'How can I prevent java.lang.NumberFormatException: For input string: "N/A"?

While running my code I am getting a NumberFormatException:

java.lang.NumberFormatException: For input string: "N/A"
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.valueOf(Unknown Source)
    at java.util.TreeMap.compare(Unknown Source)
    at java.util.TreeMap.put(Unknown Source)
    at java.util.TreeSet.add(Unknown Source)`

How can I prevent this exception from occurring?

javastringintnumberformatexception


Solution 1:[1]

"N/A" is not an integer. It must throw NumberFormatException if you try to parse it to an integer.

Check before parsing or handle Exception properly.

  1. Exception Handling

    try{
    int i = Integer.parseInt(input);
} catch(NumberFormatException ex){ // handle your exception
    ...
}

or - Integer pattern matching -

String input=...;
String pattern ="-?\\d+";
if(input.matches("-?\\d+")){ // any positive or negetive integer or not!
 ...
}

Solution 2:[2]

Integer.parseInt(str) throws NumberFormatException if the string does not contain a parsable integer. You can hadle the same as below.

int a;
String str = "N/A";

try {   
   a = Integer.parseInt(str);
} catch (NumberFormatException nfe) {
  // Handle the condition when str is not a number.
}

Solution 3:[3]

Make an exception handler like this,

private int ConvertIntoNumeric(String xVal)
{
 try
  { 
     return Integer.parseInt(xVal);
  }
 catch(Exception ex) 
  {
     return 0; 
  }
}

.
.
.
.

int xTest = ConvertIntoNumeric("N/A");  //Will return 0

Solution 4:[4]

Obviously you can't parse N/A to int value. you can do something like following to handle that NumberFormatException .

   String str="N/A";
   try {
        int val=Integer.parseInt(str);
   }catch (NumberFormatException e){
       System.out.println("not a number"); 
   }

Solution 5:[5]

"N/A" is a string and cannot be converted to a number. Catch the exception and handle it. For example:

    String text = "N/A";
    int intVal = 0;
    try {
        intVal = Integer.parseInt(text);
    } catch (NumberFormatException e) {
        //Log it if needed
        intVal = //default fallback value;
    }

Solution 6:[6]

'N/A' cannot be parsed to int and we get the exception and there might a case where the provided string could be <-2147483648 or > 2147483648 (int max and min) and in such case too we get number format exception and in such case we can try as below.

String str= "8765432198";
Long num= Long.valueOf(str);
int min = Integer.MIN_VALUE;
int max = Integer.MAX_VALUE;
Integer n=0;
if (num > max) {
        n = max;
    }
if (num < min) {
        n = min;
    }
if (num <= max && num >= min)
  n = Integer.valueOf(str);

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Jitesh Prajapati
Solution 2 Jayamohan
Solution 3 Rajaprabhu Aravindasamy
Solution 4 Ruchira Gayan Ranaweera
Solution 5 rocketboy
Solution 6 NKR

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