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

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How can I write a method to find wh开发者_如何学Pythonether a given string contains a number? The method should return true if the string contains a number and false otherwise.

---

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

---

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);
}

---

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.*"));

---

You could use a regex to find out if the String contains a number. Take a look at the matches() method.

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Why don't you try to write a function based on Integer.parseInt(String obj) ? The function could accept as parameter your String object, and then tokenize the String and use Integer.parseInt(String obj) to extract the number from the "lucky" substring...

Javadoc of Integer.parseInt(String obj)

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Looks like people like doing spoonfeeding, so I have decided to post the worst solution to an easy task:

public static boolean isNumber(String s) throws Exception {
    boolean result = false;
    byte[] bytes = s.getBytes("ASCII");
    int tmp, i = bytes.length;
    while (i >0  && (result = ((tmp = bytes[--i] - '0') >= 0) && tmp <= 9));
    return result;
}

About the worst code I could imagine, but there might be other people here who can come up with even worse solutions.

Hm, containsNumber is worse:

public static boolean containsNumber(String s) throws Exception {
    boolean result = false;
    byte[] bytes = s.getBytes("ASCII");
    int tmp, i = bytes.length;
    while (i >0  && (true | (result |= ((tmp = bytes[--i] - '0') >= 0) && tmp <= 9)));
    return result;
}

---

 try{
      Integer.parseInt(string);
      return true;
}catch (Exception  e){
      return false;
}

or you can do it himself:

 for ( int i = 0; i < string.length; ++i ) {
      if ( !( string[i] >= '0' || string[i] <= '9' ) )
           return false;
 }
 return true;

Of course is also function isDigit

---

Another possible solution is to use a Scanner object like this:

Scanner scanner = new Scanner(inputString);  
if (scanner.hasNextInt()) {  
  return true;
}
else {
  return false
}

Of course, if you are looking for a double, use hasNextDouble() method (see: Scanner javadoc)