Regex to get first number in string with other characters

Question

I\'m new to regular expressions, and was wondering how I could get only the first number in a string like 100 2011-10-20 14:28:55. In this case, I\'d want it to

Answer 1

Assuming there's always a space between the first two numbers, then

preg_match('/^(\d+)/', $number_string, $matches);
$number = $matches[1]; // 100

But for something like this, you'd be better off using simple string operations:

$space_pos = strpos($number_string, ' ');
$number = substr($number_string, 0, $space_pos);

Regexs are computationally expensive, and should be avoided if possible.

Answer 2

[0-9] means the numbers 0-9 can be used the + means 1 or more times. if you use [0-9]{3} will get you 3 numbers

Answer 3

/^[^\d]*(\d+)/

This will start at the beginning, skip any non-digits, and match the first sequence of digits it finds

EDIT: this Regex will match the first group of numbers, but, as pointed out in other answers, parseInt is a better solution if you know the number is at the beginning of the string

Answer 4

NOTE: In Java, when you define the patterns as string literals, do not forget to use double backslashes to define a regex escaping backslash (\. = "\\.").

To get the number that appears at the start or beginning of a string you may consider using

^[0-9]*\.?[0-9]+       # Float or integer, leading digit may be missing (e.g, .35)
^-?[0-9]*\.?[0-9]+     # Optional - before number (e.g. -.55, -100)
^[-+]?[0-9]*\.?[0-9]+  # Optional + or - before number (e.g. -3.5, +30)

See this regex demo.

If you want to also match numbers with scientific notation at the start of the string, use

^[0-9]*\.?[0-9]+([eE][+-]?[0-9]+)?        # Just number
^-?[0-9]*\.?[0-9]+([eE][+-]?[0-9]+)?      # Number with an optional -
^[-+]?[0-9]*\.?[0-9]+([eE][+-]?[0-9]+)?   # Number with an optional - or  +

See this regex demo.

To make sure there is no other digit on the right, add a \b word boundary, or a (?!\d) or (?!\.?\d) negative lookahead that will fail the match if there is any digit (or . and a digit) on the right.

Answer 5

the below code would do the trick.

Integer num = Integer.parseInt("100 2011-10-20 14:28:55");

Answer 6

public static void main(String []args){
        Scanner s=new Scanner(System.in);
        String str=s.nextLine();
        Pattern p=Pattern.compile("[0-9]+");
        Matcher m=p.matcher(str);
        while(m.find()){
            System.out.println(m.group()+" ");

        }