceContext;

 import org.springframework.beans.factory.annotation.Value;
 import org.springframework.stereotype.Repository;

 import com.core.model.entity.ShippingTable;

 @Repository("sTableDAO")
 public class STableDao {


 @PersistenceContext(unitName = "somaUnit")
 private EntityManager entityManager;


 public STable save(STable sTable) {    
    try{
        sTable = entityManager.merge(sTable);
    }catch(Exception e){
        e.printStackTrace();
    }
    return sTable;      
 }  
}

Answer 1

只需将此记录器的loglevel设置为致命。在代码中：

private static Logger logger = org.apache.log4j.Logger.getLogger("org.hibernate.engine.jdbc.spi.SqlExceptionHelper");
logger.setLevel(Level.FATAL);

或者：

...
<!-- ADDED THE FOLLOWING -->
<logger category="org.hibernate.engine.jdbc.spi.SqlExceptionHelper">
<level name="FATAL"/>
</logger>
...

Answer 2

尝试从DataIntegrityViolationException获取消息

catch (DataIntegrityViolationException ex) {
   handleException(ex.getRootCause().getMessage()) /*get the message and handle it*/
}

Answer 3

Level.FATAL甚至吞下了错误日志消息。设置

org.apache.log4j.Logger.getLogger("org.hibernate.engine.jdbc.spi.SqlExceptionHelper")
    .setLevel(Level.ALL);

在某个任意类的static初始化程序块中做了诀窍！

捕获hibernate异常SqlExceptionHelper

3 个答案: