is there a way to use > and < with Strings?

Question

i want to ask the user to input a number between 0 and 180. but the variable has to be a String.

// Saisie du nombre de jours de location
            System.out.println("Entrez le nombre de jours de location (supérieur à 0 et inférieur ou égal à 180) : ");
            nbJoursLouer = Clavier.lireString();
            while ( nbJoursLouer < "0" || nbJoursLouer > "180") {
                System.out.println("Entrée invalide !");
                System.out.println("Entrez le nombre de jours de location (supérieur à 0 et inférieur ou égal à 180) : ");
                nbJoursLouer = Clavier.lireString();
            }

when i run and input a letter instead of a number, i get this message:

Entrez le nombre de jours de location (supérieur à 0 et inférieur ou égal à 180) : 
d
Exception in thread "main" java.lang.NumberFormatException: For input string: "d"
    at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:67)
    at java.base/java.lang.Integer.parseInt(Integer.java:668)
    at java.base/java.lang.Integer.parseInt(Integer.java:786)
    at Tp1.main(Tp1.java:100)

Does this answer your question? [How do I convert a String to an int in Java?](https://stackoverflow.com/questions/5585779/how-do-i-convert-a-string-to-an-int-in-java) — Jesse, Feb 01 '23 at 02:56
The stack trace says that the error is on line 100 of `Tp1.java`, where you're calling `parseInt`, on a value that's not a number. Presumably, that's not the code that you've shown here. — Dawood ibn Kareem, Feb 01 '23 at 03:08
The code in your question doesn't compile. How are you getting a runtime error? — tgdavies, Feb 01 '23 at 03:12

Malte Hartwig · Answer 1 · 2023-02-01T09:59:01.003

You can not use < and > for String in Java, it's a compile error. The closest equivalent is String.compareTo(String), which won't work how you intend it to: "100" would be less than "18" because strings are ordered lexicographically and Java won't try to be "smart" like JavaScript. Hence, you must convert your string to integers explicitly. Have a look at this related question for how to use Integer.parseInt().

The error message you shared shows that Integer.parseInt() is used somewhere. It tries to convert the string to a number by reading it character by character. But when it encounters a character that is invalid for integers (letters, special characters...), it will fail with the exceptions you are seeing. By using a try-catch clause, you can react to that and handle it by returning an error message similar to the one in your code.

int lireNombre() {
  while (true) {
    System.out.println("Entrez le nombre de jours de location (supérieur à 0 et inférieur ou égal à 180) : ");
    try {
      int value = Integer.parseInt(Clavier.lireString());
      if (value > 0 || value <= 180) {
        return value;
      } else {
        System.out.println("Ce n'est pas supérieur à 0 et inférieur ou égal à 180!");
      }
    } catch (NumberFormatException e) {  
      System.out.println("Ce n'est pas un nombre!");
    }
  }
  throw new RuntimeException("We should never have gotten here");
}

Dương Thái · Answer 2 · 2023-02-01T03:16:57.923

-1

u must convert String to an Integer

Integer n = Integer.valueOf(nbJoursLouer);
while ( n < 0 || n > 180)

the nbJoursLouer variable is not a number
it's just the letter "d"
Your code is wrong.

edited Feb 01 '23 at 03:16

answered Feb 01 '23 at 03:06

Dương Thái

25
2
9

that's true, but even though his code is a complete dumpster fire it looks like his issue is handling what happens when a NumberFormatException is thrown. – Krusty the Clown Feb 01 '23 at 03:35

is there a way to use > and < with Strings?

2 Answers2