Check and extract a number from a String in Java

I'm writing a program where the user enters a String in the following format:

"What is the square of 10?" 

1. I need to check that there is a number in the String
2. and then extract just the number.
3. If i use .contains("\\d+") or .contains("[0-9]+"), the program can't find a number in the String, no matter what the input is, but .matches("\\d+")will only work when there is only numbers.

What can I use as a solution for finding and extracting?

16 Answers

try this

str.matches(".*\\d.*"); 

If you want to extract the first number out of the input string, you can do-

public static String extractNumber(final String str) { if(str == null || str.isEmpty()) return ""; StringBuilder sb = new StringBuilder(); boolean found = false; for(char c : str.toCharArray()){ if(Character.isDigit(c)){ sb.append(c); found = true; } else if(found){ // If we already found a digit before and this char is not a digit, stop looping break; } } return sb.toString(); } 

Examples:

For input "123abc", the method above will return 123.

For "abc1000def", 1000.

For "555abc45", 555.

For "abc", will return an empty string.

I think it is faster than regex .

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

s=s.replaceAll("[*a-zA-Z]", "") replaces all alphabets

s=s.replaceAll("[*0-9]", "") replaces all numerics

if you do above two replaces you will get all special charactered string

If you want to extract only integers from a String s=s.replaceAll("[^0-9]", "")

If you want to extract only Alphabets from a String s=s.replaceAll("[^a-zA-Z]", "")

Happy coding :)

The code below is enough for "Check if a String contains numbers in Java"

Pattern p = Pattern.compile("([0-9])"); Matcher m = p.matcher("Here is ur string"); if(m.find()){ System.out.println("Hello "+m.find()); } 

I could not find a single pattern correct. Please follow below guide for a small and sweet solution.

String regex = "(.)*(\\d)(.)*"; Pattern pattern = Pattern.compile(regex); String msg = "What is the square of 10?"; boolean containsNumber = pattern.matcher(msg).matches(); 

Pattern p = Pattern.compile("(([A-Z].*[0-9])"); Matcher m = p.matcher("TEST 123"); boolean b = m.find(); System.out.println(b); 

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. :)

Try the following pattern:

.matches("[a-zA-Z ]*\\d+.*") 

Below code snippet will tell whether the String contains digit or not

str.matches(".*\\d.*") or str.matches(.*[0-9].*) 

For example

String str = "abhinav123"; str.matches(".*\\d.*") or str.matches(.*[0-9].*) will return true str = "abhinav"; str.matches(".*\\d.*") or str.matches(.*[0-9].*) will return false 

public String hasNums(String str) { char[] nums = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' }; char[] toChar = new char[str.length()]; for (int i = 0; i < str.length(); i++) { toChar[i] = str.charAt(i); for (int j = 0; j < nums.length; j++) { if (toChar[i] == nums[j]) { return str; } } } return "None"; } 

As I was redirected here searching for a method to find digits in string in Kotlin language, I'll leave my findings here for other folks wanting a solution specific to Kotlin.

Finding out if a string contains digit:

val hasDigits = sampleString.any { it.isDigit() } 

Finding out if a string contains only digits:

val hasOnlyDigits = sampleString.all { it.isDigit() } 

Extract digits from string:

val onlyNumberString = sampleString.filter { it.isDigit() } 

You can try this

String text = "ddd123.0114cc"; String numOnly = text.replaceAll("\\p{Alpha}",""); try { double numVal = Double.valueOf(numOnly); System.out.println(text +" contains numbers"); } catch (NumberFormatException e){ System.out.println(text+" not contains numbers"); } 

As you don't only want to look for a number but also extract it, you should write a small function doing that for you. Go letter by letter till you spot a digit. Ah, just found the necessary code for you on stackoverflow: find integer in string. Look at the accepted answer.

.matches(".*\\d+.*") only works for numbers but not other symbols like // or * etc.

ASCII is at the start of UNICODE, so you can do something like this:

(x >= 97 && x <= 122) || (x >= 65 && x <= 90) // 97 == 'a' and 65 = 'A' 

I'm sure you can figure out the other values...