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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?

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  • BigInt and Long are not same. Try-> long id = ((Long) employeeObject[0]).longValue(); Check this page: tutorialspoint.com/java/lang/long_longvalue.htm Commented Apr 13, 2016 at 12:56
  • Can you try this. long id = Long.parseLong(String.valueOf(employeeObject[0])) ? this should work, Commented Apr 13, 2016 at 13:01
  • changed,interesting,now i got new exception is java.lang.ClassCastException: java.lang.Boolean cannot be cast to java.lang.Integer Commented Apr 13, 2016 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. Commented Apr 13, 2016 at 13:14
  • 1
    Why aren't you using HQL ? Commented Apr 13, 2016 at 13:19

2 Answers 2

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1

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; }

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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 !

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1 Comment

long id = (Long) employeeObject[0].longValue(); incorrect line, and i am repeating that me helped this long id = Long.parseLong(String.valueOf(employeeObject[0]))

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