Checking whether a String contains a number value in Java [closed]

Solution 1:

if(str.matches(".*\\d.*")){
   // contains a number
} else{
   // does not contain a number
}

Previous suggested solution, which does not work, but brought back because of @Eng.Fouad's request/suggestion.

Not working suggested solution

String strWithNumber = "This string has a 1 number";
String strWithoutNumber = "This string does not have a number";

System.out.println(strWithNumber.contains("\d"));
System.out.println(strWithoutNumber.contains("\d"));

Working solution

String strWithNumber = "This string has a 1 number";
if(strWithNumber.matches(".*\\d.*")){
    System.out.println("'"+strWithNumber+"' contains digit");
} else{
    System.out.println("'"+strWithNumber+"' does not contain a digit");
}

String strWithoutNumber = "This string does not have a number";
if(strWithoutNumber.matches(".*\\d.*")){
    System.out.println("'"+strWithoutNumber+"' contains digit");
} else{
    System.out.println("'"+strWithoutNumber+"' does not contain a digit");
}

Output

'This string has a 1 number' contains digit
'This string does not have a number' does not contain a digit

Solution 2:

Using a loop -

public static boolean containsDigit(final String aString)
{
    if (aString != null && !aString.isEmpty())
    {
        for (char c : aString.toCharArray())
        {
            if (Character.isDigit(c))
            {
                return true;
            }
        }
    }

    return false;
}

Using a stream -

public static boolean containsDigit(final String aString)
{
    return aString != null && !aString.isEmpty() &&
            aString.chars().anyMatch(Character::isDigit);
}

Solution 3:

Shef's answer doesn't compile for me. It looks like he's using RegEx in String.contains(). If you want to use RegEx use this:

String strWithNumber = "This string has a 1 number";
String strWithoutNumber = "This string has a number";

System.out.println(strWithNumber.matches(".*\\d.*"));
System.out.println(strWithoutNumber.matches(".*\\d.*"));