java.lang.ClassCastException:java.lang.String不能转换为[Ljava.lang.Object [英] java.lang.ClassCastException: java.lang.String cannot be cast to [Ljava.lang.Object
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问题描述
我正在运行以下查询。它显示一条错误消息。
I'm running the following query. It displays an error message. How to solve this error?
List<Route>routeList=null;
List<?> companyList = session.createSQLQuery ("select name " +
"from company "+
"where company_id= " + companyId).list();
if(companyList.size() <= 0){
//throw(new AppException(1018,ErrorMessages.getString("INVALID_USER_ID")));
}
routeList = new ArrayList<Route>(companyList.size());
Route vgDetails=null;
for (int i = 0; i < companyList.size(); i++) {
vgDetails = new Route();
Object[] row = (Object[])companyList.get(i);
vgDetails.setRouteName ((String)row[0]);
routeList.add(vgDetails);
}
session.getTransaction().commit();
return routeList;
}
错误的是:
The errror is:
java.lang.ClassCastException: java.lang.String cannot be cast to [Ljava.lang.Object;
at com.claystone.server.user.UserListServiceImpl.getParticipantsDestination(UserListServiceImpl.java:902)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
at java.lang.reflect.Method.invoke(Unknown Source)
at com.google.gwt.user.server.rpc.RPC.invokeAndEncodeResponse(RPC.java:527)
at com.google.gwt.user.server.rpc.RemoteServiceServlet.processCall(RemoteServiceServlet.java:166)
at com.google.gwt.user.server.rpc.RemoteServiceServlet.doPost(RemoteServiceServlet.java:86)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:637)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.mortbay.jetty.servlet.ServletHolder.handle(ServletHolder.java:487)
at org.mortbay.jetty.servlet.ServletHandler.handle(ServletHandler.java:362)
at org.mortbay.jetty.security.SecurityHandler.handle(SecurityHandler.java:216)
at org.mortbay.jetty.servlet.SessionHandler.handle(SessionHandler.java:181)
at org.mortbay.jetty.handler.ContextHandler.handle(ContextHandler.java:729)
at org.mortbay.jetty.webapp.WebAppContext.handle(WebAppContext.java:405)
at org.mortbay.jetty.handler.HandlerWrapper.handle(HandlerWrapper.java:152)
at org.mortbay.jetty.handler.RequestLogHandler.handle(RequestLogHandler.java:49)
at org.mortbay.jetty.handler.HandlerWrapper.handle(HandlerWrapper.java:152)
at org.mortbay.jetty.Server.handle(Server.java:324)
at org.mortbay.jetty.HttpConnection.handleRequest(HttpConnection.java:505)
at org.mortbay.jetty.HttpConnection$RequestHandler.content(HttpConnection.java:843)
at org.mortbay.jetty.HttpParser.parseNext(HttpParser.java:647)
at org.mortbay.jetty.HttpParser.parseAvailable(HttpParser.java:205)
at org.mortbay.jetty.HttpConnection.handle(HttpConnection.java:380)
at org.mortbay.io.nio.SelectChannelEndPoint.run(SelectChannelEndPoint.java:395)
at org.mortbay.thread.QueuedThreadPool$PoolThread.run(QueuedThreadPool.java:488)
推荐答案
你可以写:
You can just write:
vgDetails = new Route();
vgDetails.setRouteName ((String)companyList.get(i));
companyList
已经是一个字符串列表,因为您只需选择一列。
companyList
is already a list of strings because you select just one column.
此外,由于Thilo的回答也表明,您可以直接将结果转换为 List< string>
而不是使用 List<?>
。
Additionally, as Thilo's answer also suggests, you can cast your result directly to a List<string>
instead of using List<?>
.
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