我在hibernate中编写一个查询,我的方法是:
public String a(Object[] selectedAlarmId,
Object[] selecteddeviceID, String loggedInUser, String clientIp,
String role) throws Exception {
int result = 0;
b grp = new b();
try {
Session hiberSession = HibernateSessionFactory.currentSession();
Transaction transaction = hiberSession.beginTransaction();
List deviceQueryList = null;
String clauseAppender = ("select max(severity),devicenodeid from AlarmMappingBean where devicenodeid in (:devicelist) group by devicenodeid");
Query deviceQuery = hiberSession.createQuery(clauseAppender);
deviceQuery.setParameterList("devicelist", selecteddeviceID);
deviceQueryList = deviceQuery.list();
Iterator<Object[]> iter = deviceQueryList.iterator();
while(iter.hasNext()){
Object[] objAlarm = iter.next();
System.out.println(objAlarm.length);
System.out.println("Severity - > " + objAlarm[0]);
System.out.println("Device Node ID - > " + objAlarm[1]);
Query updateMangedNode = hiberSession
.createQuery("update ManagedNode set highestSeverity =? where nodeId = ?");
updateMangedNode.setParameter(0, objAlarm[0]);
updateMangedNode.setParameter(1, Long.parseLong(objAlarm[1].toString()));
//updateMangedNode.executeUpdate();
}
// Long[] deviceArray =(Long[]) selecteddeviceID;
Exception occurs here--> Object[] devArray = (Long[]) selecteddeviceID;
Query groupQuery = hiberSession
.createQuery("select groupId from b where nodeId in (:devicelist)");
groupQuery.setParameterList("devicelist", devArray);
List<ManagedNode> devicelist = new ArrayList<ManagedNode>();
devicelist = groupQuery.list();
if(!(devicelist.isEmpty() )){
Iterator<ManagedNode> itergroup =devicelist.iterator();
while(itergroup.hasNext()){
ManagedNode objgroup = itergroup.next();
grp.updateGroupHighestSeverity(objgroup.getGroupId());
}
}
transaction.commit();
} catch (Exception e) {
e.printStackTrace();
throw e;
} finally {
HibernateSessionFactory.closeSession();
}
return Integer.toString(result);
}
这里selecteddeviceId的值为[1234,12345,null,null] 并从表b中查询“从b中选择groupid,其中nodeid in(devicelist)”这里groupid的数据类型为int,nodeid为long
它给了我一个classcast异常作为Ljava.lang.Object;不可能是 施放到[Ljava.lang.Long;
我正在使用PostgresSQl
请帮忙
答案 0 :(得分:1)
从您的代码:Object[] devArray = (Long[]) selecteddeviceID;
如果需要将selecteddeviceID(已经是Object [])分配给Object [],为什么要将它转换为Long []?
尝试这样做
Long[] devArray = Arrays.copyOf(selecteddeviceID, selecteddeviceID.length, Long[].class)