CODEWITHSUNDEEP
javastringreplace
Deb
Deb

Reputation: 2972

Remove all the leading zero from the number part of a string

I am trying to remove all the leading zero from the number part of a string. I have came up with this code (below). From the given example it worked. But when I add a '0' in the begining it will not give the proper output. Anybody know how to achive this? Thanks in advance

input: (2016)abc00701def00019z -> output: (2016)abc701def19z -> resut: correct

input: 0(2016)abc00701def00019z -> output: (2016)abc71def19z -> result: wrong -> expected output: (2016)abc701def19z

EDIT: The string can contain other than english alphabet.

String localReference = "(2016)abc00701def00019z";
String localReference1 = localReference.replaceAll("[^0-9]+", " ");
List<String> lists =  Arrays.asList(localReference1.trim().split(" "));
System.out.println(lists.toString());
String[] replacedString = new String[5];
String[] searchedString = new String[5];
int counter = 0;
for (String list : lists) {
   String s = CharMatcher.is('0').trimLeadingFrom(list);
   replacedString[counter] = s;
   searchedString[counter++] = list;

   System.out.println(String.format("Search: %s, replace: %s", list,s));
}
System.out.println(StringUtils.replaceEach(localReference, searchedString, replacedString));

Upvotes: 3

Views: 2485

Answers (3)

Markus Benko
Markus Benko

Reputation: 1507

str.replaceAll("(^|[^0-9])0+", "$1");

This removes any row of zeroes after non-digit characters and at the beginning of the string.

Upvotes: 3

Avaneesh
Avaneesh

Reputation: 162

java has \P{Alpha}+, which matches any non-alphabetic character and then removing the the starting Zero's.

String stringToSearch = "0(2016)abc00701def00019z"; 
Pattern p1 = Pattern.compile("\\P{Alpha}+");
Matcher m = p1.matcher(stringToSearch);
StringBuffer sb = new StringBuffer();
while(m.find()){
    m.appendReplacement(sb,m.group().replaceAll("\\b0+",""));
}
m.appendTail(sb);
System.out.println(sb.toString());

output:

(2016)abc701def19z

Upvotes: 0

Mohit
Mohit

Reputation: 608

I tried doing the task using Regex and was able to do the required according to the two test cases you gave. Also $1 and $2 in the code below are the parts in the () brackets in preceding Regex.

Please find the code below:

    public class Demo {

        public static void main(String[] args) {

            String str = "0(2016)abc00701def00019z";

/*Below line replaces all 0's which come after any a-z or A-Z and which have any number after them from 1-9. */
            str = str.replaceAll("([a-zA-Z]+)0+([1-9]+)", "$1$2");
            //Below line only replace the 0's coming in the start of the string
            str = str.replaceAll("^0+","");
            System.out.println(str);
        }
    }

Upvotes: 0

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