How to remove leading zeros from alphanumeric text

Question

I ve seen questions on how to prefix zeros here in SO  But not the other way   Can you guys suggest me how to remove the leading zeros in alphanumeric text  Are there any built-in APIs or do I need to write a method to trim the leading zeros   Example   01234 converts to 1234 0001234a converts to 1234a 001234-a converts to 1234-a 101234 remains as 101234 2509398 remains as 2509398 123z remains as 123z 000002829839 converts to 2829839

User · Answer

You could replace   0       to   1  with regex

User · Answer

Use Apache Commons StringUtils class   StringUtils strip String str  String stripChars

User · Answer

And what about just searching for the first non-zero character    1-9  d    This regex finds the first digit between 1 and 9 followed by any number of digits  so for  00012345  it returns  12345   It can be easily adapted for alphanumeric strings

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Use this   String x    00123  replaceAll   0            - gt  123

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Regex is the best tool for the job  what it should be depends on the problem specification  The following removes leading zeroes  but leaves one if necessary  i e  it wouldn t just turn  0  to a blank string    s replaceFirst   0               The   anchor will make sure that the 0  being matched is at the beginning of the input  The       negative lookahead ensures that not the entire string will be matched   Test harness   String   in          01234                1234        0001234a             1234a        101234               101234        000002829839         2829839        0                    0        0000000              0        0000009              9        000000z              z        000000 z              z      for  String s   in        System out println       s replaceFirst   0                         See also   regular-expressions info   repetitions  lookarounds  and anchors  String replaceFirst String regex

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If you are using Kotlin This is the only code that you need   yourString trimStart  0

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Using regex as some of the answers suggest is a good way to do that  If you don t want to use regex then you can use this code   String s    00a0a121    while s length   gt 0  amp  amp  s charAt 0    0        s   s substring 1

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How about the regex way   String s    001234-a   s   s replaceFirst    0           The   anchors to the start of the string  I m assuming from context your strings are not multi-line here  otherwise you may need to look into  A for start of input rather than start of line   The 0  means zero or more 0 characters  you could use 0  as well   The replaceFirst just replaces all those 0 characters at the start with nothing   And if  like Vadzim  your definition of leading zeros doesn t include turning  0   or  000  or similar strings  into an empty string  a rational enough expectation   simply put it back if necessary   String s    00000000   s   s replaceFirst    0         if  s isEmpty    s    0

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I made some benchmark tests and found  that the fastest way  by far  is this solution       private static String removeLeadingZeros String s          try             Integer intVal   Integer parseInt s             s   intVal toString            catch  Exception ex                 whatever               return s          Especially regular expressions are very slow in a long iteration   I needed to find out the fastest way for a batchjob

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You can use the StringUtils class from Apache Commons Lang like this   StringUtils stripStart yourString  0

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Using Regexp with groups   Pattern pattern   Pattern compile   0          String result       Matcher matcher   pattern matcher content   if  matcher matches               first group contains 0  second group the remaining characters          000abcd -  gt  000  abcd       result   matcher group 2      return result

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To go with thelost s Apache Commons answer  using guava-libraries  Google s general-purpose Java utility library which I would argue should now be on the classpath of any non-trivial Java project   this would use CharMatcher   CharMatcher is  0   trimLeadingFrom inputString

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I think that it is so easy to do that  You can just loop over the string from the start and removing zeros until you found a not zero char   int lastLeadZeroIndex   0  for  int i   0  i  lt  str length    i        char c   str charAt i     if  c     0         lastLeadZeroIndex   i      else       break         str   str subString lastLeadZeroIndex 1  str length

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If you  like me  need to remove all the leading zeros from each  word  in a string  you can modify  polygenelubricants  answer to the following   String s    003 d0g 00ss 00 0 00   s replaceAll    b0      b           which results in   3 d0g ss 0 0 0

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You could just do  String s   Integer valueOf  0001007   toString

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Without using Regex or substring   function on String which will be inefficient -  public static String removeZero String str           StringBuffer sb   new StringBuffer str           while  sb length   gt 1  amp  amp  sb charAt 0      0               sb deleteCharAt 0           return sb toString        return in String

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String s  0000000000046457657772752256266542 56256010000085100000             String removeString          for int i  0 i lt s length   i           if s charAt i    0           removeString removeString  0         else          break             System out println  original string -   s        System out println  after removing 0 s -  s replaceFirst removeString

User · Answer

If you don t want to use regex or external library  You can do with  for     String input  0000008008451  String output   input trim    for   output length    gt  1  amp  amp  output charAt 0      0   output   output substring 1     System out println output    8008451

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A clear way without any need of regExp and any external libraries   public static String trimLeadingZeros String source        for  int i   0  i  lt  source length      i            char c   source charAt i           if  c     0                 return source substring i                       return        or return  0

[java] How to remove leading zeros from alphanumeric text?

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