How to parse String to BigDecimal without getting error

Question

When I want to parse value from String array to BigDecimal I have this error:

Exception in thread "main" java.text.ParseException: Unparseable number: "86400864008640086400222"

I am searching on internet for the solution how to fix this. Maybe you guys know?

It was not my idea to use BigDecimal but unfortunately I have to. I founded some code which supposed to change value from String to BigDecimal and return it:

public static BigDecimal parseCurrencyPrecise (String value) throws ParseException
{
    NumberFormat  format = NumberFormat.getCurrencyInstance ();
    if (format instanceof DecimalFormat)
        ((DecimalFormat) format).setParseBigDecimal (true);

    Number  result = format.parse (value);
    if (result instanceof BigDecimal)
        return (BigDecimal) result;
    else {
        // Oh well...
        return new BigDecimal (result.doubleValue ());
    }
}

Here is my code when try to parse:

public void Function() throws ParseException {
    String [] array;
    array=OpenFile().split("\\s");
    for(int i = 10 ;i < array.length; i+= 11) {
        BigDecimal EAE = parseCurrencyPrecise(array[i]);
        System.out.println(EAE);
    }
}

OpenFile function open the file with data and read this in this wayL temp+=line+" "; This is the reason why i split by \s. This work for me with Strings and Integers but I have got problem with BigDecimal.

Greetings,

Is there a currency sign for your locale before the really large number? I can duplicate this error with a currency sign, but it works when I use a dollar sign (currency for my locale). — rgettman, Jan 05 '15 at 21:14
What is going on with this local sign? I started using BigDecimal today and I heard some article about local signs. — Agnes, Jan 05 '15 at 21:16
Please use `BigDecimal.valueOf(result.doubleValue ())` instead of `new BigDecimal (result.doubleValue ())`. — Tom, Jan 05 '15 at 21:18
Please don't use EITHER `BigDecimal.valueOf(result.doubleValue())` OR `new BigDecimal(result.doubleValue())`, since both of them will cause you to lose precision with some very big numbers. It's best to forget about the `NumberFormat`, and just use the constructor of `BigDecimal` that takes a `String` as its argument. — Dawood ibn Kareem, Jan 05 '15 at 21:23

mkobit · Accepted Answer · 2015-01-05T21:23:35.957

3

Instead of dealing with the parsing yourself you could make use of the BigDecimal(String val) constructor. From the Javadoc

BigDecimal(String val)
Translates the string representation of a BigDecimal into a BigDecimal

For example:

BigDecimal bigDecimal = new BigDecimal("86400864008640086400222");

See the Javadoc for the formats the constructor takes.

edited Jan 05 '15 at 21:23

answered Jan 05 '15 at 21:17

mkobit

43,979
12
156
150

You guys Mike Kobit and markbernard are right it works fine for me too. Thank you all for help – Agnes Jan 05 '15 at 21:31

score 2 · Answer 2 · answered Jan 05 '15 at 21:23

2

Is there any reason you can't use the constructor? It looks like you are making it more complicated than it has to be. This worked for me:

System.out.println(new BigDecimal("86400864008640086400222"));

answered Jan 05 '15 at 21:23

markbernard

1,412
9
18

How to parse String to BigDecimal without getting error

2 Answers2