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UnoGeeks
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One Compiler Java
December 26, 2023 - In Java, you typically use the ... bytecode, which can be executed on the Java Virtual Machine (JVM). There is no single โ€œone compilerโ€ for Java; instead, you use the standard JDK tools for compiling Java progr...
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OneCompiler
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Java Online Compiler
Write, Run & Share Java code online using OneCompiler's Java online compiler for free. It's one of the robust, feature-rich online compilers for Java language, running the Java LTS version 17. Getting started with the OneCompiler's Java editor is easy and fast.
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How to make two files within one project in an online compiler?
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๐ŸŒ r/javahelp
8
1
September 27, 2022
java - Solving mismatch exceptions only appearing in one compiler - Stack Overflow
This is for an online assignment. The goal is to accept two inputs (in this code, the x and y variables) multiple times and then print the largest and smallest value of each variable. Here is the c... More on stackoverflow.com
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java - Online Compiler won't run more than one class - Stack Overflow
I am using the online compiler "onlinegdp" to run java and I am unable to use multiple files in my programming. The exact same code works on eclipse so I am unsure where the problem is. I don't know More on stackoverflow.com
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Just Some Help
## Just Some Help Okay, before I start. You might of seen this on the posts thingy (and if you did you are one of 10 so well done) and the reason is because *I More on onecompiler.com
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1
April 13, 2025
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OneCompiler
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Top answer
1 of 2
2

The error trace seems to indicate that the InputMismatchException is thrown when scan.nextInt() is called.

After your print System.out.println("Would you like to enter another location?");, if the user inputs anything other than an integer, scan.nextInt() will throw an exception. Even if you read the user's input as a string as suggested by someone else, the parseInt method in escape = Integer.parseInt(scan.nextLine()) will throw an error because the string may not be a valid integer.

I suggest adding try-catch block as follows around the scan.nextInt() call so that your program doesn't crash when a user inputs something other than a valid integer (like the number 8.238).

boolean inputValid = false;
System.out.println("Would you like to enter another location?");
while (!inputValid) {
    try {
        escape = scan.nextInt();
        inputValid = true;
    } catch (InputMismatchException e) {
        inputValid = false;
        System.out.println("Please enter a valid Integer");
    }
}

This code will continue asking the user until the user enters a valid integer.

2 of 2
1

It might be because the online scanner's implementation of nextInt() is different and is causing the problems.

Perhaps trying to parse the input as a string and the as a double/int would work:

Scanner scan = new Scanner(System.in);
double x; // the latitude input
double y; // the longitude input
double n = -90; // the maximum north value
double e = -180; // the maximum east value
double s = 90; // the maximum south value
double w = 180; // the maximum west value
int escape = 1; // the value that dictates when it's time to exit the
// while loop
while (escape == 1) {
    System.out.println("Please enter the latitude:");
    x = Double.parseDouble(scan.nextLine());
    System.out.println("Please enter the longitude:");
    y = Double.parseDouble(scan.nextLine());
    if ((x > n) && (x <= 90))
        n = x;
    if ((x < s) && (x >= -90))
        s = x;
    if ((y > e) && (y <= 180))
        e = y;
    if ((y < w) && (y >= -180))
        w = y;
    if ((x > 90) || (x < -90) || (y > 180) || (y < -180))
        System.out.println("Incorrect Latitude or Longitude");
    System.out.println("Would you like to enter another location?");
    escape = Integer.parseInt(scan.nextLine());
        }
System.out.println("Farthest North: " + n);
System.out.println("Farthest South: " + s);
System.out.println("Farthest East: " + e);
System.out.println("Farthest West: " + w);

Again, this depends on the online scanner's way of doing things.

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