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spliting a string by space and dot and comma at the same time

How can I split a string by space, dot and comma at the same time? I want to get rid of them and get words only.
My code for space:
str=array.get(0).split(" ");
After advices i wrote this
str=array.get(0).split("[ ]|[.]|[,]|[ \t]");
but i see a new problem
String
New problem

The method split can be used with a Regex pattern, so you can match more elaborated cases to split your string.
A matching pattern for your case would be:
[ \.,]+
Regex Exaplanation:
[ .,]+ - The brackets create Character Set, that will match any character in the set.
[ .,]+ - The plus sign is a Quantifier, it will match the previous token (the character set) one or more times, this solves the problem where the tokens are following one another, creating empty strings in the array.
You can test it with the following code:
class Main {
public static void main(String[] args) {
String str = "Hello, World!, StackOverflow. Test Regex";
String[] split = str.split("[ .,]+");
for(String s : split){
System.out.println(s);
}
}
}
The output is:
Hello
World!
StackOverflow
Test
Regex

Using .split() can lead to having empty entries in your array.
Try this:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
String text = "This is... a real sentence, actually.";
Pattern reg = Pattern.compile("\\w+");
Matcher m = reg.matcher(text);
while (m.find()) {
System.out.println(m.group());
}

Print out the last match of a regex

I have this code:
String responseData = "http://xxxxx-f.frehd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/.m3u8";
"http://xxxxx-f.frehd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/.m3u8";
String pattern = ^(https://.*\.54325)$;
Pattern pr = Pattern.compile(pattern);
Matcher math = pr.matcher(responseData);
if (math.find()) {
// print the url
}
else {
System.out.println("No Math");
}
I want to print out the last string that starts with http and ends with .m3u8. How do I do this? I'm stuck. All help is appreciated.
The problem I have now is that when I find a math and what to print out the string, I get everything from responseData.

In case you need to get some substring at the end that is preceded by similar substrings, you need to make sure the regex engine has already consumed as many characters before your required match as possible.
Also, you have a ^ in your pattern that means beginning of a string. Thus, it starts matching from the very beginning.
You can achieve what you want with just lastIndexOf and substring:
System.out.println(str.substring(str.lastIndexOf("http://")));
Or, if you need a regex, you'll need to use
String pattern = ".*(http://.*?\\.m3u8)$";
and use math.group(1) to print the value.
Sample code:
import java.util.regex.*;
public class HelloWorld{
public static void main(String []args){
String str = "http://xxxxx-f.akamaihd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/index_0_av.m3u8" +
"EXT-X-STREAM-INF:PROGRAM-ID=1,BANDWIDTH=2795000,RESOLUTION=1280x720,CODECS=avc1.64001f, mp4a.40.2" +
"http://xxxxx-f.akamaihd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/index_6_av.m3u8";
String rx = ".*(http://.*?\\.m3u8)$";
Pattern ptrn = Pattern.compile(rx);
Matcher m = ptrn.matcher(str);
while (m.find()) {
System.out.println(m.group(1));
}
}
}
Output:
http://xxxxx-f.akamaihd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/index_6_av.m3u8
Also tested on RegexPlanet

regex last word in a sentence ending with punctuation (period)

I'm looking for the regex pattern, not the Java code, to match the last word in an English (or European language) sentence. If the last word is, in this case, "hi" then I want to match "hi" and not "hi."
The regex (\w+)\.$ will match "hi.", whereas the output should be just "hi". What's the correct regex?
thufir#dur:~/NetBeansProjects/regex$
thufir#dur:~/NetBeansProjects/regex$ java -jar dist/regex.jar
trying
a b cd efg hi
matches:
hi
trying
a b cd efg hi.
matches:
thufir#dur:~/NetBeansProjects/regex$
code:
package regex;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static void main(String[] args) {
String matchesLastWordFine = "a b cd efg hi";
lastWord(matchesLastWordFine);
String noMatchFound = matchesLastWordFine + ".";
lastWord(noMatchFound);
}
private static void lastWord(String sentence) {
System.out.println("\n\ntrying\n" + sentence + "\nmatches:");
Pattern pattern = Pattern.compile("(\\w+)$");
Matcher matcher = pattern.matcher(sentence);
String match = null;
while (matcher.find()) {
match = matcher.group();
System.out.println(match);
}
}
}
My code is in Java, but that's neither here nor there. I'm strictly looking for the regex, not the Java code. (Yes, I know it's possible to strip out the last character with Java.)
What regex should I put in the pattern?

You can use lookahead asserion. For example to match sentence without period:
[\w\s]+(?=\.)
and
[\w]+(?=\.)
For just last word (word before ".")

If you need to have the whole match be the last word you can use lookahead.
\w+(?=(\.))
This matches a set of word characters that are followed by a period, without matching the period.
If you want the last word in the line, regardless of wether the line ends on the end of a sentence or not you can use:
\w+(?=(\.?$))
Or if you want to also include ,!;: etc then
\w+(?=(\p{Punct}?$))

You can use matcher.group(1) to get the content of the first capturing group ((\w+) in your case). To say a little more, matcher.group(0) would return you the full match. So your regex is almost correct. An improvement is related to your use of $, which would catch the end of the line. Use this only if your sentence fill exactly the line!

With this regular expression (\w+)\p{Punct} you get a group count of 1, means you get one group with punctionation at matcher.group(0) and one without the punctuation at matcher.group(1).
To write the regular expression in Java, use: "(\\w+)\\p{Punct}"
To test your regular expressions online with Java (and actually a lot of other languages) see RegexPlanet

By using the $ operator you will only get a match at the end of a line. So if you have multiple sentences on one line you will not get a match in the middle one.
So you should just use:
(\w+)\.
the capture group will give the correct match.
You can see an example here

I don't understand why really, but this works:
package regex;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static void main(String[] args) {
String matchesLastWordFine = "a b cd efg hi";
lastWord(matchesLastWordFine);
String noMatchFound = matchesLastWordFine + ".";
lastWord(noMatchFound);
}
private static void lastWord(String sentence) {
System.out.println("\n\ntrying\n" + sentence + "\nmatches:");
Pattern pattern = Pattern.compile("(\\w+)"); //(\w+)\.
Matcher matcher = pattern.matcher(sentence);
String match = null;
while (matcher.find()) {
match = matcher.group();
}
System.out.println(match);
}
}
I guess regex \w+ will match all the words (doh). Then the last word is what I was after. Too simple, really, I was trying to exclude punctuation, but I guess regex does that automagically for you..?

Java: How to extract substring between two characters from a string?

I am relatively new to Java and I need some help to extract multiple substrings from a string. An example of a string is as given below:
String = "How/WRB can/MD I/PRP find/VB a/DT list/NN of/IN celebrities/NNS '/POS real/JJ names/NNS ?/."
Desired result: WRB MD PRP VB DT NN IN NNS POS JJ NNS
I have a text file with possibly thousands of similar POS-tagged lines that I need to extract the POS tags from and do some calculation based on the POS tags.
I have tried using tokenizer but didn't really get the result I wanted. I even tried using split() and saving to arrays because I need to store it and use it later and that still didn't work.
Lastly, I tried using Pattern Matcher and I am having problems with the regex as it return the word with the forward slash.
Regex: [\/](.*?)\s\b
Result: /WRB /MD ....
If there's a better way to do this, please let me know or if anyone can help me figure out what's wrong with my regex.

This should work:
String string = "How/WRB can/MD I/PRP find/VB a/DT list/NN of/IN celebrities/NNS '/POS real/JJ names/NNS ?/.";
System.out.println(string.replaceAll("[^/]+/([^ ]+ ?)", "$1"));
Prints: WRB MD PRP VB DT NN IN NNS POS JJ NNS .

If you still wanted to use pattern matching, look at positive lookbehinds. It will allow you to match a word that begins with a slash, but not actually match the slash itself.
An example would be something like this:
(?<=/).+?(?= |$)
Matches anything that starts with a slash, and is followed by a space OR the end of the string
Here is a working example written in Java:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
import java.util.LinkedList;
public class SO {
public static void main(String[] args) {
String string = "How/WRB can/MD I/PRP find/VB a/DT list/NN of/IN celebrities/NNS '/POS real/JJ names/NNS ?/.";
Pattern pattern = Pattern.compile("(?<=/).+?(?= |$)");
Matcher matcher = pattern.matcher(string);
LinkedList<String> list = new LinkedList<String>();
// Loop through and find all matches and store them into the List
while(matcher.find()) {
list.add(matcher.group());
}
// Print out the contents of this List
for(String match : list) {
System.out.println(match);
}
}
}

String string = "How/WRB can/MD I/PRP find/VB a/DT list/NN of/IN celebrities/NNS '/POS real/JJ names/NNS ?/.";
string = string .replaceAll("\\S+/", "").replace(".", "");
System.out.println(string );

What about str = str.repalceAll("\\S+/", "")? It will replace remove non-whitespace characters followed by slash.

How to extract uppercase substrings from a String in Java?

I need a piece of code with which I can extract the substrings that are in uppercase from a string in Java.
For example:
"a:[AAAA|0.1;BBBBBBB|-1.90824;CC|0.0]"
I need to extract CC BBBBBBB and AAAA

You can do it with String[] split(String regex). The only problem can be with empty strings, but it's easy to filter them out:
String str = "a:[AAAA|0.1;BBBBBBB|-1.90824;CC|0.0]";
String[] substrings = str.split("[^A-Z]+");
for (String s : substrings)
{
if (!s.isEmpty())
{
System.out.println(s);
}
}
Output:
AAAA
BBBBBBB
CC

This should demonstrate the proper syntax and method. More details can be found here http://docs.oracle.com/javase/1.5.0/docs/api/java/util/regex/Pattern.html and http://docs.oracle.com/javase/1.5.0/docs/api/java/util/regex/Matcher.html
String myStr = "a:[AAAA|0.1;BBBBBBB|-1.90824;CC|0.0]";
Pattern upperCase = Pattern.compile("[A-Z]+");
Matcher matcher = upperCase.matcher(myStr);
List<String> results = new ArrayList<String>();
while (matcher.find()) {
results.add(matcher.group());
}
for (String s : results) {
System.out.println(s);
}
The [A-Z]+ part is the regular expression which does most of the work. There are a lot of strong regular expression tutorials if you want to look more into it.

If you want just to extract all the uppercase letter use [A-Z]+, if you want just uppercase substring, meaning that if you have lowercase letters you don't need it (HELLO is ok but Hello is not) then use \b[A-Z]+\b

I think you should do a replace all regular expression to turn the character you don't want into a delimiter, perhaps something like this:
str.replaceAll("[^A-Z]+", " ")
Trim any leading or trailing spaces.
Then, if you wish, you can call str.split(" ")

This is probably what you're looking for:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class MatcherDemo {
private static final String REGEX = "[A-Z]+";
private static final String INPUT = "a:[AAAA|0.1;BBBBBBB|-1.90824;CC|0.0]";
public static void main(String[] args) {
Pattern p = Pattern.compile(REGEX);
// get a matcher object
Matcher m = p.matcher(INPUT);
List<String> sequences = new Vector<String>();
while(m.find()) {
sequences.add(INPUT.substring(m.start(), m.end()));
}
}
}