[regex] Regex: Check if string contains at least one digit

I have got a text string like this:

test1test

I want to check if it contains at least one digit using a regex.

What would this regex look like?

This question is related to regex

The answer is


In perl:

if($testString =~ /\d/) 
{
    print "This string contains at least one digit"
}

where \d matches to a digit.


This:

\d+

should work

Edit, no clue why I added the "+", without it works just as fine.

\d

In Java:

public boolean containsNumber(String string)
{
    return string.matches(".*\\d+.*");
}  

Another possible solution, in case you're looking for all the words in a given string, which contain a number

Pattern

\w*\d{1,}\w*
  • \w* - Matches 0 or more instances of [A-Za-z0-9_]
  • \d{1,} - Matches 1 or more instances of a number
  • \w* - Matches 0 or more instances of [A-Za-z0-9_]

The whole point in \w* is to allow not having a character in the beginning or at the end of a word. This enables capturing the first and last words in the text.

If the goal here is to only get the numbers without the words, you can omit both \w*.

Given the string

Hello world 1 4m very happy to be here, my name is W1lly W0nk4

Matches

1 4m W1lly W0nk4

Check this example in regexr - https://regexr.com/5ff7q


you could use look-ahead assertion for this:

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

Ref this

SELECT * FROM product WHERE name REGEXP '[0-9]'

If anybody falls here from google, this is how to check if string contains at least one digit in C#:

With Regex

if (Regex.IsMatch(myString, @"\d"))
{
  // do something
}

Info: necessary to add using System.Text.RegularExpressions

With linq:

if (myString.Any(c => Char.IsDigit(c)))
{
  // do something
}

Info: necessary to add using System.Linq


The regular expression you are looking for is simply this:

[0-9]

You do not mention what language you are using. If your regular expression evaluator forces REs to be anchored, you need this:

.*[0-9].*

Some RE engines (modern ones!) also allow you to write the first as \d (mnemonically: digit) and the second would then become .*\d.*.