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Add integer to condition to getTypeValidator #347

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@mkfreeman

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@mkfreeman

The getTypeValidator function doesn't have a case for integer. As a result, invalid results are formatted in the same way as valid integers. Perhaps:

const isValidInteger = (value) =>  Number.isInteger(value) && !Number.isNaN(value);

Screen Shot 2023-02-03 at 11 20 36 AM

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