How to check if array[] has a null value in java?

Question

I am reading a csv file. One of the requirements is to check if a certain column has a value or not. In this case I want to check the value in array[18]. However, I am getting

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 18

Is there any other way to check the array index if it has a value or empty?

My code:

while ((sCurrentLine = br.readLine())!= null) { 

    String[] record = sCurrentLine.split(",");   

    if(record.length > 0){ // checking if the current line is not empty on the file
       if(record[18] != null){ // as per console, I am getting the error in this line
          String readDateRecord = record[18];
          // other processing here
       }
    }
}

get the length of the array first and you can avoid IndexOutfBoundsException.Or else catch the ArrayOutofBoundsException — Renjith
– Renjith, Commented Jul 21, 2015 at 5:55
The exception says there is no element at this position in array. Show us the code where you add elements to array. — Kiki
– Kiki, Commented Jul 21, 2015 at 5:55
Is there a reason you arbitrarily accessed the 18 element? (need more code) — ydobonebi
– ydobonebi, Commented Jul 21, 2015 at 5:57

mazhar islam · Accepted Answer · 2015-07-21 06:10:08Z

2

Look, according to JavaSE7:

ArrayIndexOutOfBoundsException thrown to indicate that an array has been accessed with an illegal index. (In your case) The index is greater than or equal to the size of the array.

Means, the index 18 is not legal for array record in your code. Moreover if the array record is null then you will get another exception called NullPointerException.

To overcome your problem, there are many solutions, one can be

//assuming that record is initialized 
if(record.length > 18){
      String readDateRecord = record[18];
      ...
   }

answered Jul 21, 2015 at 6:10

mazhar islam

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Comments

Subodh Joshi · Accepted Answer · 2015-07-21 05:55:41Z

0

Here one approach is this

Object array[] = new Object[18];
boolean isempty = true;
for (int i=0; i<arr.length; i++) {
  if (arr[i] != null) {
    isempty = false;
    break;
  }
}

answered Jul 21, 2015 at 5:55

Subodh Joshi

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Razib · Accepted Answer · 2015-07-21 06:01:49Z

0

You may try the following code snippet -

int length = record.length;   
if( (n>0 && n<length-1) && record[n] != null ){

   String readDateRecord = record[n];
   //other processing here

}

answered Jul 21, 2015 at 6:01

Razib

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Sunil · Accepted Answer · 2015-07-21 06:08:29Z

0

Size is fixed in array after creation .if you get index beyond of size then it's generated ArrayIndexOutOfBoundException .So first you need to get the size of array then retrive the value from array

 int size=record.length;
 for(int i=0;i<size;i++)
  {
   if(record[i] != null){
     // other processing here
   }
 }

Declare an array with size “n” and access the n-th element. However, as we already mentioned, the indices of an array with size “n”, reside in the interval [0, n – 1].

answered Jul 21, 2015 at 6:08

Sunil

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xrcwrn · Accepted Answer · 2015-07-21 07:07:17Z

0

public static void main (String[] args) {
    Integer [] record = {1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,1};
    for(int i = 0; i < record.length; i++)
        if(record[i] == null) {
            System.out.println(i+1 + " position is null");
        }
}

edited Jul 21, 2015 at 7:07

answered Jul 21, 2015 at 6:04

xrcwrn

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1 Comment

Ian2thedv Over a year ago

This will simply print "null is null", which seems pretty silly. Better add the index i to that println

Akash Yellappa · Accepted Answer · 2019-05-30 15:50:03Z

-2

If its an arrayList then check if it contains null

array.contains(null));

answered May 30, 2019 at 15:50

Akash Yellappa

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