We Keep Coding

Using regular expression, how to remove matching sequence at the beginning and ending of the text but keeping what's in the middle?

my problem is very simple but I can't figure out the correct regular expression I should use.
I have the following variable (Java) :
String text = "\033[1mYO\033[0m"; // this is ANSI for bold text in the Terminal
My goal is to remove the ANSI codes with a single regular expression (I just want to keep the plain text at the middle). I cannot modify the text in any way and those ANSI codes will always be at the same place (so one at the beginning, one at the end, though sometimes it's possible that there is none).
With this regular expression, I will remove them using replaceAll method :
String plainText = text.replaceAll(unknownRegex, "");
Any idea on what the unknown regex could be?

Well, you use a single regex that has the ansi codes optionally at the beginning and end, captures anything in between and replaces the entire string with the value of the group: text.replaceAll("^(?:\\\\\\d+\\[1m)?(.*?)(?:\\\\\\d+\\[0m)?$", "$1"). (this might not capture every ansi code - adjust if needed).
Breaking the expression down (note that the example above escapes backslashes for Java strings so they are doubled):
^ is the start of the string
(?:\\\d+\[1m)? matches an optional \<at least 1 digit>[1m
(.*?) matches any text but as little as possible, and captures it into group 1
(?:\\\d+\[0m)? atches an optional \<at least 1 digit>[0m
$ is the end of the input
In the replacement $1 refers to the value of capturing group 1 which is (.*?) in the expression.

Found the answer thanks to a comment that disappeared.
Actually, i just need to make a group to get what's in the middle of the string and using it ($1) to replace the whole thing :
String plainText = text.replaceAll("\\033\\[.*m(.+)\\033\\[.*m", "$1")
Not sure if this will remove every ANSI codes but that is enough for what I want to do.

How to match ^(d+) in a particular text using regex

For example I have text like below :
case1:
(1) Hello, how are you?
case2:
Hi. (1) How're you doing?
Now I want to match the text which starts with (\d+).
I have tried the following regex but nothing is working.
^[\(\d+\)], ^\(\d+\).

[] are used to match any of the things you specify inside the brackets, and are to be followed by a quantifier.
The second regexp will work: ^\(\d+\), so check your code.
Check also so there's no space in front of the first parenthesis, or add \s* in front.
EDIT: Also, java can be tricky with escapes depending on if the regexp you type is directly translated to a regexp or is first a string literal. You may need to double escape your escapes.

In Java you have to escape parenthesis, so "\\(\\d+\\)" should match (1) in case one and two. Adding ^ as you did "^\\(\\d+\\)" will match only case1.
You have to use double back slashes within java string. Consider this
"\n" give you [line break]
"\\n" give you [backslash][n]

If you are going to downvote my post, at least comment to tell me WHY it's not useful.
I believe Java's Regex Engine supports Positive Lookbehind, in which case you can use the following regex:
(?<=[(][0-9]{1,9999}[)]\s?)\b.*$
Which matches:
The literal text (
Any digit [0-9], between 1 and 9999 times {1,9999}
The literal text )
A space, between 0 and 1 times \s?
A word boundary \b
Any character, between 0 and unlimited times .*
The end of a string $

regex certain character can exist or not but nothing after that

I'm new to regex and I'm trying to do a search on a couple of string.
I wanted to check if a certain character, in this case its ":" (without the quote) exist on the strings.
If : does not exist in the string it would still match, but if : exist there should be nothing after that only space and new line will be allowed.
I have this pattern, but it does not seem to work as I want it.
(.*)(:?\s*\n*)
Thank you.

If I understand your question correctly, ^[^:]*(:\s*)?$
Let's break this down a bit:
^ Starting anchor; without this, the match can restart itself every time it sees another colon, or non-whitespace following a colon.
[^:]* Match any number of characters that AREN'T colon characters; this way, if the entire string is non-colon characters, the string is treated as a valid match.
(:\s*)? If at any point we do see a colon, all following characters must be white space until the end of the string; the grouping parens and following ? act to make this an all-or-nothing conditional statement.
$ Ending anchor; without this, the regex won't know that if it sees a colon the following whitespace MUST persist until the end of the string.

here is a pattern which should work
/^([^:]*|([^:]*:\s*))$/
you can use the pipe to manage alternatives

Another way is :
^[^:]*(|:[\n]*)$
^[^:]* => starts with anything except :
(|:[\n]*)$ => ends either with exactly nothing OR ':' followed by line breaks

Formulating a regex with a single dot

I am trying to formulate a regex for the following scenario :
The String to match : mName87.com
So, the string may consist of any number of alpha numeric characters , but can contain only a single dot anywhere in the string .
I formulated this regex : [a-zA-Z0-9.], but it matches even multiple dots(.)
What am i doing wrong here ?

The regex you provided matches only a single character in the whole string you're trying to validate. There are a few things to take care of in your scenario
You want to match over the whole string, so your regex must start with ^ (beginning of the string) and end with $ (end of the string).
Then you want to accept any number of alpha-numeric characters, this is done with [a-zA-Z0-9]+, here the + means one or more characters.
Then match the point: \. (you must escape it here)
Finally accept more characters again.
All together the regex would then be:
^[a-zA-Z0-9]+\.[a-zA-Z0-9]+$

You can use this regex:
\\w*\\.\\w*
You can try here

Try with:
^([a-zA-Z0-9]+\.)+[a-zA-Z]$

use this regular expression ^[a-zA-Z0-9]*\.[a-zA-Z0-9.]*$

EDITED:
Try
([a-zA-Z0-9]+\.[a-zA-Z0-9]+)|(\.[a-zA-Z0-9]+)|([a-zA-Z0-9]+\.)
That is: [a word that ends with a dot] OR [two words and the dot in the middle] OR [a word that starts with a dot]

Java - Unknown characters passing as [a-zA-z0-9]*?

I'm no expert in regex but I need to parse some input I have no control over, and make sure I filter away any strings that don't have A-z and/or 0-9.
When I run this,
Pattern p = Pattern.compile("^[a-zA-Z0-9]*$"); //fixed typo
if(!p.matcher(gottenData).matches())
System.out.println(someData); //someData contains gottenData
certain spaces + an unknown symbol somehow slip through the filter (gottenData is the red rectangle):
In case you're wondering, it DOES also display Text, it's not all like that.
For now, I don't mind the [?] as long as it also contains some string along with it.
Please help.
[EDIT] as far as I can tell from the (very large) input, the [?]'s are either white spaces either nothing at all; maybe there's some sort of encoding issue, also perhaps something to do with #text nodes (input is xml)

The * quantifier matches "zero or more", which means it will match a string that does not contain any of the characters in your class. Try the + quantifier, which means "One or more": ^[a-zA-Z0-9]+$ will match strings made up of alphanumeric characters only. ^.*[a-zA-Z0-9]+.*$ will match any string containing one or more alphanumeric characters, although the leading .* will make it much slower. If you use Matcher.lookingAt() instead of Matcher.matches, it will not require a full string match and you can use the regex [a-zA-Z0-9]+.

You have an error in your regex: instead of [a-zA-z0-9]* it should be [a-zA-Z0-9]*.
You don't need ^ and $ around the regex.
Matcher.matches() always matches the complete string.
String gottenData = "a ";
Pattern p = Pattern.compile("[a-zA-z0-9]*");
if (!p.matcher(gottenData).matches())
System.out.println("doesn't match.");
this prints "doesn't match."

The correct answer is a combination of the above answers. First I imagine your intended character match is [a-zA-Z0-9]. Note that A-z isn't as bad as you might think it include all characters in the ASCII range between A and z, which is the letters plus a few extra (specifically [,\,],^,_,`).
A second potential problem as Martin mentioned is you may need to put in the start and end qualifiers, if you want the string to only consists of letters and numbers.
Finally you use the * operator which means 0 or more, therefore you can match 0 characters and matches will return true, so effectively your pattern will match any input. What you need is the + quantifier. So I will submit the pattern you are most likely looking for is:
^[a-zA-Z0-9]+$

You have to change the regexp to "^[a-zA-Z0-9]*$" to ensure that you are matching the entire string

Looks like it should be "a-zA-Z0-9", not "a-zA-z0-9", try correcting that...

Did anyone consider adding space to the regex [a-zA-Z0-9 ]*. this should match any normal text with chars, number and spaces. If you want quotes and other special chars add them to the regex too.
You can quickly test your regex at http://www.regexplanet.com/simple/

You can check input value is contained string and numbers? by using regex ^[a-zA-Z0-9]*$
if your value just contained numberString than its show match i.e, riz99, riz99z
else it will show not match i.e, 99z., riz99.z, riz99.9
Example code:
if(e.target.value.match('^[a-zA-Z0-9]*$')){
console.log('match')
}
else{
console.log('not match')
}
}
online working example