convert string line 3.14M into Long

Question

I have a method which tries to convert strings into long:

public static Long getLong(String str) { long multiplier = 1; try { Long value = -1L; str = str.replaceAll("\\s", ""); // remove whitespace str = str.replaceAll("%", ""); str = str.replaceAll(",", ""); if (str.contains("M")) { str = str.replaceAll("M", ""); multiplier = 1000000; } else if (str.contains("B")) { str = str.replaceAll("B", ""); multiplier = 1000000000; } else if (str.contains("K")){ str = str.replaceAll("K", ""); multiplier = 1000; } // an exception is thrown for values like 199.00, so removing // decimals if any str = str.contains(".") ? str.substring(0, str.lastIndexOf(".")) : str; value = Long.valueOf(str); value = value * multiplier; return value; } catch (Exception e) { throw new RuntimeException(str + " cannot be parsed into long value"); }

This method works fine values like "1K", "300M", "2B", "0" but not for the values like 3.14M or 1.1K. As mentioned in above comment Long.valueOf(str) throws exception if it has to take a input like 1.24 so I have to remove the digits after decimal (which I don't want to)

Parse it as Double, then multiply and finally round to integer — Leos Literak
– Leos Literak, Commented May 11, 2014 at 9:46

metacubed · Accepted Answer · 2014-05-11 09:50:12Z

6

Use a java.math.BigDecimal instead of a Long. This will preserve the decimal portion.

Note: DO NOT use float or double. You may run into weird precision issues while parsing the string values.

answered May 11, 2014 at 9:50

metacubed

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4 Comments

Leos Literak Over a year ago

I agree, this is better solution

Peter Lawrey Over a year ago

You will not get "weird" precision issues for up to 15 digits of accuracy. The problem with using double is that developers don't understand it, not a problem with double itself.

metacubed Over a year ago

@PeterLawrey It is better to stick to data types which were designed for the scenario in question. Float and Double are not meant for representing arbitrary decimal numbers, because of the inherent nature of floating-point formats.

Peter Lawrey Over a year ago

@metacubed I totally agree that double is not appropriate for arbitrary precision, however there is little in real life which conforms to arbitrary precision.

Jurgen Camilleri · Accepted Answer · 2014-05-11 09:49:20Z

Instead of

value = Long.valueOf(str);

use:

value = Float.valueOf(str); or value = Double.valueOf(str);

Obviously change the type of value to match which one you choose. The reason behind the is that long represents a long integer, and cannot contain numbers with decimal points.

Here's a link describing the primitive Java data types.

user207421 · Accepted Answer · 2014-05-11 09:58:15Z

if (str.contains("M")) { str = str.replaceAll("M", ""); multiplier = 1000000;

The multiplier here is a million.

 } else if (str.contains("B")) { str = str.replaceAll("B", ""); multiplier = 1000000000;

So the multiplier here should be 1, not 1,000,000,000.

 } else if (str.contains("K")){ str = str.replaceAll("K", ""); multiplier = 1000; }

This multiplier is correct.

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