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!array[0].isEmpty() returning NullPointer?

Anonymous User 1856 18-Nov-2014

Here's my code:

if (!quizDescs[0].isEmpty()) {
    mDescText.setText(quizDescs[0]);
} else {
    mDescText.setVisibility(View.INVISIBLE);
}

So, when this code runs, and the if condition returns true, everything is fine and dandy, however, if it returns false, it says there's a NullPointerException, and points me to the line of code containing the if statement.

Am I checking the condition right? Why is it returning a NullPointer?!

ANSWER:

    if (quizDescs[0] == null) {
       mDescText.setVisibility(View.INVISIBLE);
    } else {
       mDescText.setText(quizDescs[0]);
    }

android java  nullpointerexception 
Updated on 19-Nov-2014
Anonymous User

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1 Answers

Tom Cruser

Tom Cruser

19-Nov-2014

if quizDesc[0] is String, you can do

if(!StringUtility.isEmptyOrNull(quizDesc[0])){
 mDescText.setText(quizDescs[0]);
}else {
    mDescText.setVisibility(View.INVISIBLE);
}

By the way, Null and being empty is not same

Consider

String s; //Initialize to null
String a =""; //A blank string
Its always a good practise to use
try{
   //Your code here..
}catch(Exception e){
    e.printStacktrace();
}

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