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Форум программистов > Java: Общие вопросы > Ошибка при компиляции illegal start of type


Автор: lv151 23.2.2008, 15:21
Код

Buildfile: build.xml

clean:
   [delete] Deleting directory D:\java\Login\WEB-INF\classes
    [mkdir] Created dir: D:\java\Login\WEB-INF\classes

copy-resources:
     [copy] Copying 8 files to D:\java\Login\WEB-INF\classes

compile:
    [javac] Compiling 8 source files to D:\java\Login\WEB-INF\classes
    [javac] D:\java\Login\source\dummies\struts\LoginAction.java:37: illegal start of type
    [javac]                  if (loginStatus==true){
    [javac]                            ^
    [javac] D:\java\Login\source\dummies\struts\LoginAction.java:61: <identifier> expected
    [javac]           }
    [javac]                ^
    [javac] D:\java\Login\source\dummies\struts\LoginAction.java:62: 'class' or 'interface' expected
    [javac]    }
    [javac]         ^
    [javac] D:\java\Login\source\dummies\struts\LoginAction.java:69: 'class' or 'interface' expected
    [javac] ^
    [javac] 4 errors



Сам код :
Код

package dummies.struts;

import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.apache.struts.action.Action;
import org.apache.struts.action.ActionError;
import org.apache.struts.action.ActionErrors;
import org.apache.struts.action.ActionForm;
import org.apache.struts.action.ActionForward;
import org.apache.struts.action.ActionMapping;

import dummies.struts.ServiceFinder;
import dummies.struts.UserLoginForm;


/**
 * @author Mike Robinson
 *
 */
public class LoginAction extends Action {       //Retrieve the DAO Reference
             
             dummies.struts.SpringHibernateDAO springHibernateDAO = (dummies.struts.SpringHibernateDAO) ServiceFinder.getContext(request).getBean("SpringHibernateDao");
              
              
//            Create object of ActionMesssages
                ActionMessages errors = new ActionMessages();
              
                LoginForm objForm = (LoginForm) form;
                 
                   String strUserid=objForm.getUserid();
                 
                  String strPassword=objForm.getPassword();
                  
                  boolean loginStatus = springHibernateDAO.checkUserLogin(strUserid,strPassword);
                  
                  if (loginStatus==true){
                      
                      HttpSession  session = request.getSession();
                       session.setAttribute("userID", strUserid); 
                       
                       //int id =springHibernateDAO.getUserId(strUserid);
                       
                       String id=String.valueOf(springHibernateDAO.getUserId(strUserid));
                        
                       //Integer idvalue=new Integer(id);
                        
                       session.setAttribute("ID", id); 
                       
                       System.out.println("Session value:"+session.getAttribute("userID"));

                     return mapping.findForward("success");
                 } else {
                     // not allowed
                  
                  errors.add("login",new ActionMessage("error.login.invalid"));
                     saveErrors(request,errors);
                  
                     return mapping.findForward("failure");
                 }
           }
    }



    
    

    


Автор: dorogoyIV 23.2.2008, 15:58
Цитата

    [javac] D:\java\Login\source\dummies\struts\LoginAction.java:37: illegal start of type
    [javac]                  if (loginStatus==true)

переменная loginStatus не инициализирована

Автор: Kangaroo 23.2.2008, 16:02
lv151,
код в какой-нибудь метод засунь  smile 

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