Remove trailing zero in Java

Question

I have Strings (from DB), which may contain numeric values. If it contains numeric values, I'd like to remove trailing zeros such as:

10.0000
10.234000

str.replaceAll("\\.0*$", ""), works on the first one, but not the second one.

A lot of the answers point to use BigDecimal, but the String I get may not be numeric. So I think a better solution probably is through the Regex.

Just convert the string to an int and convert it back to a string. — David, Feb 20 '13 at 16:17

score 103 · Accepted Answer · edited Oct 16 '19 at 11:11

103

there are possibilities:

1000    -> 1000
10.000  -> 10 (without point in result)
10.0100 -> 10.01 
10.1234 -> 10.1234

I am lazy and stupid, just

s = s.indexOf(".") < 0 ? s : s.replaceAll("0*$", "").replaceAll("\\.$", "");

Same solution using contains instead of indexOf as mentioned in some of the comments for easy understanding

 s = s.contains(".") ? s.replaceAll("0*$","").replaceAll("\\.$","") : s

edited Oct 16 '19 at 11:11

Markus

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answered Feb 20 '13 at 16:42

Kent

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A single regex suffices, see my answer. – maaartinus Sep 23 '14 at 14:00
7

You can replace `s.indexOf(".") < 0` with `!s.contains(".")` for clarity. – Bob Liberatore Dec 01 '16 at 01:15
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@Project `contains()` and `indexOf()` are doing exactly same thing. You can check the implementation of `contains()` – Kent Dec 01 '16 at 10:15
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@Kent "For clarity" – Bob Liberatore Dec 01 '16 at 21:33
@Sergey I run the code in my Eye-JRE, and it works for `0.0` and `6.0` – Kent Feb 07 '18 at 13:32
but in Android does not, too strange(( – Serg Burlaka Feb 07 '18 at 16:49

score 27 · Answer 2 · answered Feb 20 '13 at 19:50

27

Use DecimalFormat, its cleanest way

String s = "10.1200";
DecimalFormat decimalFormat = new DecimalFormat("0.#####");
String result = decimalFormat.format(Double.valueOf(s));
System.out.println(result);

answered Feb 20 '13 at 19:50

Srujan Kumar Gulla

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And surely also the slowest (and maybe even the most complicated?). – maaartinus Sep 23 '14 at 13:59
@maaartinus why is this complicated? As someone who doesn't know regex very well, I can understand this code and adapt it to what I need. Regex may be faster, but syntax-wise, it's just a mess to me. – TheWanderer Aug 29 '18 at 20:41
@TheWanderer On the second thought, I mostly agree with you. The regex is actually pretty simple (and I'd recommend to learn regexes at this level), but more complicated regexes get too messy too fast. – maaartinus Aug 30 '18 at 15:20

score 17 · Answer 3 · answered Jun 16 '17 at 12:47

Kent's string manipulation answer magically works and also caters for precision loss, But here's a cleaner solution using BigDecimal

String value = "10.234000";
BigDecimal stripedVal = new BigDecimal(value).stripTrailingZeros();

You can then convert to other types

String stringValue = stripedVal.toPlainString();
double doubleValue = stripedVal.doubleValue();
long longValue = stripedVal.longValue();

If precision loss is an ultimate concern for you, then obtain the exact primitive value. This would throw ArithmeticException if there'll be any precision loss for the primitive. See below

int intValue = stripedVal.intValueExact();

Not exactly what OP asked (does not work for non-numeric values) but exactly what I needed - thank you! — domids, Nov 25 '21 at 11:23

score 11 · Answer 4 · answered Apr 13 '15 at 10:49

11

String value = "10.010"
String s = new DecimalFormat("0.####").format(Double.parseDouble(value));
System.out.println(s);

Output:

   10.01

answered Apr 13 '15 at 10:49

PAC

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Hiya there, while this may well answer the question, please be aware that other users might not be as knowledgeable as you. Why don't you add a little explanation as to why this code works? Thanks! – Vogel612 Apr 13 '15 at 11:11

score 7 · Answer 5 · edited Jun 20 '20 at 09:12

7

I find all the other solution too complicated. Simply

s.replaceFirst("\\.0*$|(\\.\\d*?)0+$", "$1");

does the job. It tries the first alternative first, so that dot followed by all zeros gets replaced by nothing (as the group doesn't get set). Otherwise, if it finds a dot followed by some digits (as few as possible due to the lazy quantifier *?) followed by some zeros, the zeros get discarded as they're not included in the group. It works.

Warning

My code relies on my assumption that appending a unmatched group does nothing. This is true for the Oracle implementation, but not for others, including Android, which seem to append the string "null". I'd call the such implementations broken as it just may no sense, but they're correct according to the Javadoc.

edited Jun 20 '20 at 09:12

Community

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answered Jun 13 '14 at 20:04

maaartinus

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What is $1? It added null to the end of my String – Rohan Kandwal Oct 23 '14 at 05:53
@RohanKandwal `$1` is `matcher.group(1)`, but what is your test case? I have quite a few test cases as you can see and they all work. – maaartinus Oct 25 '14 at 18:19
@Everett Is there any input, too? It may be wrong, but the output is pretty useless for finding the problem. – maaartinus Dec 12 '14 at 08:06
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@maaartinus: Looking at their profiles, it is possible that they execute the code on Android. – nhahtdh Dec 12 '14 at 10:18

score 3 · Answer 6 · edited Mar 08 '16 at 14:24

3

The following works for all the following examples:

"1" -> "1"
"1.0" -> "1"
"1.01500" -> "1.015"
"1.103" -> "1.103"

s = s.replaceAll("()\\.0+$|(\\..+?)0+$", "$2");

edited Mar 08 '16 at 14:24

Dipen Shah

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answered Mar 08 '16 at 14:07

Pit

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above solution is not working for me for 890.0 I am getting 890null – Vignesh Jan 29 '19 at 13:08

score 0 · Answer 7 · answered Feb 20 '13 at 16:20

0

What about replacing

(\d*\.\d*)0*$

by

\1

?

answered Feb 20 '13 at 16:20

alestanis

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Any java built in functions to use? – fivelements Feb 20 '13 at 16:32
this has problem with the 1st example. – Kent Feb 20 '13 at 16:38

mdm · Answer 8 · 2013-02-20T17:23:37.707

0

You could replace with:

String result = (str.indexOf(".")>=0?str.replaceAll("\\.?0+$",""):str);

To keep the Regex as simple as possible. (And account for inputs like 1000 as pointed out in comments)

edited Feb 20 '13 at 17:23

answered Feb 20 '13 at 16:23

mdm

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Right. I just thought of "trailing zeros". Not to the mathematical representation of the number. My bad. In this case, Kent's answer (checking for "." before applying Regex) is more correct. – mdm Feb 20 '13 at 17:19

score 0 · Answer 9 · answered Oct 12 '18 at 12:54

My implementation with possibility to select numbers of digits after divider:

public static String removeTrailingZero(String number, int minPrecise, char divider) {
    int dividerIndex = number.indexOf(divider);

    if (dividerIndex == -1) {
        return number;
    }

    int removeCount = 0;
    for (int i = dividerIndex + 1; i < number.length(); i++) {
        if (number.charAt(i) == '0') {
            removeCount++;
        } else {
            removeCount = 0;
        }
    }

    int fracLen = number.length() - dividerIndex - 1;

    if (fracLen - removeCount < minPrecise) {
        removeCount = fracLen - minPrecise;
    }

    if (removeCount < 0) {
        return number;
    }

    String result = number.substring(0, number.length() - removeCount);

    if (result.endsWith(String.valueOf(divider))) {
        return result.substring(0, result.length() - 1);
    }

    return result;
}

score 0 · Answer 10 · answered Apr 25 '20 at 06:54

0

In addition to Kent's answer.

Be careful with regex in Kotlin. You have to manually write Regex() constructor instead of a simple string!

s = if (s.contains(".")) 
        s.replace(Regex("0*\$"),"").replace(Regex("\\.\$"),"") 
    else s

answered Apr 25 '20 at 06:54

coldembrace

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score 0 · Answer 11 · answered Dec 27 '22 at 05:25

Try to use this code:

DecimalFormat df = new DecimalFormat("#0.#####");

String value1 = df.format(101.00000);
String value2 = df.format(102.02000);
String value3 = df.format(103.20000);
String value4 = df.format(104.30020);

Output:

score -1 · Answer 12 · answered Aug 19 '14 at 21:59

Separate out the fraction part first. Then you can use the below logic.

BigDecimal value = BigDecimal.valueOf(345000);
BigDecimal div = new BigDecimal(10).pow(Integer.numberOfTrailingZeros(value.intValue()));
System.out.println(value.divide(div).intValue());

Remove trailing zero in Java

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