I'm writing a program where the user enters a String in the following format:
"What is the square of 10?"
.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?
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() }
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.
.matches(".*\\d+.*")
only works for numbers but not other symbols like //
or *
etc.
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;
}
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...
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.
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
Try the following pattern:
.matches("[a-zA-Z ]*\\d+.*")
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 :)
try this
str.matches(".*\\d.*");
Pattern p = Pattern.compile("(([A-Z].*[0-9])");
Matcher m = p.matcher("TEST 123");
boolean b = m.find();
System.out.println(b);
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");
}
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();
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());
}
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";
}
Source: Stackoverflow.com