Question

This part of my code checks to see that when the user clicks on a contact, the contact actually has a phone number, otherwise a message is displayed and nothing else is done.

I am confused as to where my null pointer exception is coming from….

Uri contactData = data.getData();
     Cursor c =  managedQuery(contactData, null, null, null, null);
     if (c.moveToFirst()) 
     {
       String stringName = c.getString(c.getColumnIndexOrThrow(People.NAME));
       if(c.getColumnIndex(People.NUMBER)!= -1 )
       {
       stringNumber = c.getString(c.getColumnIndexOrThrow(People.NUMBER));
       }
        if((!stringNumber.equals(null)) && (stringNumber.length() >= 12))
        {
         name.setText(stringName);
         number.setText(stringNumber);
        }
        else
        {
         Context context = getApplicationContext();
         CharSequence text = "That contact has an invalid phone number.";
         int duration = Toast.LENGTH_SHORT;

         Toast toast = Toast.makeText(context, text, duration);
         toast.show();
        }

     }

my error is “failure delivering result RESULTINFO” , and it also says it’s a nullpointer exception error in this line:

if((!stringNumber.equals(null)) && (stringNumber.length() >= 12))

earlier in the code, I declare stringNumber to initialize to null. I thought that since the first part of the if evals to false, it wouldn’t even check the second part (which is where I think the error is coming from).

Answer 1

If stringNumber is null, then stringNumber.equals(... will throw a null pointer exception.

Use if ((stringNumber != null) ... instead.