如何删除长字符串的前导零

Question

I'm trying to remove leading zeros from a string where there are multiple other strings and numbers in java

so ow -   00250 =     00000000 ]

for the above input I'm trying to get the output to be

so ow - 250 = 0 ]

I have tried

removeLeadingZeros(str.trim().replaceAll(" +", " ")

where removeLeadingZeros is

    // Regex to remove leading
    // zeros from a string
    String regex = "^0+(?!$)";

    // Replaces the matched
    // value with given string
    str = str.replaceAll(regex, "");

    return str;

but I am getting the output

so ow -   00250 =     00000000 ]

which is the same as the original input.

These methods seem to only work for the first element in the string if it is a number such as

00015039 so ow + - 003948 83

which returns

15039 so ow + - 003948 83

but want to return

15039 so ow + - 3948 83

thanks!

Answer 1

You can split the string at each space character, then process each substring through your function, and then join them together again.

String[] substrings = myString.split(" ", 0);
for (int i = 0; i < substring.length; i++) {
    substrings[i] = removeLeadingZeros(substrings[i]);
}
myString = String.Join(" ", substrings);

Please excuse any typos. I'm writing this from my phone, so I haven't been able to put it through its paces. Hopefully the general idea comes through.

Answer 2

By splitting the given string into array and then remove the leading zeroes, is working fine.

where removeLeadingZeros is:

public static  String removeLeading(String str) {
String regex = "^0+(?!$)";
String x[] = str.split(" ");
StringBuilder z = new StringBuilder();
    
for(String y:x) {
    str = y.replaceAll(regex, "");
    z.append(" " + str);
}
return z.toString();
}

Answer 3

tl;dr

Arrays
.stream( input.split( " " ) )
.filter( s -> ! s.isBlank() ) 
.map(
    s -> {
        try{
            return String.valueOf( Integer.parseInt( s ) ) ; 
        } catch ( NumberFormatException e ) {
            return s ; 
        }
    }
)
.collect( Collectors.joining( " " ) )

Split into parts, parsing each as an int

I would split the string into parts, delimiting by SPACE character.

input.split( " " )

Splitting results in an array. Make a stream of that array.

String input = "so ow -   00250 =     00000000 ]" ;
String output = 
    Arrays
    .stream( input.split( " " ) )
…

All those extra spaces will result in a bunch of chunks that are empty strings, no text within. So filter those out.

And we want to eliminate whitespace as well as empty strings, so call String#isBlank rather than String#isEmpty .

.filter( s -> ! s.isBlank() ) 

Try to parse each part as an integer number. If the parsing fails, an exception is thrown. Trap for that exception. If caught, simply return the text we tried and failed to parse. If no exception, return a string of the newly minted int value.

.map(
    s -> {
        try{
            int x = Integer.parseInt( s ) ;
            return String.valueOf( x ) ; 
        } catch ( NumberFormatException e ) {
            return s ;  // Swallow the exception.
        }
    }
)

Collect all our remaining and newly-created parts with a SPACE as their delimiter.

.collect( Collectors.joining( " " ) )

Code example

Pulling all that code together looks like this:

String input = "so ow -   00250 =     00000000 ]" ;
String output = 
    Arrays
    .stream( input.split( " " ) )
    .filter( s -> ! s.isBlank() ) 
    .map(
        s -> {
            try{
                int x = Integer.parseInt( s ) ;
                return String.valueOf( x ) ; 
            } catch ( NumberFormatException e ) {
                return s ;  // Swallow the exception.
            }
        }
    )
    .collect( Collectors.joining( " " ) )
;
System.out.println( output ) ;

See that code run live at IdeOne.com .

so ow - 250 = 0 ]

Answer 4

You can use

String result = text.replaceAll("\\b(?<!\\d[,.])(?:(0)+\\b|0+(?=[1-9]))", "$1").replaceAll("(?U)\\s+", " ").trim();

See the regex demo .

Depending on how messy your input is you might need to adjust or even expand the solution. Details :

• \b(?<,\d[..]) - a word boundary, there must be start of string or a non-word char immediately to the right, and there can be no digit + a dot or a comma immediately to the left of the current location (that is, we exclude the decimal digit matching)
• (?:(0)+\b|0+(?=[1-9])) - either of the two patterns:
  • (0)+\b - 0 , one or more times, the last 0 is saved in Group 1, followed with a word boundary
  • | - or
  • 0+(?=[1-9]) - one or more 0 chars followed with a non-zero digit.

.replaceAll("(?U)\\s+", " ") shrinks any Unicode whitespace character chunks into a single ASCII space.