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Lookahead and lookbehind with regex

I am trying to build a regex pattern and I'm a beginner.
The string looks like this
INITIAL TEXT\KEYWORD1\TEXT1\KEYWORD2\TEXT2\KEYWORD3\TEXT3
The string starts with initial text but the keywords with their texts could be in any order or may not be present.
The initial text could contain any character including backslashes.
I want to capture the initial text so I tried something like this
(?<=(.*)(?=\KEYWORD1\|\KEYWORD2\|KEYWORD3).*)
I am able to capture it on regex101 in group1 but my java code doesn't recognize the group 1.
Thanks for helping.

If the string starts with the text you want to capture, then you can use a start-of-string anchor followed by a lazy match on any character, terminating with a forward lookahead to one of the keywords (or end-of-string, to allow for the case with no keywords):
^.*?(?=\\(?:KEYWORD1|KEYWORD2|KEYWORD3)\\|$)
This will match only the INITIAL TEXT
Demo on regex101
Note that in Java you will need to double the backslash characters in the regex string. Demo on ideone

Regex-How to prevent repeated special characters?

I don't have an experience on Regular Expressions. I need to a regular expression which doesn't allow to repeat of special characters (+-*/& etc.)
The string can contain digits, alphanumerics, and special characters.
This should be valid : abc,df
This should be invalid : abc-,df
i will be really appreciated if you can help me ! Thanks for advance.

Two solutions presented so far match a string that is not allowed.
But the tilte is How to prevent..., so I assume that the regex
should match the allowed string. It means that the regex should:
match the whole string if it does not contain 2
consecutive special characters,
not match otherwise.
You can achieve this putting together the following parts:
^ - start of string anchor,
(?!.*[...]{2}) - a negative lookahead for 2 consecutive special
characters (marked here as ...), in any place,
a regex matching the whole (non-empty) string,
$ - end of string anchor.
So the whole regex should be:
^(?!.*[!##$%^&*()\-_+={}[\]|\\;:'",<.>\/?]{2}).+$
Note that within a char class (between [ and ]) a backslash
escaping the following char should be placed before - (if in
the middle of the sequence), closing square bracket,
a backslash itself and / (regex terminator).
Or if you want to apply the regex to individual words (not the whole
string), then the regex should be:
\b(?!\S*[!##$%^&*()\-_+={}[\]|\\;:'",<.>\/?]{2})\S+

[\,\+\-\*\/\&]{2,} Add more characters in the square bracket if you want.
Demo https://regex101.com/r/CBrldL/2

Use the following regex to match the invalid string.
[^A-Za-z0-9]{2,}

[^\w!\s]{2,} This would be a shortest version to match any two consecutive special characters (ignoring space)
If you want to consider space, please use [^\w]{2,}

Java regex for matching certain parts of JSON substrings

I am trying to write a regex that looks for strings with the following pattern:
Begin with an opening bracket { followed by a double-quote "
Then allows for a string of 1+ alphanumeric characters a-zA-Z0-9
Then another double-quote " followed by a colon : and an opening brace [
Then allows for any string of 0+ alphanumeric characters a-zA-Z0-9
So some strings that would match the regex:
{"hello":[blah
{"hello":[
{"1":[
And some strings that would not match:
{hello:[blah
hello":[
{"2:[
So far, the best I've been able to come up with is:
String regex = "{\"[a-zA-Z0-9]+\":\[[a-zA-Z0-9]*";
if(myString.matches(regex))
// do something
But I know I'm way off base. Can any regex gurus help reel me in? Thanks in advance!

String regex = "{\"[a-zA-Z0-9]+\":\[[a-zA-Z0-9]*";
The problem here is that you need an extra backslash before the square bracket. This is because you need the regex to contain \[ in order to match a square bracket, which means the string literal needs to contain \\[ to escape the backslash for the Java code parser. Similarly, you may also need to escape the { in the regex as it is a metacharacter (for bounded repetition counts)
String regex = "\\{\"[a-zA-Z0-9]+\":\\[[a-zA-Z0-9]*";

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 to match \Q and \E in Java regex?

I want to match \Q and \E in a Java regex.
I am writing a program which will compute the length of the string, matching to the pattern (this program assumes that there is no any quantifier in regex except {some number}, that's why the length of the string is uniquely defined) and I want at first delete all expressions like \Qsome text\E.
But regex like this:
"\\Q\\Q\\E\\Q\\E\\E"
obviously doesn't work.

Use Pattern.quote(...):
String s = "\\Q\\Q\\E\\Q\\E\\E";
String escaped = Pattern.quote(s);

Just escape the backslashes. The sequence \\\\ matches a literal backslash, so to match a literal \Q:
"\\\\Q"
and to match a literal \E:
"\\\\E"
You can make it more readable for a maintainer by making it obvious that each sequence matches a single character using [...] as in:
"[\\\\][Q]"