Checking If a String Contains an Integer

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

How to determine whether a string contains an integer?

If you want to make sure that it is only an integer and convert it to one, I would use parseInt in a try/catch. However, if you want to check if the string contains a number then you would be better to use the String.matches with Regular Expressions: stringVariable.matches("\\d")

check if string contains an integer python

Your code works fine if you unindent the else to make it part of the for:

for num in password:
    if num.isdigit():
        break
else:
    print("Your password must contain a number.")

If it's part of the if, the else happens for every character that's not a digit; if it's part of the for, it happens at the end of the loop if the loop was never broken, which is the behavior you want.

An easier way of writing the same check is with the any function ("if there aren't any digits..."):

if not any(num.isdigit() for num in password):
    print("Your password must contain a number.")

or equivalently with all ("if all the characters aren't digits..."):

if all(not num.isdigit() for num in password):
    print("Your password must contain a number.")

In C#, how to check whether a string contains an integer?

The answer seems to be just no.

Although there are many good other answers, they either just hide the uglyness (which I did not ask for) or introduce new problems (edge cases).

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.

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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).

Checking if a string contains an int

int.ParseInt will pass only when name is an int, and has no other characters.

You can check if a string contains a number anywhere in it with LINQ using Any:

if (name.Any(Char.IsDigit)) {
    ...
}

Check and extract a number from a String in Java

The solution I went with looks like this:

Pattern numberPat = Pattern.compile("\\d+");
Matcher matcher1 = numberPat.matcher(line);

Pattern stringPat = Pattern.compile("What is the square of", Pattern.CASE_INSENSITIVE);
Matcher matcher2 = stringPat.matcher(line);

if (matcher1.find() && matcher2.find())
{
    int number = Integer.parseInt(matcher1.group());                    
    pw.println(number + " squared = " + (number * number));
}

I'm sure it's not a perfect solution, but it suited my needs. Thank you all for the help. :)

check if string contains number and return the number

1. List Comprehension : Return the digits

   To return the digits, use a list comprehension with if

   def hashnumbers(inputString):
       return [char for char in inputString if char.isdigit()]
   
   print(hashnumbers("super string"))    # []
   print(hashnumbers("super 2 string"))  # ['2']
   print(hashnumbers("super 2 3 string")) # ['2', '3']
   

2. Return a default value if no digits found (empty list is evaluated as False)

   return [char for char in inputString if char.isdigit()] or None
   

3. Regex version with re.findall

   return re.findall(r"\d", inputString) 
   return re.findall(r"\d", inputString) or None
   

4. Return first one only

   def hashnumbers(inputString):
       return next((char for char in inputString if char.isdigit()), None)
   
   print(hashnumbers("super string"))    # None
   print(hashnumbers("super 2 string"))  # 2
   print(hashnumbers("super 2 3string")) # 2
   

How can I check if a string represents an int, without using try/except?

If you're really just annoyed at using try/excepts all over the place, please just write a helper function:

def RepresentsInt(s):
    try: 
        int(s)
        return True
    except ValueError:
        return False

>>> print RepresentsInt("+123")
True
>>> print RepresentsInt("10.0")
False

It's going to be WAY more code to exactly cover all the strings that Python considers integers. I say just be pythonic on this one.

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