I have got a String for example.
"Quantity : 2.85 kg"
Here I only need to extract the number with exact configuration 2.85.
String.replaceAll("[^0-9]",""); only extracts number as 285, but I need 2.85.
Kindly help.
2 Answers
You must be kidding guys, the pattern for a digit is
p = "/d*(./d+)?"
and extract it is :
Matcher m = Pattern.compile(p)
while(m.find())
res = m.group(0)
3 Comments
The fourth bird
No it is not. Your pattern actually does not work. You have to escape the d with a backslash, you have to escape the dot to match it literally and because everything is optional it would also match an empty string.
user8426627
ok its then "((//d*([.]//d+) | //d+)" , forgot that in java you cant control how greedy * is. Other patterns are worse in this answer
The fourth bird
I am afraid that is also not correct.
((//d*([.]//d+) | //d+) is missing a parenthesis, you have to use a backslash instead of a forward slash. With those corrections these are the matches which I think is not what OP is meaning to match. regex101.com/r/x2akHU/1Regex is any character, then one or more digits with a dot and following two digits, space and kg. A capturing group is used to capture the part thas is to be extracted.
String test = "Quantity : 2.85 kg";
Pattern pattern = Pattern.compile(".*(\\d+\\.\\d\\d) kg");
Matcher matcher = pattern.matcher(test);
if(matcher.matches()){
System.out.println(matcher.group(1));
}else{
System.out.println("no match");
}
\d+(\.\d+)?to extract the number