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how to read string upto certain comma with java

i need to read a file upto certain comma,for example;
String s=hii,lol,wow,and,finally
need output as hii,lol,wow,and
Dont want last comma followed with characters
As my code is reading last comma string
Example:iam getting my code out put as: finally
Below is my code
please guide me
File file =new File("C:/Users/xyz.txt");
FileInputStream inputStream = new FileInputStream(file);
String filke = IOUtils.toString(inputStream);
String[] pieces = filke.split("(?=,)");
String answer = Arrays.stream(pieces).skip(pieces.length - 1).collect(Collectors.joining());
String www=answer.substring(1);
System.out.format("Answer = \"%s\"%n", www);

You don't necessarily need to use regex for this. Just get the index of the last ',' and get the substring from 0 to that index:
String answer = "hii,lol,wow,and,finally";
String www = answer.substring(0, answer.lastIndexOf(','));
System.out.println(www); // prints hii,lol,wow,and

String in Java has a method called lastIndexOf(String str). That might come in handy for you.
Say your input is String s = "hii,lol,wow,and,finally";
You can do a String operation like:
String s = "hii,lol,wow,and,finally";
s = s.substring(0, s.lastIndexOf(","));
This gives you the output: hii,lol,wow,and

If you want to use java 8 stream to do it for you maybe try filter ?
String answer = Arrays.stream(pieces).filter(p -> !Objects.equals(p, pieces[pieces.length-1])).collect(Collectors.joining());
this will print Answer = "hii,lol,wow,and"

To have stricly regex you can use the Pattern.compile and Matcher
Pattern.compile("\w+(?=,)");
Matcher matcher = pattern.matcher(filke);
while (matcher.find()) {
System.out.println(matcher.group(1) + ","); // regex not good enough, maybe someone can edit it to include , (comma)
}
Will match hii, lol, wow, and,
See the regex example here https://regex101.com/r/1iZDjg/1

Using regex and android for categorizing different fields

I am currently trying do a business name card scanner app. The idea here is to take a picture of a name card and it would extract the text and categorize the text into different EditText.
I have already completed the OCR part which extract out all the text from a name card image.
What I am missing now is to make a regex method which can take this entire text extracted from OCR and categorize the name, email address, phone number into their respective fields in EditText.
Through some googling I have already found the regex formulas below:
private static final String EMAIL_PATTERN =
"[a-zA-Z0-9\\+\\.\\_\\%\\-\\+]{1,256}" +
"\\#" +
"[a-zA-Z0-9][a-zA-Z0-9\\-]{0,64}" +
"(" +
"\\." +
"[a-zA-Z0-9][a-zA-Z0-9\\-]{0,25}" +
")+";
private static final String PHONE_PATTERN =
"^[89]\\d{7}$";
private static final String NAME_PATTERN =
"/^[a-z ,.'-]+$/i";
Currently I am just able to extract out the email address using the below method:
public String EmailValidator(String email) {
Pattern pattern = Pattern.compile(EMAIL_PATTERN);
Matcher matcher = pattern.matcher(email);
if (matcher.find()) {
return email.substring(matcher.start(), matcher.end());
} else {
// TODO handle condition when input doesn't have an email address
}
return email;
}
I am unsure of how to edit the ^above method^ to include using all the 3 regex patterns at once and display them to different EditText fields like (name, email address, phone number).
--------------------------------------------EDIT-------------------------------------------------
After using #Styx answer,
it has a problem with the parameter whereby how I used to pass the text "textToUse" to the method as shown below:
I have also tried passing the text into all three parameters. But since the method is void, it cannot be done. Or if I change the method to a String instead of void, it would require a return value.

Try this code. The function takes in the recognize text and split it using break line symbol. Then run a loop and determine the type of content by running a pattern check. Whenever a pattern is determined then the loop will go into next iteration using continue keyword. This piece of code also able to handle situation where 1 or more email and phone number appear on a single business card. Hope it helps. Cheers!
public void validator(String recognizeText) {
Pattern emailPattern = Pattern.compile(EMAIL_PATTERN);
Pattern phonePattern = Pattern.compile(PHONE_PATTERN);
Pattern namePattern = Pattern.compile(NAME_PATTERN);
String possibleEmail, possiblePhone, possibleName;
possibleEmail = possiblePhone = possibleName = "";
Matcher matcher;
String[] words = recognizeText.split("\\r?\\n");
for (String word : words) {
//try to determine is the word an email by running a pattern check.
matcher = emailPattern.matcher(word);
if (matcher.find()) {
possibleEmail = possibleEmail + word + " ";
continue;
}
//try to determine is the word a phone number by running a pattern check.
matcher = phonePattern.matcher(word);
if (matcher.find()) {
possiblePhone = possiblePhone + word + " ";
continue;
}
//try to determine is the word a name by running a pattern check.
matcher = namePattern.matcher(word);
if (matcher.find()) {
possibleName = possibleName + word + " ";
continue;
}
}
//after the loop then only set possibleEmail, possiblePhone, and possibleName into
//their respective EditText here.
}

How to trim specific chars inside a string

I have this string for example:
Username: tester1tt8e677 Password: b6a492e14c
I need to get out the username and the password only.
The password and user name are dynamically changing.
What is the best way doing it with Java, and how?
Thanks.

Do yourself a massive favour and get started on regular expressions. It will take more time than blindly copy-pasting an answer, but once you grasp the concept you can solve a huge number of problems with it.
You might want to check out this tutorial, which is java-specifiy and looks quite solid. Or just go ahead and google, you will find tons of information out there.
Once again - please do learn about regular expressions. You will not regret it.

Try using a Matcher with a regex:
String pattern = "Username: (\\.+?) Password: (\\.+?)";
Matcher matcher = Pattern.compile( pattern ).matcher();
matcher.find();
You can then get your username and password from the first and second group:
String u = matcher.group(1);
String p = matcher.group(2);
However this does not sound like a good way to do whatever you are doing and you might want to consider another approach.

You could use substring:
String String1 = "Username: testter1tt8e677";
System.out.println(String1.substring(0,10));
This should return "Username:"
So:
System.out.println(String1.substring(10));
returns "testter1tt8e677".

Try with below code:
String s1 = "Username:tester1tt8e677 Password:b6a492e14c";
// splitting String
String[] splitString = s1.split(" ");
for (String sp: splitString) {
// Again Splitting
String[] s2 = sp.split(":");
System.out.println(s2[1]);
}

Rather than think of how you can work your solution around your problem, see if it can be broken up.
What do we want to do? We want to get values out of a String.
How? Rather than removing what we don't want, let's take out what we do want.
You then make the solution a lot simpler for yourself.
str = "Username: tester1tt8e677 Password: b6a492e14c";
String[] splitStrings = str.split("\\s+");
System.out.println(splitStrings[1]); //Username
System.out.println(splitStrings[3]); //Password

String str="Username: tester1tt8e677 Password: b6a492e14c";
System.out.println(str.substring(0,str.indexOf(" ", str.indexOf(" ") + 1)).split(": ")[1]);
System.out.println(str.substring(str.indexOf(" ", str.indexOf(" ") + 1)+1,str.length()).split(": ")[1]);
Output:
tester1tt8e677
b6a492e14c

Well I have found my answer and it was pretty easy:
StringBuilder pass1 = new StringBuilder(pass).delete(0, 35);
String resultString = pass1.toString();
StringBuilder user1 = new StringBuilder(user).delete(0, 10);
StringBuilder user2 = new StringBuilder(user1).delete(14, 35);
String resultString = user2.toString();
Thanks anyway (-:

I want to check if a word or a set of words exists in a String

My requirement is to check if a group of words or a single word is present in a larger string. I tried using String.contains() method but this fails in case the larger string has new line character. Currently I am using a regex mentioned below. But this works for only one word. The searched text is a user entered value and can contain more than one word. This is an android application.
String regex = ".*.{0}" + searchText + ".{0}.*";
Pattern pattern = Pattern.compile(regex);
pattern.matcher(largerString).find();
Sample String
String largerString ="John writes about this, and John writes about that," +
" and John writes about everything. ";
String searchText = "about this";

Why not just replace line breaks with spaces, and on top of that, convert it all to lower case?
String s = "hello";
String originalString = "Does this contain \n Hello?";
String formattedString = originalString.toLowerCase().replace("\n", " ");
System.out.println(formattedString.contains(s));
Edit: Thinking about it, I don't really understand how line breaks make a difference...
Edit 2: I was right. Line breaks don't matter.
String s = "hello";
String originalString = "Does this contain \nHello?";
String formattedString = originalString.toLowerCase();
System.out.println(formattedString.contains(s));

here is code not using regex.
String largerString = "John writes about this, and John writes about that," +" and John writes about everything. ";
String searchText = "about this";
Pattern pattern = Pattern.compile(searchText);
Matcher m = pattern.matcher(largerString);
if(m.find()){
System.out.println(m.group().toString());
}
Result:
about this
I hope it will help you.

Replace String in Java with regex and replaceAll

Is there a simple solution to parse a String by using regex in Java?
I have to adapt a HTML page. Therefore I have to parse several strings, e.g.:
href="/browse/PJBUGS-911"
=>
href="PJBUGS-911.html"
The pattern of the strings is only different corresponding to the ID (e.g. 911). My first idea looks like this:
String input = "";
String output = input.replaceAll("href=\"/browse/PJBUGS\\-[0-9]*\"", "href=\"PJBUGS-???.html\"");
I want to replace everything except the ID. How can I do this?
Would be nice if someone can help me :)

You can capture substrings that were matched by your pattern, using parentheses. And then you can use the captured things in the replacement with $n where n is the number of the set of parentheses (counting opening parentheses from left to right). For your example:
String output = input.replaceAll("href=\"/browse/PJBUGS-([0-9]*)\"", "href=\"PJBUGS-$1.html\"");
Or if you want:
String output = input.replaceAll("href=\"/browse/(PJBUGS-[0-9]*)\"", "href=\"$1.html\"");

This does not use regexp. But maybe it still solves your problem.
output = "href=\"" + input.substring(input.lastIndexOf("/")) + ".html\"";

This is how I would do it:
public static void main(String[] args)
{
String text = "href=\"/browse/PJBUGS-911\" blahblah href=\"/browse/PJBUGS-111\" " +
"blahblah href=\"/browse/PJBUGS-34234\"";
Pattern ptrn = Pattern.compile("href=\"/browse/(PJBUGS-[0-9]+?)\"");
Matcher mtchr = ptrn.matcher(text);
while(mtchr.find())
{
String match = mtchr.group(0);
String insMatch = mtchr.group(1);
String repl = match.replaceFirst(match, "href=\"" + insMatch + ".html\"");
System.out.println("orig = <" + match + "> repl = <" + repl + ">");
}
}
This just shows the regex and replacements, not the final formatted text, which you can get by using Matcher.replaceAll:
String allRepl = mtchr.replaceAll("href=\"$1.html\"");
If just interested in replacing all, you don't need the loop -- I used it just for debugging/showing how regex does business.