Java: java.lang.NumberFormatException

Question

import java.io.*;

public class tempdetection {
public static int celciustofarenheit(int temp){
    int fTemp = ((9 * temp)/5) + 32;
    return fTemp;
}


public static void examineTemperature(int temp){
    System.out.println("\nTemperature is " + temp + " in celcius. Hmmm...");

    int fTemp = celciustofarenheit(temp);
    System.out.println("\nThats " + fTemp + " in Farenheit...");

    if(fTemp<20)
        System.out.println("\n***Burrrr. Its cold...***\n\n");
    else if(fTemp>20 || fTemp<50)
        System.out.println("\n***The weather is niether too hot nor too cold***\n\n");
    else if(fTemp>50)
        System.out.println("\n***Holy cow.. Its scorching.. Too hot***\n\n");
}


public static void main(String[] args) throws IOException {
    int temperature;
    char c;

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    do{
        System.out.println("Input:\n(Consider the input is from the sensor)\n");

        temperature = Integer.parseInt(br.readLine());

        examineTemperature(temperature);
        System.out.println("Does the sensor wanna continue giving input?:(y/n)\n");
        c = (char) br.read();
    }while(c!= 'N' && c!='n');          //if c==N that is no then stop



}

}

This is the complete code guys.. I still din get my answer.. I've searched a lot on net but to no avail.. Also thanks for who've already helped but that din resolve my problem.. Temperature is int.. SO why i should convert to string.?? Also i tried try catch as specified by one of the member but then examineTemperature(temperature) throws n error saying its not initialized..

Input:
(Consider the input is from the sensor)

45

Temperature is 45 in celcius. Hmmm...

Thats 113 in Farenheit...

***The weather is niether too hot nor too cold***


Does the sensor wanna continue giving input?:(y/n)

N
Input:
(Consider the input is from the sensor)

Exception in thread "main" java.lang.NumberFormatException: For input string: ""
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at tempdetection.main(tempdetection.java:33)

Also it works fyn until it reaches while loop..

Answer 1

If you have seen Number format exception like this:

     java.lang.NumberFormatException: For input string: ""
     java.lang.Integer.parseInt(Integer.java:627)
     com.tejveer.hiandroiddevelopers.MainActivity.onCreate(MainActivity.java:24)
     android.app.Activity.performCreate(Activity.java:7802)    

this type of error is created then application will be crashed.

then resolved this problem by writing the separate method

like this

public void thisIsMethod(View textView){
    EditText edtFn = findViewById(R.id.efn);
    EditText edtSN = findViewById(R.id.esn);
     int mResult = Integer.parseInt(edtFn.getText().toString())*
     Integer.parseInt(edtSN.getText().toString());
}

This method write outside this block

protected void onCreate(Bundle savedInstanceState){
 super.onCreate(savedInstanceState);
 setContentView(R.layout.activity_main);  }

Answer 2

The error in the line

temperature = Integer.parseInt(br.readLine());

This is reading in the input and trying to parse it as an Integer. As the exception suggests the input is not a number i.e. NumberFormatException as Integer.parseInt() expects the argument to be a number.

There are multiple ways to fix this:

One way (I personally believe not the best) is to catch the exception and just do nothing and continue

try
{
    temperature = Integer.parseInt(br.readLine());
    // and do any code that uses temperature
    //if you don't then temperature will not be assigned
}
catch (NumberFormatException nfex)
{}

A better way would be to check the input string is a number before trying to parse

String input = br.readLine();
if(input.matches("\\d+"))  // Only ints so don't actually need to consider decimals
    //is a number... do relevant code
    temperature = Integer.parseInt(input);
else
    //not a number
    System.out.println("Not a number that was input");

Answer 3

You can't parse a string that can't convert to a integer number.

Integer.parseInt(String s) throw this exception;

Answer 4

In your do/while loop's condition, the || should be an &&:

do{
    System.out.println("Input:\n(Consider the input is from the sensor)\n");
    temperature = Integer.parseInt(br.readLine());
    examineTemperature(temperature);
    System.out.println("Does the sensor wanna continue giving input?:(y/n)\n");
    c = (char) br.read();
} while (c != 'N' && c != 'n');