CODEWITHSUNDEEP
java
ehlert5290
ehlert5290

Reputation: 33

String to Int evaluation using parse:

I'm very new to java and trying to write a program for the user to enter a number and compare it to a set of rules. I have used parse.Int after JOptionPane, but when I try an 'if' statement I recieve a bad operand type; first type-string, second type-int. I don't understand since I used parse why it will not work. I know I'm missing something.

    answer = 0;
    while(answer == JOptionPane.YES_OPTION)

  JOptionPane.showMessageDialog(null, "  enter a number between 1-50", "by ...`enter code here`",
                    JOptionPane.PLAIN_MESSAGE);
      String inputNumber; 
      inputNumber = JOptionPane.showInputDialog("Input an integer from one to 50: ");

      Integer.parseInt(inputNumber);

      if (inputNumber == 0 || inputNumber >= 51);

Upvotes: 0

Views: 125

Answers (4)

kai
kai

Reputation: 905

See comments in the code. Additionally you need a try/catch or a throws arround the Integer.parseInt (don't know where you want to have it):

answer = 0;
while(answer == JOptionPane.YES_OPTION)
  ...
  String inputNumber; 
  inputNumber = JOptionPane.showInputDialog("Input an integer from one to 50: ");

  //replace:
  //Integer.parseInt(inputNumber);
  int number = Integer.parseInt(inputNumber);

  //replace:
  //if (inputNumber == 0 || inputNumber >= 51);
  if (number == 0 || number >= 51);

Upvotes: 0

LWChris
LWChris

Reputation: 4201

Integer.parseInt is a function, it will return the int. The variable you passed as an argument, inputNumber in this case, will remain a String variable however. So you need to assing the result of Integer.parseInt to a new variable of type int and then use that variable for the comparisons.

Upvotes: 0

nullPointer
nullPointer

Reputation: 4574

It's because you're comparing a String (inputNumber) against an int value. Try this instead

int myint = Integer.parseInt(inputNumber);
if (myint == 0 || myint >= 51) {
...
}

Upvotes: 0

Jon Skeet
Jon Skeet

Reputation: 1500465

You're ignoring the return value of Integer.parseInt. The inputNumber is still a String variable, not a number. I would rename it, to be clearer:

// This variable is the string representation...
String inputText = JOptionPane.showInputDialog("Input an integer from one to 50: ");

// ... and now inputNumber is the integer value parsed from inputText
int inputNumber = Integer.parseInt(inputText);

if (inputNumber < 1 || inputNumber > 50)
{
    ...
}

Note that Integer.parseInt will throw an exception if the user doesn't enter an integer, but that's probably something to fix in a separate step...

Upvotes: 3

Related Questions