CODEWITHSUNDEEP
javathrownumberformatexception
Bek
Bek

Reputation: 45

How to print custom message using problematic input string when NumberFormatException

Exception in thread "main" java.lang.NumberFormatException: For input string: "a"

How can I access the input string which is causing NumberFormatException to print a custom error message in this form:

try {
    /* some code which includes many Integer.parseInt(some_string); */
}
catch (NumberFormatException nfe) {
    System.out.println("This is not a number: " + input_string_which_causing_the_error);
}

Then I should get:

This is not a number: a

Upvotes: 0

Views: 1698

Answers (4)

gaelle3182
gaelle3182

Reputation: 450

You can try/catch NumberFormatException where this can occur and propagate exception to "main" with a throw that contains custom message. 

try { 
    combinationLength = Integer.parseInt(strCombinationLength);
} catch (NumberFormatException e) {
    throw new NumberFormatException("The parameter \"combinationLength \" must be a number");
}

In the main 

try {
    //some code
} catch (NumberFormatException e) {
    System.out.println(e.getMessage());
}

Upvotes: 0

Ingo
Ingo

Reputation: 36339

You can extract it from the exception:

} catch (NumberFormatException e) {
    System.err.println(e.getMessage().replaceFirst(".*For input string: ", "This is not a number"));
}

Upvotes: 2

harsh
harsh

Reputation: 7692

It should be straight forward:

String some_string = "12";//retrieval logic
try {
    int num = Integer.parseInt(some_string);
} catch (NumberFormatException nfe) {
    System.out.println("This is not a number: " + some_string);
}

Upvotes: 0

Ruchira Gayan Ranaweera
Ruchira Gayan Ranaweera

Reputation: 35557

You can try something like this

    String input="abc"; // declare input such that visible to both try and catch
    try {
        int a = Integer.parseInt(input);
    } catch (NumberFormatException e) {
        System.out.println("This is not a number: " + input);
    }

Upvotes: 0

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