How to define if greater then in java string

Question

In the following method I need to define that only set the value if its not null and if the value is greater than "0.00" How do I define that.

 String postagePaid1 = trackInfo.getPostage();
                    if(postagePaid1 != null && ????) {
                        claim.setClPtsPostageFee(new BigDecimal(postagePaid1));
                    }

What if the string doesn't contain digits at all, or doesn't contain just digits? E.g., "foo"? Or "1.17€"? — T.J. Crowder
– T.J. Crowder, Commented Sep 22, 2014 at 16:52

Hovercraft Full Of Eels · Accepted Answer · 2014-09-22 16:53:02Z

3

You can't do numeric comparisons with Strings as it doesn't make sense in the String world, only with numbers. So first check that it's not null, and if not, convert it to a double via Double.parseDouble(...)

if (someStr != null && !someStr.trim().isEmpty()) {
  try {
    double someNumber = Double.parseDouble(someStr.trim());
    if (someNumber > 0.0) {
       // ***** here you are ****
    }
  } catch (NumberFormatException nfe) { 
      // some action if String is not numeric
  }
}

answered Sep 22, 2014 at 16:53

Hovercraft Full Of Eels

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Comments

TheLostMind · Accepted Answer · 2014-09-22 16:53:26Z

1

Don't use String values as doubles during comparison. String comparison is done in natural order, so "1111" will be lesser than "9".

In your case, use Double.parseDouble(doubleString) and then compare values.

answered Sep 22, 2014 at 16:53

TheLostMind

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Kick Buttowski · Accepted Answer · 2014-09-22 17:01:21Z

1

if postagePaid1 contains only double type value ,you have to convert the string to double type by using following code in order to be able to do comparison

Double.parseDouble(postagePaid1)

so if statement become :

 if(postagePaid1 != null && Double.parseDouble(postagePaid1)>0.00) {

edited Sep 22, 2014 at 17:01

answered Sep 22, 2014 at 16:53

Kick Buttowski

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