Is there a way to find out how many numbers are at the end of a string without knowing the exact index?

user_10224292 0
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I have a method that extracts a certain substring from a string. This substring consists of the numbers in the string. Then this is parsed to an integer.

Method:

protected int startIndex() throws Exception {
    String str = getWorkBook().getDefinedName("XYZ");
    String sStr = str.substring(10,13);

    return Integer.parseInt(sStr) - 1;
}

Example:

String :

'0 DB'!$B$460

subString :

460

Well, I manually entered the index range for the substring. But I would like to automate it.

My approach:

String str = getWorkBook().getDefinedName("XYZ");
int length = str.length();
String sStr = str.substring(length - 3, length);

This works well for this example.

Now there is the problem that the numbers at the end of the string can also be 4 or 5 digits. If that is the case, I naturally get a NullPointerException.

Is there a way or another approach to find out how many numbers are at the end of the string?

substring java string
5 Answers

You can use the regex, (?<=\D)\d+$ which means one or more digits (i.e. \d+) from the end of the string, preceded by non-digits (i.e. \D).

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {
    public static void main(String[] args) {
        // Test
        System.out.println(getNumber("'0 DB'!$B$460"));
    }

    static String getNumber(String str) {
        Matcher matcher = Pattern.compile("(?<=\\D)\\d+$").matcher(str);
        if (matcher.find()) {
            return matcher.group();
        }

        // If no match is found, return the string itself
        return str;
    }
}
user_108195730

In your case I would recommend to use regex with replaceAll like this:

String sStr = str.replaceAll(".*?([0-9]+)$", "$1");

This will extract the all the digits in the end or your String or any length.

Also I think you are missing the case when there are no digit in your String, for that I would recommend to check your string before you convert it to an Integer.

String sStr = str.replaceAll(".*?([0-9]+)$", "$1");
if (!sStr.isEmpty()) {
    return Integer.parseInt(sStr) - 1;    
}
return 0; // or any default value
user_55580720

If you just want to get the last number, you can go through the entire string on revert and get the start index:

protected static int startIndex() {
    String str = getWorkBook().getDefinedName("XYZ");

    if(Character.isDigit(str.charAt(str.length() - 1))) {
        for(int i = str.length() - 1; i >= 0; i--){
            if(!Character.isDigit(str.charAt(i)))
                return i+1;
        }
    }
    return -1;
}

and then print it:

public static void main(String[] args)  {
    int start = startIndex();
    if(start != -1)
       System.out.println(getWorkBook().getDefinedName("XYZ").substring(start));
   else 
      System.out.println("No Number found");
}

You will have to add the

user_13668710

Simple and fast solution without RegEx:

public class Main
{
    public static int getLastNumber(String str) {
        int index = str.length() - 1;
        while (index > 0 && Character.isDigit(str.charAt(index)))
            index--;
            
        return Integer.parseInt(str.substring(index + 1));        
    }
    
    public static void main(String[] args) {
        final String text = "'0 DB'!$B$460";
        System.out.println(getLastNumber(text));
    }
}

The output will be:

460
user_3653850

If I were going to do this I just search from the end. This is quite efficient. It returns -1 if no positive number is found. Other return options and the use of an OptionalInt could also be used.

String s = "'0 DB'!$B$460";
int i;
for (i = s.length(); i > 0 && Character.isDigit(s.charAt(i-1)); i--);
int vv = (i < s.length()) ? Integer.valueOf(s.substring(i)) : -1;
System.out.println(vv);

Prints

460

If you know that there will always be a number at the end you can forget the ternary (?:) above and just do the following:

int vv = Integer.valueOf(s.substring(i));
user_15525340
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