Convert an array of string to an array of BigDecimals

Question

I'm trying to convert a String[] to BigDecimal[] but i'm getting java.lang.NumberFormatException

this is my code

BigDecimal montanttt[] = new BigDecimal[montantt.length];
        for (int i = 0; i < montantt.length; i++) {
            montanttt[i] = new BigDecimal(montantt[i]);
            System.out.println(montanttt[i]);
        }

String[] montantt = request.getParameterValues("montant"); — Wassim Abbas
– Wassim Abbas, Commented Mar 31, 2016 at 9:39
What is the content of request.getParameterValues("montant");? — PendingValue
– PendingValue, Commented Mar 31, 2016 at 9:40
I think he means what is the actual string content. For example ['James', 'Harry', '2']. — Tiz
– Tiz, Commented Mar 31, 2016 at 9:40
Please also give your variables better names.. Now it so easy to confuse montanttt and montantt — Rob Audenaerde
– Rob Audenaerde, Commented Mar 31, 2016 at 9:41

Cootri · Accepted Answer · 2016-03-31 09:43:23Z

3

You should check for exception using try/catch in the for loop:

for (int i = 0; i < montantt.length; i++) {
    try {
        montanttt[i] = new BigDecimal(montantt[i]);
        System.out.println(montanttt[i]);
    } catch (NumberFormatException e) {
        System.out.println("Exception while parsing: " + montantt[i]);
    }
}

answered Mar 31, 2016 at 9:43

Cootri

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Comments

Andrew · Accepted Answer · 2016-03-31 09:57:30Z

2

The compact solution on Java 8 is here.

Arrays.stream(montantt).map(s -> {
        try { 
            return new BigDecimal(s);
        } catch (NumberFormatException e) {
            return BigDecimal.ZERO;
        }
}).toArray(BigDecimal[]::new);

You will get NumberFormatException if String cannot be converted to BigDecimal. But you can return some constant if the exception was thrown, for example BigDecimal.ZERO.

It works fine for your input ["0.50","0.20"].

edited Mar 31, 2016 at 9:57

answered Mar 31, 2016 at 9:42

Andrew

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1 Comment

Rudy Velthuis Over a year ago

If montantt contains invalid strings, that won't help much.

Alex Salauyou · Accepted Answer · 2016-03-31 10:08:07Z

1

Try to clear all chars that are not part of number and normalize decimal point before parsing, like:

String normalized = montantt[i].replace(",", ".").replaceAll("[^-.0-9eE]", "");
BigDecimal d = new BigDecimal(normalized);

edited Mar 31, 2016 at 10:08

answered Mar 31, 2016 at 10:00

Alex Salauyou

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Comments

Charif DZ · Accepted Answer · 2016-03-31 10:16:18Z

1

try this:

BigDecimal montanttt[] = new BigDecimal[montantt.length];
    for (int i = 0; i < montantt.length; i++) {
        try {
         montanttt[i] =BigDecimal.valueOf(Double.valueOf(montantt[i]))
        System.out.println(montanttt[i]);
    } catch (Exception e) {
      e.printStackTrace();
    }           
    }

answered Mar 31, 2016 at 10:16

Charif DZ

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