I am trying to split an inputted number such as (123) 456-7890.
String [] split = s.split(delimiters);
I have been searching the web for ways of delimiting the area code inside the set of the parentheses but I haven't found anything that works for my case. I do not know if the array is messing up with it printing either. The array is not required but I did not know what else to do since it is required to use the split method.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class HelloWorld{
public static void main(String[] args){
String phoneNumber = "(123)-456-7890";
String pattern = "\\((\\d+)\\)-(\\d+)-(\\d+)";
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(phoneNumber);
if (m.find())
System.out.println(m.group(1) + " " + m.group(2) + " " + m.group(3));
}
}
You can try it here.
If I understand your question, you could use a Pattern like (XXX) XXX-XXXX where X is a digit. You can also use {n} to require n occurences. You group with (). Something like,
String str = "(123) 456-7890";
Pattern p = Pattern.compile("\\((\\d{3})\\) (\\d{3})-(\\d{4})");
Matcher m = p.matcher(str);
if (m.matches()) {
String areaCode = m.group(1);
String first3digits = m.group(2);
String last4digits = m.group(3);
System.out.printf("(%s) %s-%s%n", areaCode, first3digits,
last4digits);
}
Gives your requested output of
(123) 456-7890
or, if you must use split you might first remove the ( and ) with a call to replaceAll and something like
String str = "(123) 456-7890";
String[] arr = str.replaceAll("[()]", "").split("[ -]");
System.out.printf("(%s) %s-%s%n", arr[0], arr[1], arr[2]);
which also gives your requested output of
(123) 456-7890
this works:
String s= "(123) 456-7890";
String[] parts = s.split("[()\\- ]");
System.out.println("(" + parts[1] + ") " + parts[3] + "-" + parts[4]);
If you must use the split method:
String s= "(123) 456-7890"
String[] split = s.split("[()-]");
System.out.println("(" + split[1] + ")" + split[2] + "-" + split[3]);
Related
My String:
BByTTheWay .I want to split the string as B By T The Way BByTheWay .That means I want to split string if I get any capital letters and last put the main string as it is. As far I tried in java:
public String breakWord(String fileAsString) throws FileNotFoundException, IOException {
String allWord = "";
String allmethod = "";
String[] splitString = fileAsString.split(" ");
for (int i = 0; i < splitString.length; i++) {
String k = splitString[i].replaceAll("([A-Z])(?![A-Z])", " $1").trim();
allWord = k.concat(" " + splitString[i]);
allWord = Arrays.stream(allWord.split("\\s+")).distinct().collect(Collectors.joining(" "));
allmethod = allmethod + " " + allWord;
// System.out.print(allmethod);
}
return allmethod;
}
It givs me the output: B ByT The Way BByTTheWay . I think stackoverflow community help me to solve this.
You may use this code:
Code 1
String s = "BByTTheWay";
Pattern p = Pattern.compile("\\p{Lu}\\p{Ll}*");
String out = p.matcher(s)
.results()
.map(MatchResult::group)
.collect(Collectors.joining(" "))
+ " " + s;
//=> "B By T The Way BByTTheWay"
RegEx \\p{Lu}\\p{Ll}* matches any unicode upper case letter followed by 0 or more lowercase letters.
CODE DEMO
Or use String.split using same regex and join it back later:
Code 2
String out = Arrays.stream(s.split("(?=\\p{Lu})"))
.collect(Collectors.joining(" ")) + " " + s;
//=> "B By T The Way BByTTheWay"
Use
String s = "BByTTheWay";
Pattern p = Pattern.compile("[A-Z][a-z]*");
Matcher m = p.matcher(s);
String r = "";
while (m.find()) {
r = r + m.group(0) + " ";
}
System.out.println(r + s);
See Java proof.
Results: B By T The Way BByTTheWay
EXPLANATION
--------------------------------------------------------------------------------
[A-Z] any character of: 'A' to 'Z'
--------------------------------------------------------------------------------
[a-z]* any character of: 'a' to 'z' (0 or more
times (matching the most amount possible))
As per requirements, you can write in this way checking if a character is an alphabet or not:
char[] chars = fileAsString.toCharArray();
StringBuilder fragment = new StringBuilder();
for (char ch : chars) {
if (Character.isLetter(ch) && Character.isUpperCase(ch)) { // it works as internationalized check
fragment.append(" ");
}
fragment.append(ch);
}
String.join(" ", fragment).concat(" " + fileAsString).trim(); // B By T The Way BByTTheWay
I have this function written in Java-Selenium and wondering how I can pass a variable(key) in place of evar90? Your support will be appreciated a lot.
public void verifyXyz(String key,String expectedResult) {
selenium.waitForPageLoad();
String resultString = selenium.executeJSCall("localStorage.getItem('s.tl-data')");
String matchedString = "";
String variable = "key";
Pattern pattern = Pattern.compile("(eVar90=[^:]*)");
Matcher matcher = pattern.matcher(resultString);
if (matcher.find()) {
matchedString = matcher.group();
}
Assert.assertTrue("Product variable does not match expected value \n"
+ "Expected: " + expectedResult + " \n"
+ "Actual: " + matchedString
, matchedString.equals(expectedResult));
}
Thank you in Advance!
I have to find the size of matched substring.
For example
string s="2---3"
Pattern p=Pattern.compile("-+");
Matcher m=p.matcher(lst_str.get(i));
if(m.find()) // answer is 3*
if String s="2--2" // then answer is 2
How can I find the size of that substring which is matched?
Just use the length property of each string match:
String s = "2---3";
Pattern p = Pattern.compile("-+");
Matcher m = p.matcher(s);
while (m.find()) {
System.out.println("MATCH: " + m.group(0) + ", with length = " + m.group(0).length());
}
If you only have to do this one string at a time, and you want a one-liner, here is a way:
String s = "2---3";
int lengthOfMatch = s.length() - s.replaceAll("-+", "").length();
I have a String
String str = (a AND b) OR (c AND d)
I tokenise with the help of code below
String delims = "AND|OR|NOT|[!&|()]+"; // Regular expression syntax
String newstr = str.replaceAll(delims, " ");
String[] tokens = newstr.trim().split("[ ]+");
and get String[] below
[a, b, c, d]
To each element of the array I add " =1" so it becomes
[a=1, b=1, c=1, d=1]
NOW I need to replace these values to the initial string making it
(a=1 AND b=1) OR (c=1 AND d=1)
Can someone help or guide me ? The initial String str is arbitrary!
This answer is based on #Michael's idea (BIG +1 for him) of searching words containing only lowercase characters and adding =1 to them :)
String addstr = "=1";
String str = "(a AND b) OR (c AND d) ";
StringBuffer sb = new StringBuffer();
Pattern pattern = Pattern.compile("[a-z]+");
Matcher m = pattern.matcher(str);
while (m.find()) {
m.appendReplacement(sb, m.group() + addstr);
}
m.appendTail(sb);
System.out.println(sb);
output
(a=1 AND b=1) OR (c=1 AND d=1)
Given:
String str = (a AND b) OR (c AND d);
String[] tokened = [a, b, c, d]
String[] edited = [a=1, b=1, c=1, d=1]
Simply:
for (int i=0; i<tokened.length; i++)
str.replaceAll(tokened[i], edited[i]);
Edit:
String addstr = "=1";
String str = "(a AND b) OR (c AND d) ";
String delims = "AND|OR|NOT|[!&|() ]+"; // Regular expression syntax
String[] tokens = str.trim().split( delims );
String[] delimiters = str.trim().split( "[a-z]+"); //remove all lower case (these are the characters you wish to edit)
String newstr = "";
for (int i = 0; i < delimiters.length-1; i++)
newstr += delimiters[i] + tokens[i] + addstr;
newstr += delimiters[delimiters.length-1];
OK now the explanation:
tokens = [a, b, c, d]
delimiters = [ "(" , " AND " , ") OR (" , " AND " , ") " ]
When iterating through delimiters, we take "(" + "a" + "=1".
From there we have "(a=1" += " AND " + "b" + "=1".
And on: "(a=1 AND b=1" += ") OR (" + "c" + "=1".
Again : "(a=1 AND b=1) OR (c=1" += " AND " + "d" + "=1"
Finally (outside the for loop): "(a=1 AND b=1) OR (c=1 AND d=1" += ")"
There we have: "(a=1 AND b=1) OR (c=1 AND d=1)"
How long is str allowed to be? If the answer is "relatively short", you could simply do a "replace all" for every element in the array. This obviously is not the most performance-friendly solution, so if performance is an issue, a different solution would be desireable.
Pattern p = Pattern.compile("(ma)|([a-zA-Z_]+)");
Matcher m = p.matcher("ma");
m.find();
System.out.println("1 " + m.group(1) + ""); //ma
System.out.println("2 " + m.group(2)); // null
Matcher m = p.matcher("mad");
m.find();
System.out.println("1 " + m.group(1) + ""); //ma
System.out.println("2 " + m.group(2)); // null
But I need that the string "mad" would be in the 2nd group.
I think what you are looking for is something like:
(ma(?!d))|([a-zA-Z_]+)
from "perldoc perlre":
"(?!pattern)"
A zero-width negative look-ahead assertion. For
example
"/foo(?!bar)/" matches any occurrence of "foo" that
isn't
followed by "bar".
the only thing I'm not sure about is whether Java supports this syntax, but I think it does.
If you use matches instead of find, it will try to match the entire string against that pattern, which it can only do by putting mad in the second group:
import java.util.regex.*;
public class Test {
public static void main(String[] args) {
Pattern p = Pattern.compile("(ma)|([a-zA-Z_]+)");
Matcher m = p.matcher("ma");
m.matches();
System.out.println("1 " + m.group(1)); // ma
System.out.println("2 " + m.group(2)); // null
m = p.matcher("mad");
m.matches();
System.out.println("1 " + m.group(1)); // null
System.out.println("2 " + m.group(2)); // mad
}
}