java.lang.ClassCastException: java.lang.Integer cannot be cast to java.math.BigInteger HTTP Status 500

Question

I'm getting the following exception.

HTTP Status 500 - Request processing failed; nested exception is java.lang.ClassCastException: java.lang.Integer cannot be cast to java.math.BigInteger

with the following code

Emploee.Contoller.java

@RequestMapping("searchEmployee")
    public ModelAndView searchEmployee(@RequestParam("searchName") String searchName) {  
        logger.info("Searching the Employee. Employee Names: "+searchName);
        List<Employee> employeeList = employeeService.getAllEmployees(searchName);
        return new ModelAndView("employeeList", "employeeList", employeeList);      
    }

EmployeeDAOImpl.java

@Override
    public List<Employee> getAllEmployees(String employeeName) { 
        String query = "SELECT e.* FROM Employees e WHERE e.name like '%"+ employeeName +"%'";
        List<Object[]> employeeObjects = hibernateUtil.fetchAll(query);
        List<Employee> employees = new ArrayList<Employee>();
        for(Object[] employeeObject: employeeObjects) {
            Employee employee = new Employee();
            long id = ((BigInteger) employeeObject[0]).longValue();
            String name = (String) employeeObject[1];
            int age = (int) employeeObject[2];
            int admin = (int) employeeObject[3];
            boolean isAdmin=false;
            if(admin==1)
            isAdmin=true;
            Date createdDate = (Date) employeeObject[4];
            employee.setId(id);
            employee.setName(name);
            employee.setAge(age);
            employee.setAdmin(isAdmin);
            employee.setCreatedDate(createdDate);
            employees.add(employee);
        }
        System.out.println(employees);
        return employees;
    }

at this line

long id = ((BigInteger) employeeObject[0]).longValue();

Does anybody have any idea?

BigInt and Long are not same. Try-> long id = ((Long) employeeObject[0]).longValue(); Check this page: http://www.tutorialspoint.com/java/lang/long_longvalue.htm — Eray Balkanli, Apr 13 '16 at 12:56
Can you try this. `long id = Long.parseLong(String.valueOf(employeeObject[0]))` ? this should work, — Deendayal Garg, Apr 13 '16 at 13:01
changed,interesting,now i got new exception is java.lang.ClassCastException: java.lang.Boolean cannot be cast to java.lang.Integer — Александр, Apr 13 '16 at 13:05
Deendayal Garg,long id = Long.parseLong(String.valueOf(employeeObject[0])) it helped, you can duplicate the answer, so that I can choose it as a right. — Александр, Apr 13 '16 at 13:14

score 1 · Accepted Answer · edited May 23 '17 at 12:24

you are executing sql statement and creating objects manually. If you use HQL or criteria Hibernate does it for you, and simplifies things. Use parametrized query, is a good practice, helps in preventing SQL injection

@Override
        public List<Employee> getAllEmployees(String employeeName) { 
            String query = "SELECT e.* FROM Employees e WHERE e.name like '%"+ employeeName +"%'";
            List<Object[]> employeeObjects = hibernateUtil.fetchAll(query);
            List<Employee> employees = new ArrayList<Employee>();
            for(Object[] employeeObject: employeeObjects) {
                Employee employee = new Employee();
                long id = ((BigInteger) employeeObject[0]).longValue();
                String name = (String) employeeObject[1];
                int age = (int) employeeObject[2];
                int admin = (int) employeeObject[3];
                boolean isAdmin=false;
                if(admin==1)
                isAdmin=true;
                Date createdDate = (Date) employeeObject[4];
                employee.setId(id);
                employee.setName(name);
                employee.setAge(age);
                employee.setAdmin(isAdmin);
                employee.setCreatedDate(createdDate);
                employees.add(employee);
            }
            System.out.println(employees);
            return employees;
        }

When you use HQL it looks like this

    @Override
        public List<Employee> getAllEmployees(String employeeName) { 
        Session session = //initialize session            
        Query query = session.createQuery("FROM Employees e WHERE e.name like '%"+ ? + "%'");
           query.setParameter(0, "%"+employeeName+"%");
           List<Employee>  employees = query.list();
           System.out.println(employees);
           return employees;
        }

Check This Ans

score 0 · Answer 2 · answered Apr 13 '16 at 13:03

0

Your id is of type long, so try to cast it to Long instead of BigInteger.

Like this: long id = (Long) employeeObject[0].longValue();

Hope this helps !

answered Apr 13 '16 at 13:03

serendipity

103
1
10

long id = (Long) employeeObject[0].longValue(); incorrect line, and i am repeating that me helped this long id = Long.parseLong(String.valueOf(employeeObject[0])) – Александр Apr 13 '16 at 13:28

java.lang.ClassCastException: java.lang.Integer cannot be cast to java.math.BigInteger HTTP Status 500

2 Answers2