正则表达式Java正则表达式以提取除最后一个数字以外的字符串

Question

How to extract all characters from a string without the last number (if exist ) in Java, I found how to extract the last number in a string using this regex [0-9.]+$ , however I want the opposite.

Examples :

abd_12df1231 => abd_12df

abcd => abcd

abcd12a => abcd12a

abcd12a1 => abcd12a

Answer 1

What you might do is match from the start of the string ^ one or more word characters \\w+ followed by not a digit using \\D

^\\w+\\D

As suggested in the comments, you could expand the characters you want to match using a character class ^[\\w-]+\\D or if you want to match any character you could use a dot ^.+\\D

Answer 2

If you want to remove one or more digits at the end of the string, you may use

s = s.replaceFirst("[0-9]+$", ""); 

See the regex demo

To also remove floats, use

s = s.replaceFirst("[0-9]*\\.?[0-9]+$", ""); 

See another regex demo

Details

• (?s) - a Pattern.DOTALL inline modifier
• ^ - start of string
• (.*?) - Capturing group #1: any 0+ chars other than line break chars as few as possible
• \\\\d*\\\\.?\\\\d+ - an integer or float value
• $ - end of string.

Java demo :

List<String> strs = Arrays.asList("abd_12df1231", "abcd", "abcd12a", "abcd12a1", "abcd12a1.34567");
for (String str : strs)
    System.out.println(str + " => \"" + str.replaceFirst("[0-9]*\\.?[0-9]+$", "") + "\"");

Output:

abd_12df1231 => "abd_12df"
abcd => "abcd"
abcd12a => "abcd12a"
abcd12a1 => "abcd12a"
abcd12a1.34567 => "abcd12a"

To actually match a substring from start till the last number, you may use

(?s)^(.*?)\d*\.?\d+$

See the regex demo

Java code:

String s = "abc234 def1.566";
Pattern pattern = Pattern.compile("(?s)^(.*?)\\d*\\.?\\d+$");
Matcher matcher = pattern.matcher(s);
if (matcher.find()){
    System.out.println(matcher.group(1)); 
} 

Answer 3

With this Regex you could capture the last digit(s)

\d+$

You could save that digit and do a string.replace(lastDigit,"");