Kotlin substring until specific character - string

Question

I have a string that starts with 7 characters (let's say XXXXXXX) and after that there are 8 characters that represnet a decimal number.
For example: XXXXXXX30.00000 would be 30.00 (I want 2 decimal places)
So it should start on index 7 read until dot (.) + 2 decimal places. It should be a string, not a number. I tried with string.substring() but got stuck here.

Answer 1

First you can remove the first 7 characters by
var newString = yourString.removeRange(0,6)
then you can cast to a double if you're certain it will always be a number
var yourNumber = newString.ToDouble()
If you're not sure you can wrap in a try/catch eg:
try{
var yourNumber = newString.ToDouble()
}catch(e:TypeCastException){
println("Failed to cast to double - "+ e.message)
}
additionally, to round to a 2 decimal places:
val number2digits:Double = String.format("%.2f", yourNumber).toDouble()

Answer 2

I suggest you do it using the regex, [A-Za-z]{8}(\d+.\d{2})
Explanation of the regex:
[A-Za-z]{8} specifies 8 characters. If you want to support characters other than English alphabets, you can replace [A-Za-z] with \p{L} which specifies unicode letters.
(\d+.\d{2}) specifies a capturing group consisting of digits followed by . which in turn should be followed by 2 digits as per your requirement. A regex pattern can have more than one capturing groups. Since it is the first capturing group in this pattern, it can be accessed by group(1).
A test code:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static void main(String[] args) {
// Some test strings
String[] arr = { "abcdefgh30.00000", "abcdefgh300.0000", "abcdefgh3000.00", "abcdefgh3.00000",
"abcdefgh0.05000" };
Pattern pattern = Pattern.compile("[A-Za-z]{8}(\\d+.\\d{2})");
for (String s : arr) {
Matcher matcher = pattern.matcher(s);
if (matcher.find()) {
String str = matcher.group(1);
System.out.println(str);
}
}
}
}
Output:
30.00
300.00
3000.00
3.00
0.05