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up vote -2 down vote favorite

i have stats method using JPQL in jpa maven project , i have got that error while testing in the client project.

java.lang.ClassCastException: java.lang.Class cannot be cast to java.lang.String

public int StatDoctorBySpeciality()
 {

    int  count = 0 ;
    Query query = em.createQuery("SELECT  COUNT(u)  FROM User AS u where u.role like 'doctor'  GROUP BY u.specialite ");

    List<Object[]> results = query.getResultList();

        for (int i = 0; i < results.size(); i++) {
             Object[] arr =(Object[]) results.get(i);
            for (int j = 0; j < arr.length; j++) {
                System.out.print(arr[j] + " ");
                count= (int) arr[j] ;

            }
            }
        return count;
}
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2  
arr[j].toString() – Compass Nov 2 '16 at 15:22
    
remove the + " " would work too System.out.print(arr[j]); , an exception is an object, so the toString() of Object will be call. PS : I don't see why you cast the get result since this is the type of your list. – AxelH Nov 2 '16 at 15:25
    
the return type is Int – hela Nov 2 '16 at 15:26
    
Wow, you use a lot of casting here then ... First, your exception say this is a Class type, not a String because it try to cast in String during the System.out.println() because you concat it with a String + " ". Then you will have an exception because you try to cast into int a Class instance – AxelH Nov 2 '16 at 15:28
    
the error persists even if i moved the System.out.println() and it still cannaot be casted to String while i don't even have a string now ! – hela Nov 2 '16 at 15:30
up vote 0 down vote accepted

Finally it was a casting exception because getResultList() returns just List <Long> instead of List<Long[]>

Ljava.lang.Long is a list of java.lang.Longs 

public int StatDoctorBySpeciality() {
       int count =0;
    List<Long> results = em
            .createQuery("SELECT  COUNT(u)  FROM User AS u where u.role like 'doctor'  GROUP BY u.specialite").getResultList();
    for (Long result : results) {

         count = ((Number) result).intValue();
    }  
    return count ;
}
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