Develop Reference

use regex to map TY_111.22-L007-C010 from a text

I want to get TY_111.22-L007-C010,Tzo11-L010-C100 and Tff-L010-C110 from this string with regex
"12.5*MAX(\"TY_111.22-L007-C010\";\"Tzo11-L010-C100\";\"Tff-L010-C110\")
I tested this T.*-L\d*-C\d* but it don't give the result I want :
My code java for test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "T.*-L\\d*-C\\d*";
final String string = "\"12.5*MAX(\\\"TY_111.22-L007-C010\\\";\\\"Tzo11-L010-C100\\\";\\\"Tff-L010-C110\\\"";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}

You need to use this regex T.*?\-L\d*?\-C\d*
final String regex = "T.*?\\-L\\d*?\\-C\\d*";
Note: you need to escape the hyphens \- and use non-greedy quantifier .*? instead of .*, also you can use only matcher.group() instead of matcher.group(0), in your regex you don't have any groups, so the 0 is useless.
Outputs
Full match: TC_24.00-L010-C090
Full match: TC_24.00-L010-C100
Full match: TC_24.00-L010-C110

Why use a verbose regex pattern matcher when you can handle the problem with one line of code:
String input = "12.5*MAX(\"Txxxx-L007-C010\";\"Txxxx-L010-C100\";\"Txxxx-L010-C110\")";
String[] matches = input.replaceAll("^.*?\"|\"[^\"]*$", "")
.split("\";\"");
System.out.println(Arrays.toString(matches));
This prints:
[Txxxx-L007-C010, Txxxx-L010-C100, Txxxx-L010-C110]
OK...I used three lines of code, but the first and third are just for setting up the data and printing it.

RegEx for matching mp3 URLs

How can I get an mp3 url with REGEX?
This mp3 url, for example:
https://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3
This is a what I've tried so far but I want it to only accept a url with '.mp3' on the end.
(https?|ftp|file)://[-a-zA-Z0-9+&##/%?=~_|!:,.;]*[-a-zA-Z0-9+&##/%=~_|]

This expression would likely pass your desired inputs:
^(https?|ftp|file):\/\/(www.)?(.*?)\.(mp3)$
If you wish to add more boundaries to it, you can do that. For instance, you can add a list of chars instead of .*.
I have added several capturing groups, just to be simple to call, if necessary.
RegEx
If this wasn't your desired expression, you can modify/change your expressions in regex101.com.
RegEx Circuit
You can also visualize your expressions in jex.im:
const regex = /^(https?|ftp|file):\/\/(www.)?(.*?)\.(mp3)$/gm;
const str = `https://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3
http://soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3
http://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3
ftp://soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3
file://localhost/examples/mp3/SoundHelix-Song-1.mp3
file://localhost/examples/mp3/SoundHelix-Song-1.wav
file://localhost/examples/mp3/SoundHelix-Song-1.avi
file://localhost/examples/mp3/SoundHelix-Song-1.m4a`;
let m;
while ((m = regex.exec(str)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`Found match, group ${groupIndex}: ${match}`);
});
}
Java Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "^(https?|ftp|file):\\/\\/(www.)?(.*?)\\.(mp3)$";
final String string = "https://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3\n"
+ "http://soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3\n"
+ "http://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3\n"
+ "ftp://soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3\n"
+ "file://localhost/examples/mp3/SoundHelix-Song-1.mp3\n"
+ "file://localhost/examples/mp3/SoundHelix-Song-1.wav\n"
+ "file://localhost/examples/mp3/SoundHelix-Song-1.avi\n"
+ "file://localhost/examples/mp3/SoundHelix-Song-1.m4a";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}

If you want it to match inputs ending with '.mp3' you should add \.mp3$ at the end of your regex.
$ indicates the end of your expression
(https?|ftp|file):\/\/[-a-zA-Z0-9+&##\/%?=~_|!:,.;]*[-a-zA-Z0-9+&##\/%=~_|]\.mp3$
Matching:
https://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3 **=> Match**
https://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp4 **=> No Match**

You could use anchors to assert the start ^ and the end $ of the string and end the pattern with .mp3:
^https?://\S+\.mp3$
Explanation
^ Assert start of string
https?:// Match http with optional s and ://
\S+ Match 1+ times a non whitespace char
\.mp3 Match .mp3
$ Assert end of string
Regex demo | Java demo
For example:
String regex = "^https?://\\S+\\.mp3$";
String[] strings = {
"https://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3",
"https://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp4"
};
Pattern pattern = Pattern.compile(regex);
for (String s : strings) {
Matcher matcher = pattern.matcher(s);
if (matcher.find()) {
System.out.println(matcher.group(0));
}
}
Result
https://www.soundhelix.com/examples/mp3/SoundHelix-Song-1.mp3

How to use REGEX to validate EditText?

I want to validate a username against these requirements:
Just accept character or digital
At least one character
I tried with
public boolean validateFormat(String input){
return Pattern.compile("^[A-Za-z0-9]+$").matcher(input).matches();
}
How can I do this one?

Try with this regex:
^(\w|\d)+$
^ indicates the start of the string
$ indicates the end of the string
\w means any word character
\d means any digit
| is the logical OR operator
Anyway, i suggest you to use an online regex tester like regex101.com .It is very helpful to quickly test regular expressions.
Hope it can help!
== UPDATE ==
In Java code:
final String regex = "^(\\w|\\d)+$";
final String string = "myCoolUsername12";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
if(matcher.matches()) {
// if you are interested only in matching the full regex
}
// Otherwise, you can iterate over the matched groups (including the full match)
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}

/^[A-Za-z0-9]+(?:[ _-][A-Za-z0-9]+)*$/

Get an array of Strings matching a pattern from a String

I have a long string let's say
I like this #computer and I want to buy it from #XXXMall.
I know the regular expression pattern is
Pattern tagMatcher = Pattern.compile("[#]+[A-Za-z0-9-_]+\\b");
Now i want to get all the hashtags in an array. How can i use this expression to get array of all hash tags from string something like
ArrayList hashtags = getArray(pattern, str)

You can write like?
private static List<String> getArray(Pattern tagMatcher, String str) {
Matcher m = tagMatcher.matcher(str);
List<String> l = new ArrayList<String>();
while(m.find()) {
String s = m.group(); //will give you "#computer"
s = s.substring(1); // will give you just "computer"
l.add(s);
}
return l;
}
Also you can use \\w- instead of A-Za-z0-9-_ making the regex [#]+[\\w]+\\b

This link would surely be helpful for achieving what you want.
It says:
The find() method searches for occurrences of the regular expressions
in the text passed to the Pattern.matcher(text) method, when the
Matcher was created. If multiple matches can be found in the text, the
find() method will find the first, and then for each subsequent call
to find() it will move to the next match.
The methods start() and end() will give the indexes into the text
where the found match starts and ends.
Example:
String text =
"This is the text which is to be searched " +
"for occurrences of the word 'is'.";
String patternString = "is";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(text);
int count = 0;
while(matcher.find()) {
count++;
System.out.println("found: " + count + " : "
+ matcher.start() + " - " + matcher.end());
}
You got the hint now.

Here is one way, using Matcher
Pattern tagMatcher = Pattern.compile("#+[-\\w]+\\b");
Matcher m = tagMatcher.matcher(stringToMatch);
ArrayList<String> hashtags = new ArrayList<>();
while (m.find()) {
hashtags.add(m.group());
}
I took the liberty of simplifying your regex. # does not need to be in a character class. [A-Za-z0-9_] is the same as \w, so [A-Za-z0-9-_] is the same as [-\w]

You can use :
String val="I like this #computer and I want to buy it from #XXXMall.";
String REGEX = "(?<=#)[A-Za-z0-9-_]+";
List<String> list = new ArrayList<String>();
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(val);
while(matcher.find()){
list.add(matcher.group());
}
(?<=#) Positive Lookbehind - Assert that the character # literally be matched.

you can use the following code for getting the names
String saa = "#{akka}nikhil#{kumar}aaaaa";
Pattern regex = Pattern.compile("#\\{(.*?)\\}");
Matcher m = regex.matcher(saa);
while(m.find()) {
String s = m.group(1);
System.out.println(s);
}
It will print
akka
kumar

Regexp matching in group with special characters

I have a String that is something like this:
A20130122.0000+0000-0015+0000_name
Then I would like to extract this information:
The 20130122.0000+0000-0015+0000 that will be parsed to a date later on.
And the final part which is name.
So I am using in Java something like this:
String regexpOfdate = "[0-9]{8}\\.[0-9]{4}\\+[0-9]{4}-[0-9]{4}\\+[0-9]{4}";
String regexpOfName = "\\w+";
Pattern p = Pattern.compile(String.format("A(%s)_(%s)", regexpOfdate, regexpOfName));
Matcher m = p.matcher(theString);
String date = m.group(0);
String name = m.group(1);
But I am getting a java.lang.IllegalStateException: No match found
Do you know what I am doing wrong?

You aren't calling Matcher#find or Matcher#matches methods after this line:
Matcher m = p.matcher(theString);
Try this code:
Matcher m = p.matcher(theString);
if (m.find()) {
String date = m.group(1);
String name = m.group(2);
System.out.println("Date: " + date + ", name: " + name);
}

Matcher#group will throw IllegalStateException if the matcher's regex hasn't yet
been applied to its target text, or if the previous application was not successful.
Matcher#find applies the matcher's regex to the current region of the matcher's target text, returning a Boolean indicating whether a match is found.
Refer
You can try this :
String theString="A20130122.0000+0000-0015+0000_name";
String regexpOfdate = "([0-9]{8})\\.[0-9]{4}\\+[0-9]{4}-[0-9]{4}\\+[0-9]{4}";
String regexpOfName = "(\\w+)";
Pattern p = Pattern.compile(String.format("A(%s)_(%s)", regexpOfdate, regexpOfName));
Matcher m = p.matcher(theString);
if(m.find()){
String date = m.group(2);
String name = m.group(3);
System.out.println("date: "+date);
System.out.println("name: "+name);
}
OUTPUT
date: 20130122
name: name
Refer Grouping in REGEX