Problematic decimalFormat.parse() in Java

Question

There is a requirement that if user enters a number, parse it and doSomething(). If user enters a mixture of number and string, then doSomethingElse()

So, I wrote the code as under:

String userInput = getWhatUserEntered();
try {
   DecimalFormat decimalFormat = (DecimalFormat)     
   NumberFormat.getNumberInstance(<LocaleHere>);
   Number number = decimalFormat.parse(userInput);
   doSomething(number);    // If I reach here, I will doSomething

   return;
}
catch(Exception e)  {
  // Oh.. user has entered mixture of alpha and number
}

doSomethingElse(userInput);  // If I reach here, I will doSomethingElse
return;

The function getWhatUserEntered() looks as under

String getWhatUserEntered()
{
  return "1923";
  //return "Oh God 1923";
  //return "1923 Oh God";
}

But, there is a problem.

When user enters 1923 --> doSomething() is hit
When user enters Oh God 1923 --> doSomethingElse() is hit
When user enters 1923 Oh God --> doSomething() is hit. This is wrong Here I need that doSomethingElse() should be hit.

Is there any inbuilt (better) function to the thing I want to achieve ? Can my code be modified to suit needs ?

The problem is probably in getWhatUserEntered(). It should return a full line and not a single token. If you are using Scanner.next, change it to nextLine. — Eran
– Eran, Commented Oct 19, 2015 at 8:41
At present getWhatUserEntered() is hard-coded with values I gave in the code snippet. So no problem there — SimpleGuy
– SimpleGuy, Commented Oct 19, 2015 at 8:44
@UmNyobe It is returning harcoded values as String. The values are same as I pasted in the code snippet — SimpleGuy
– SimpleGuy, Commented Oct 19, 2015 at 8:44
@HRgiger, Yeah, I agree with you that one should not use try catch for logic implementation — SimpleGuy
– SimpleGuy, Commented Oct 19, 2015 at 9:24

babanin · Accepted Answer · 2015-10-19 08:48:44Z

Everything is OK due to specific DecimalFormat implementation. JavaDoc says:

Parses text from the beginning of the given string to produce a number. The method may not use the entire text of the given string.

So you have to fix your code to something like this:

  String userInput = getWhatUserEntered();
    try {
        NumberFormat formatter = NumberFormat.getInstance();
        ParsePosition position = new ParsePosition(0);
        Number number = formatter.parse(userInput, position);
        if (position.getIndex() != userInput.length())
            throw new ParseException("failed to parse entire string: " + userInput, position.getIndex());
        doSomething(number);    // If I reach here, I will doSomething

        return;
    }
    catch(Exception e)  {
        // Oh.. user has entered mixture of alpha and number
    }

    doSomethingElse(userInput);  // If I reach here, I will doSomethingElse
    return;

gapvision · Accepted Answer · 2015-10-19 08:56:52Z

3

You better use some regex e.g. userInput.matches("[0-9]+") for matching numbers only

edited Oct 19, 2015 at 8:56

answered Oct 19, 2015 at 8:49

gapvision

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5 Comments

Olivier Grégoire Over a year ago

The question says any number can be entered. This regex doesn't check for +, -, e, ., etc.

pinkfloydx33 Over a year ago

What if the number contains a decimal marker or thousands separator?

gapvision Over a year ago

well that can be fixed... But regex is an solution to the problem, isn't it? However, "-?\d*\.?\d+e[+-]?\d+" may solve the problem for scientific notation

Olivier Grégoire Over a year ago

No. What if the number is entered with a Locale you're not even aware of? The question specifying <LocaleHere> even hints that you should not use regexes.

gapvision Over a year ago

indeed, there might be cases where regexs are not feasible

Olivier Grégoire · Accepted Answer · 2015-10-19 08:49:45Z

2

DecimalFormat accepts any string if it starts with a number.

What you can do is perform an additional check.

try {
  DecimalFormat decimalFormat = (DecimalFormat)     
  NumberFormat.getNumberInstance(<LocaleHere>);
  Number number = decimalFormat.parse(userInput);
  if (number.toString().equals(userInput)) {
    doSomething(number);    // If I reach here, I will doSomething   
    return;
  }
}

edited Oct 19, 2015 at 8:49

answered Oct 19, 2015 at 8:47

Olivier Grégoire

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4 Comments

SimpleGuy Over a year ago

Ok. I asked, is there is way for me to achieve the thing in need, using some inbuilt function ?

Olivier Grégoire Over a year ago

I initially wrote the very basics of the answer. Now, I edited it to add a bit more.

SimpleGuy Over a year ago

Thanks. Since @usual developer gives more information in answer and has answered before you, so I would accept his as an answer. Thanks for your help

Olivier Grégoire Over a year ago

Indeed, his answer is better than mine.

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