throw simple exception with message

Question

There is simple way to throw exception with message in java ? In the following method I check for types and if the type doesn't exist i want to throw message that the type is not supported ,what is the simplest way to do that ?

public static SwitchType<?> switchInput(final String typeName) {

    if (typeName.equals("java.lang.String")) {

    }
    else if (typeName.equals("Binary")) {

    }
    else if (typeName.equals("Decimal")) {

    }

    return null;
}

Simplest way is to learn java Exception handling – Renjith Jan 28 '13 at 12:09 — Renjith, Jan 28 '13 at 12:09

score 3 · Accepted Answer · edited Feb 20 '15 at 11:51

3

Use the Exception Constructor which takes a String as parameter:

        if (typeName.equals("java.lang.String")) {

        }
        else if (typeName.equals("Binary")) {

        }
        else if (typeName.equals("Decimal")) {

        }
        else {
           throw new IllegalArgumentException("Wrong type passed");
        }

edited Feb 20 '15 at 11:51

Simon A. Eugster

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answered Jan 28 '13 at 12:09

PermGenError

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score 2 · Answer 2 · edited May 23 '17 at 12:06

2

The standard way to handle an illegal argument is to throw an IllegalArgumentException:

} else {
    throw new IllegalArgumentException("This type is not supported: " + typeName);
}

And try not to return null if you can avoid it.

edited May 23 '17 at 12:06

Community

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answered Jan 28 '13 at 12:09

assylias

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score 0 · Answer 3 · edited Jan 28 '13 at 12:33

0

this method cannot throw an exception really
because typeName in input parameter of function is a String already..

edited Jan 28 '13 at 12:33

Mxyk

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answered Jan 28 '13 at 12:14

Raghavan

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throw simple exception with message

3 Answers3