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Regular expression for matching texts before and after string

I would like to match URL strings which can be specified in the following manner.
xxx.yyy.com (For example, the regular expression should match all strings like 4xxx.yyy.com, xxx4.yyy.com, xxx.yyy.com, 4xxx4.yyy.com, 444xxx666.yyy.com, abcxxxdef.yyy.com etc).
I have tried to use
([a-zA-Z0-9]+$)xxx([a-zA-Z0-9]+$).yyy.com
([a-zA-Z0-9]*)xxx([a-zA-Z0-9]*).yyy.com
But they don't work. Please help me write a correct regular expression. Thanks in advance.
Note: I'm trying to do this in Java.

If you want to make sure there is xxx and you want to allow all non whitespace chars before and after. If you want to match the whole string, you could add anchors at the start and end.
Note to escape the dot to match it literally.
^\S*xxx\S*\.yyy\.com$
^ Start of string
\S*xxx\S* Match xxx between optional non whitespace chars
\.yyy Match .yyy
\.com Match .com
$ End of string
Regex demo
In Java double escape the backslash
String regex = "^\\S*xxx\\S*\\.yyy\\.com$";
Or specify the characters on the left and right that you would allow to match in the character class:
^[0-9A-Za-z!##$%^&*()_+]*xxx[0-9A-Za-z!##$%^&*()_+]*\.yyy\.com$
Regex demo

Java String Regex replacement

Sample Input:
a:b
a.in:b
asds.sdsd:b
a:b___a.sds:bc___ab:bd
Sample Output:
a:replaced
a.in:replaced
asds.sdsd:replaced
a:replaced___a.sds:replaced___ab:replaced
String which comes after : should be replaced with custom function.
I have done the same without Regex. I feel it can be replaced with regex as we are trying to extract string out of specific pattern.
For first three cases, it's simple enough to extract String after :, but I couldn't find a way to deal with third case, unless I split the string ___ and apply the approach for first type of pattern and again concatenate them.

Just replace only the letters with exists next to : with the string replaced.
string.replaceAll("(?<=:)[A-Za-z]+", "replaced");
DEMO
or
If you also want to deal with digits, then add \d inside the char class.
string.replaceAll("(?<=:)[A-Za-z\\d]+", "replaced");

(:)[a-zA-Z]+
You can simply do this with string.replaceAll.Replace by $1replaced.See demo.
https://regex101.com/r/fX3oF6/18

Remove everything from a string upto a certain character and optionally a string if it follows too

I am looking to write a regex that can remove any characters upto the first &emsp and if there is a (new section) following &emsp then remove that as well. But the following regex doesn't seem to work. Why? How do I correct this?
String removeEmsp =" “[<centd>[</centd>]§ 431:10A–126 (new section)[<centd>]Chemotherapy services.</centd>] <centa>Cancer treatment.</centa>test snl.";
Pattern removeEmspPattern1 = Pattern.compile("(.*( (\\(new section\\)))?)(.*)", Pattern.MULTILINE);
System.out.println(removeEmspPattern1.matcher(removeEmsp).replaceAll("$2"));

Have you tried String Split? This creates an array of strings from a string, based on a deliminator.
Once you have the string split, just select the elements of the array that you need for print statement.
Read more here

Your regex is very long and I do not want to debug it. However the tip is that some characters have special meaning in regular expressions. For example & means "and". Squire brackets allow defining characters groups etc. Such characters must be escaped if you want them to be interpreted as just characters and not regex commands. To escape special character you have to write \ in front of it. But \ is escape character for java too, so it should be duplicate.
For example to replace ampersand by letter A you should write str.replaceAll("\\&", "A")
Now you have all information you need. Try to start from simpler regex and then expand it to what you need. Good luck.
EDIT
BTW parsing XML and/or HTML using regular expressions is possible but is highly not recommended. Use special parser for such formats.

Try this:
String removeEmsp =" “[<centd>[</centd>]§ 431:10A–126 (new section)[<centd>]Chemotherapy services.</centd>] <centa>Cancer treatment.</centa>test snl.";
System.out.println(removeEmsp.replaceFirst("^.*?\\ (\\(new\\ssection\\))?", ""));
System.out.println(removeEmsp.replaceAll("^.*?\\ (\\(new\\ssection\\))?", ""));
Output:
[<centd>]Chemotherapy services.</centd>] <centa>Cancer treatment.</centa>test snl.
[<centd>]Chemotherapy services.</centd>] <centa>Cancer treatment.</centa>test snl.
It will remove everything up to " " and optionally, the following "(new section)" text if any.

Java String validation only one alphanumeric with Regex

I want to do validation for a String which can only contains alphanumeric and only one special character. I tried with (\\W).{1,1}(\\w+).
But it is true only when I start with a special character. But I can have one special character at any place in String.

Use the ^ and $ anchors to instruct the regex engine to start matching from the beginning of the string and stop matching at the end of the string, so taking your regex:
^(\\W).{1,1}(\\w+)$
Please take a look at this Oracle (Java) tutorial on regular expressions.

Try this regexp: \w*\W?\w* (Java string: "\\w*\\W?\\w*")
This expression has a drawback of matching zero-length strings. If your input must have exactly one special character, remove the question mark ? from the expression.

use matcher.find() and not matcher.match() and search for \\w and remove plus (+) because it will match all alphanumeric characters sequence in your string.If your string contains only them, your regex will match whole string.

if I understand your regex correctly, this could solve your problem:
([\w]+)([^\w])([\w]+)

How do I write a regular expression to find the following pattern?

I am trying to write a regular expression to do a find and replace operation. Assume Java regex syntax. Below are examples of what I am trying to find:
12341+1
12241+1R1
100001+1R2
So, I am searching for a string beginning with one or more digits, followed by a "1+1" substring, followed by 0 or more characters. I have the following regex:
^(\d+)(1\\+1).*
This regex will successfully find the examples above, however, my goal is to replace the strings with everything before "1+1". So, 12341+1 would become 1234, and 12241+1R1 would become 1224. If I use the first grouped expression $1 to replace the pattern, I get the wrong result as follows:
12341+1 becomes 12341
12241+1R1 becomes 12241
100001+1R2 becomes 100001
Any ideas?

Your existing regex works fine, just that you are missing a \ before \d
String str = "100001+1R2";
str = str.replaceAll("^(\\d+)(1\\+1).*","$1");
Working link

IMHO, the regex is correct.
Perhaps you wrote it wrong in the code. If you want to code the regex ^(\d+)(1\+1).* in a string, you have to write something like String regex = "^(\\d+)(1\\+1).*".
Your output is the result of ^(\d+)(1+1).* replacement, as you miss some backslash in the string (e.g. "^(\\d+)(1\+1).*").

Your regex looks fine to me - I don't have access to java but in JavaScript the code..
"12341+1".replace(/(\d+)(1\+1)/g, "$1");
Returns 1234 as you'd expect. This works on a string with many 'codes' in too e.g.
"12341+1 54321+1".replace(/(\d+)(1\+1)/g, "$1");
gives 1234 5432.

Personally, I wouldn't use a Regex at all (it'd be like using a hammer on a thumbtack), I'd just create a substring from (Pseudocode)
stringName.substring(0, stringName.indexOf("1+1"))
But it looks like other posters have already mentioned the non-greedy operator.
In most Regex Syntaxes you can add a '?' after a '+' or '*' to indicate that you want it to match as little as possible before moving on in the pattern. (Thus: ^(\d+?)(1+1) matches any number of digits until it finds "1+1" and then, NOT INCLUDING the "1+1" it continues matching, whereas your original would see the 1 and match it as well).