regex getting the first instance of a number from a string

Question

I return the following string from a webpage

Order Number: 1509596 Customer Number: 8

but it could also be

Order ID 1509596 Customer 8

I want to use regex so in my method I just return the order number, I was doing

orderNumber.replaceAll("[^0-9]", "")

but that obviously doesn't work because I then have an 8 on the end of my order number. Can somebody help me how I would just the order number as I'm rubbish at reg ex!

The string will always be 50 characters long but the order number is dynamic so I don't know how long it will be, is it possible to just return the first number in the string?

You should be able to use the "[0-9]+" regex to match the first number in your string. — Nikolaos Georgiou
– Nikolaos Georgiou, Commented Jan 29, 2014 at 10:04
Would be better to do a regex match on sequences of numbers, and select the index of the one you want. If you continue the way you are doing it, this regex could work... Customer.+|[^0-9] — Billy Moon
– Billy Moon, Commented Jan 29, 2014 at 10:05

Bohemian · Accepted Answer · 2023-07-10 21:35:50Z

8

Use a regex that matches the whole input, capturing the part you want, then replace everything captured to "extract" what you want.

String firstNumber = str.replaceAll("^\\D*(\\d+).*", "$1");

See live demo.

The regex matches all leading non-digits (if any), then as many digits as are found next as capture group 1, then matches to the end. $1 is a back reference to capture group 1.

If newlines (which dot does not match by default) are in the input, enable DOTALL flag by adding (?s) to the regex:

String firstNumber = str.replaceAll("(?s)^\\D*(\\d+).*", "$1");

edited Jul 10, 2023 at 21:35

answered Jan 29, 2014 at 10:06

Bohemian♦

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2 Comments

Saeid Z Over a year ago

When new line exist on string, this method not work correctly.

Bohemian Over a year ago

@SaeidZ See edited answer for how to cater for newline characters.

score 1 · Accepted Answer · 2014-01-29 10:11:40Z

1

I can't think of a single line that would accomplish this, but the following should work:

String orderString = "Order Number: 1509596 Customer Number: 8";
Pattern p = Pattern.compile("[0-9]+");
Matcher m = p.matcher(orderString);
if (m.find())
{
  String orderNr = m.group();
  System.out.println(orderNr);
}

edited Jan 29, 2014 at 10:11

answered Jan 29, 2014 at 10:05

user330315

3 Comments

Bohemian Over a year ago

Can't think of how to do it in one line? See this... muhahahaha

BackSlash Over a year ago

You can simply use String orderNr = m.group()

user330315 Over a year ago

@BackSlash: thank, I thought group() could only be used when there were explicit groups (using (..)) in the regex.

Pulah Nandha · Accepted Answer · 2014-01-29 10:05:41Z

0

if you change orderNumber.replaceAll("[^0-9]", "")

to

orderNumber.replaceAll("[^0-9]", ",") hence

you will get ,1509596,8 as an answer,

I think this is probably help you to solve ans.

answered Jan 29, 2014 at 10:05

Pulah Nandha

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