How to Detect If a String Contains at Least a Number

Check if a string contains a number

You can use any function, with the str.isdigit function, like this

def has_numbers(inputString):
    return any(char.isdigit() for char in inputString)

has_numbers("I own 1 dog")
# True
has_numbers("I own no dog")
# False

Alternatively you can use a Regular Expression, like this

import re
def has_numbers(inputString):
    return bool(re.search(r'\d', inputString))

has_numbers("I own 1 dog")
# True
has_numbers("I own no dog")
# False

Regex: Check if string contains at least one digit

I'm surprised nobody has mentioned the simplest version:

\d

This will match any digit. If your regular expression engine is Unicode-aware, this means it will match anything that's defined as a digit in any language, not just the Arabic numerals 0-9.

There's no need to put it in [square brackets] to define it as a character class, as one of the other answers did; \d works fine by itself.

Since it's not anchored with ^ or $, it will match any subset of the string, so if the string contains at least one digit, this will match.

And there's no need for the added complexity of +, since the goal is just to determine whether there's at least one digit. If there's at least one digit, this will match; and it will do so with a minimum of overhead.

How to detect if a string contains at least a number?

Use this:

SELECT * FROM Table WHERE Column LIKE '%[0-9]%'

MSDN - LIKE (Transact-SQL)

How do I check if a string contains at least one number, letter, and character that is neither a number or letter?

You can use Regex:

I took it from here: Regex for Password

var checkPassword = function(password){
    return !!password.match(/^(?=.*[A-Za-z])(?=.*\d)(?=.*[$@$!%* #+=\(\)\^?&])[A-Za-z\d$@$!%* #+=\(\)\^?&]{3,}$/);
};

I use this Regex:

Minimum 3 characters at least 1 Alphabet, 1 Number and 1 Special Character:

"^(?=.*[A-Za-z])(?=.*\d)(?=.*[$@$!%* #=+\(\)\^?&])[A-Za-z\d$@$!%* #=+\(\)\^?&]{3,}$"

This regex will enforce these rules:

At least one English letter, (?=.*?[A-Za-z])

At least one digit, (?=.*\d)

At least one special character, (?=.[$@$!% #+=()\^?&]) Add more if you like...

Minimum length of 3 characters (?=.[$@$!% #?&])[A-Za-z\d$@$!%* #+=()\^?&]{3,} include spaces

If you want to add more special characters, you can add it to the Regex like I have added '(' (you need to add it in two places).

And for those of you who ask yourself what are those two exclamation points, here is the answer: What is the !! (not not) operator in JavaScript?

Java String - See if a string contains only numbers and not letters

If you'll be processing the number as text, then change:

if (text.contains("[a-zA-Z]+") == false && text.length() > 2){

to:

if (text.matches("[0-9]+") && text.length() > 2) {

Instead of checking that the string doesn't contain alphabetic characters, check to be sure it contains only numerics.

If you actually want to use the numeric value, use Integer.parseInt() or Double.parseDouble() as others have explained below.

As a side note, it's generally considered bad practice to compare boolean values to true or false. Just use if (condition) or if (!condition).

Regex to determine if string contains at least 2 numbers

Try this:

.* matches everything (from 0 to n times)
[0-9] is number from 0-9

Pattern pattern = Pattern.compile("[0-9].*[0-9]");

Matcher matcher = pattern.matcher(string);

if (matcher.find()) {

   return true;

}