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validate special characters by negating unicode letters with regex pattern?

This regex: \p{L}+ matches these characters "ASKJKSDJKDSJÄÖÅüé" of the example string "ASKJKSDJK_-.;,DSJÄÖÅ!”#€%&/()=?`¨’<>üé" which is great but is the exact opposite of what I want. Which leads me to negating regexes.
Goal:
I want to match any and all characters that are not a letter nor a number in multiple languages.
Could a negative regex be a natural direction for this?
I should mention one intended use for the regex I'd like to find is to validate passwords for the rule:
that it needs to contain at least one special character, which I
define as not being a number nor a letter.
It would seem defining ranges of special characters should be avoided if possible, because why limit the possibilities? Thus my definition. I assume there could be some problems with such a wide definition, but it is a first step.
If you have some suggestions for a better solution I'm giving below or just have some thoughts on the subject, I'm sure I'm not the only one that would like to learn about it. Thanks.
Note I'm using double \\ in the Java code. Platform is Java 11.

You can shove those \\p things in []. And thus, use the fact that you can negate chargroups. This is all you need:
Pattern p = Pattern.compile("[^\\p{L}]");
Matcher m = p.matcher("ASKJKSDJK_-.;,DSJÄÖÅ!”#€%&/()=?`¨’<>üé");
while (m.find()) System.out.print(m.group(0));
That prints:
_-.;,!”#€%&/()=?`¨’<>
Which is exactly what you're looking for, no?
No need to mess with lookaheads here.

So after having read similar, though not identical questions and some equally great answers, I came up with this solution: (?=\P{L})(?=\P{N}) meaning match both not letters and not numbers. Even if I'm asserting numbers separately I need to negate both to meet the specification of special characters (See question).
This is making use of a non-consuming regular expression with the parentheses and the?=, first matching the expression in the first parenthesis and after that continue to match the whole in the second. Thanks to #Jason Cohen for this detail in the Regular Expressions: Is there an AND operator? discussion.
The upper case P in \P{L} and \P{N} expresses the "not belonging to a category" in Unicode Categories, where the uppercase P means "not", i e the opposite of a lowercase p.
It's not perfect for a real world solution, but works as a starting point at least. Note I'm using double \\ in the Java code. Platform is Java 11.

Extract last occurrence of a string [duplicate]

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I don't really understand regular expressions. Can you explain them to me in an easy-to-follow manner? If there are any online tools or books, could you also link to them?

The most important part is the concepts. Once you understand how the building blocks work, differences in syntax amount to little more than mild dialects. A layer on top of your regular expression engine's syntax is the syntax of the programming language you're using. Languages such as Perl remove most of this complication, but you'll have to keep in mind other considerations if you're using regular expressions in a C program.
If you think of regular expressions as building blocks that you can mix and match as you please, it helps you learn how to write and debug your own patterns but also how to understand patterns written by others.
Start simple
Conceptually, the simplest regular expressions are literal characters. The pattern N matches the character 'N'.
Regular expressions next to each other match sequences. For example, the pattern Nick matches the sequence 'N' followed by 'i' followed by 'c' followed by 'k'.
If you've ever used grep on Unix—even if only to search for ordinary looking strings—you've already been using regular expressions! (The re in grep refers to regular expressions.)
Order from the menu
Adding just a little complexity, you can match either 'Nick' or 'nick' with the pattern [Nn]ick. The part in square brackets is a character class, which means it matches exactly one of the enclosed characters. You can also use ranges in character classes, so [a-c] matches either 'a' or 'b' or 'c'.
The pattern . is special: rather than matching a literal dot only, it matches any character†. It's the same conceptually as the really big character class [-.?+%$A-Za-z0-9...].
Think of character classes as menus: pick just one.
Helpful shortcuts
Using . can save you lots of typing, and there are other shortcuts for common patterns. Say you want to match a digit: one way to write that is [0-9]. Digits are a frequent match target, so you could instead use the shortcut \d. Others are \s (whitespace) and \w (word characters: alphanumerics or underscore).
The uppercased variants are their complements, so \S matches any non-whitespace character, for example.
Once is not enough
From there, you can repeat parts of your pattern with quantifiers. For example, the pattern ab?c matches 'abc' or 'ac' because the ? quantifier makes the subpattern it modifies optional. Other quantifiers are
* (zero or more times)
+ (one or more times)
{n} (exactly n times)
{n,} (at least n times)
{n,m} (at least n times but no more than m times)
Putting some of these blocks together, the pattern [Nn]*ick matches all of
ick
Nick
nick
Nnick
nNick
nnick
(and so on)
The first match demonstrates an important lesson: * always succeeds! Any pattern can match zero times.
A few other useful examples:
[0-9]+ (and its equivalent \d+) matches any non-negative integer
\d{4}-\d{2}-\d{2} matches dates formatted like 2019-01-01
Grouping
A quantifier modifies the pattern to its immediate left. You might expect 0abc+0 to match '0abc0', '0abcabc0', and so forth, but the pattern immediately to the left of the plus quantifier is c. This means 0abc+0 matches '0abc0', '0abcc0', '0abccc0', and so on.
To match one or more sequences of 'abc' with zeros on the ends, use 0(abc)+0. The parentheses denote a subpattern that can be quantified as a unit. It's also common for regular expression engines to save or "capture" the portion of the input text that matches a parenthesized group. Extracting bits this way is much more flexible and less error-prone than counting indices and substr.
Alternation
Earlier, we saw one way to match either 'Nick' or 'nick'. Another is with alternation as in Nick|nick. Remember that alternation includes everything to its left and everything to its right. Use grouping parentheses to limit the scope of |, e.g., (Nick|nick).
For another example, you could equivalently write [a-c] as a|b|c, but this is likely to be suboptimal because many implementations assume alternatives will have lengths greater than 1.
Escaping
Although some characters match themselves, others have special meanings. The pattern \d+ doesn't match backslash followed by lowercase D followed by a plus sign: to get that, we'd use \\d\+. A backslash removes the special meaning from the following character.
Greediness
Regular expression quantifiers are greedy. This means they match as much text as they possibly can while allowing the entire pattern to match successfully.
For example, say the input is
"Hello," she said, "How are you?"
You might expect ".+" to match only 'Hello,' and will then be surprised when you see that it matched from 'Hello' all the way through 'you?'.
To switch from greedy to what you might think of as cautious, add an extra ? to the quantifier. Now you understand how \((.+?)\), the example from your question works. It matches the sequence of a literal left-parenthesis, followed by one or more characters, and terminated by a right-parenthesis.
If your input is '(123) (456)', then the first capture will be '123'. Non-greedy quantifiers want to allow the rest of the pattern to start matching as soon as possible.
(As to your confusion, I don't know of any regular-expression dialect where ((.+?)) would do the same thing. I suspect something got lost in transmission somewhere along the way.)
Anchors
Use the special pattern ^ to match only at the beginning of your input and $ to match only at the end. Making "bookends" with your patterns where you say, "I know what's at the front and back, but give me everything between" is a useful technique.
Say you want to match comments of the form
-- This is a comment --
you'd write ^--\s+(.+)\s+--$.
Build your own
Regular expressions are recursive, so now that you understand these basic rules, you can combine them however you like.
Tools for writing and debugging regexes:
RegExr (for JavaScript)
Perl: YAPE: Regex Explain
Regex Coach (engine backed by CL-PPCRE)
RegexPal (for JavaScript)
Regular Expressions Online Tester
Regex Buddy
Regex 101 (for PCRE, JavaScript, Python, Golang, Java 8)
I Hate Regex
Visual RegExp
Expresso (for .NET)
Rubular (for Ruby)
Regular Expression Library (Predefined Regexes for common scenarios)
Txt2RE
Regex Tester (for JavaScript)
Regex Storm (for .NET)
Debuggex (visual regex tester and helper)
Books
Mastering Regular Expressions, the 2nd Edition, and the 3rd edition.
Regular Expressions Cheat Sheet
Regex Cookbook
Teach Yourself Regular Expressions
Free resources
RegexOne - Learn with simple, interactive exercises.
Regular Expressions - Everything you should know (PDF Series)
Regex Syntax Summary
How Regexes Work
JavaScript Regular Expressions
Footnote
†: The statement above that . matches any character is a simplification for pedagogical purposes that is not strictly true. Dot matches any character except newline, "\n", but in practice you rarely expect a pattern such as .+ to cross a newline boundary. Perl regexes have a /s switch and Java Pattern.DOTALL, for example, to make . match any character at all. For languages that don't have such a feature, you can use something like [\s\S] to match "any whitespace or any non-whitespace", in other words anything.

Java RegEx removing "/*....*/" and "{.....}" [duplicate]

This question already has an answer here:
Java - Regex - Remove comments
(1 answer)
Closed 7 years ago.
I get the error: "Invalid ecape sequence on this regex:
(\/\*[^/*]*(?:(?!\/\*|\*\/)[/*][^/*]*)*\*\/)|(\{.*?\})
Are there any other regexes that are more suitable or what can I do to fix this regex?

You need to escape the backslashes one more time. This is a "feature" of Java's strings. Java "consumes" the backslashes that you have written because it recognizes special characters like '\t'. When it sees, for example '\/' at the beginning of your regex, it thinks you're asking for a special character, and it complains because this sequence is not valid for that purpose. To get the backslash considered in the regex, you need '\\'.
That being said, this entire approach to handling comments and braces is not going to work generally because it's going to have trouble with a variety of cases like nested blocks in braces. (Just to name one of many.)

(\/\*[^\/*]*(?:(?!\/\*|\*\/)[\/*][^\/*]*)*\*\/)|(\{.*?\})
This is the correct regex, you missed escaping the forward slashes which represent the start and end of a regex sequence.
Here is a simplified version (\/\*.*\*\/|\{.*\})

Identify and escape double quote, single quote and comma in Java [duplicate]

This question already has answers here:
Java, escaping (using) quotes in a regex
(2 answers)
Closed 8 years ago.
Sorry I could not find anything that works and hence I am asking this question. I have a basic string that could have feet("), inch(') or comma(,). All I want to do is identify those and escape them before further processing. Not having any luck with Regex, as you can tell I am not good with it yet. Need help. Thanks much!

Someone hinted at it in your comments, but its not entirely correct since String#replace only takes a single character, and you want to provide more than one for the replacement.
Say you have some function foo() that returns some regular expression that isn't escaped properly, with respect to the "\"" char, or the "\'" char:
String regexp = Bar.foo();
regexp = regexp.replaceAll("(\\\"|\\\')", "\\\\$0");
Pattern yourPatternName = Pattern.compile(regexp);
A little explanation: In Java, you need to escape certain special characters, such as n to mean newline ('\n'), or t to mean tab ('t'). Since you are already escaping them, they are no longer the literal characters '\' + 'n', for example. So, you need to escape them a second time, so that way when the regular expression is compiled, Pattern#compiler will see the two characters "\n" rather than the single character, which is the newline. To escape the '\n' character, you need to, of course, place in a new '\' character. Since we are doing a java.lang.String, we need to still escape that slash one more time.
As for the comma, you don't need to escape that. You only need to escape special characters. For a list of the ones that Pattern recognizes, you can check here:
http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html

Regex conversion from java to php

I have a regular expression in php and I need to convert it to java.
Is it possible to do so? If yes how can i do?
Thanks in advance
$region_pattern = "/<a href=\"#\"><img src=\"images\/ponto_[^\.]+\.gif\"[^>]*>[ ]*<strong>(?P<neighborhood>[^\(<]+)\((?P<region>[^\)]+)\)<\/strong><\/a>/i" ;

A typical conversion from any regex to java is to:
Exclude pattern delimiters => remove starting and trailing /
Remove flags, these are applied to the Pattern object, this is the trailing i. You should either put it in the initialisation of your Pattern object or prepend it to the regex like (?i)<regex>
Replace all \ with \\, \ has a meaning already in java(escape in strings), to use a backslash inside a regex in java you have to use \\ instead of \, so \w becomes \\w. and \\ becomes \\\\
Above regex would become
Pattern.compile("<a href=\"#\"><img src=\"images\\/ponto_[^\\.]+\\.gif\"[^>]*>[ ]*<strong>(?P<neighborhood>[^\\(<]+)\\((?P<region>[^\\)]+)\\)<\\/strong><\\/a>", Pattern.CASE_INSENSITIVE);
This will fail however, I think it is because ?P is a modifier, not one I know exists in Java so ye it is a invalid regex.

There are some problems with the original regex that have to be cleared away first. First, there's [ ], which matches one of the characters &, n, b, s, p or ;. To match an actual non-breaking space character, you should use \xA0.
You also have a lot of unneeded backslashes in there. You can get rid of some by changing the regex delimiter to something other than /; others aren't needed because they're inside character classes, where most metacharacters lose their special meanings. That leaves you with this PHP regex:
"~<img src=\"images/ponto_[^.]+\.gif\"[^>]*>\xA0*<strong>(?P<neighborhood>[^(<]+)\((?P<region>[^)]+)\)</strong>~i"
There are three things that make this regex incompatible with Java. One is the delimiters (/ originally, ~ in the version above) along with the trailing i modifier. Java doesn't use regex delimiters at all, so just drop those. The modifier can be moved into the regex itself by using the inline form, (?i), at the beginning of the regex. (That will work in PHP too, by the way.)
Next is the backslashes. The ones that are used to escape quotation marks remain as they are, but all the others get doubled because Java is more strict about escape sequences in string literals.
Finally, there are the named groups. Up until Java 6, named groups weren't supported at all; Java 7 supports them, but they use the shorter (?<name>...) syntax favored by .NET,
not the Pythonesque (?P<name>...) syntax. (By the way, the shorter (?<name>...) version should work in PHP, too (as should (?'name'...), also introduced by .NET).
So the Java 7 version of your regex would be:
"(?i)<img src=\"images/ponto_[^.]+\\.gif\"[^>]*>\\xA0*<strong>(?<neighborhood>[^(<]+)\\((?<region>[^)]+)\\)</strong>"
For Java 6 or earlier you would use:
"(?i)<img src=\"images/ponto_[^.]+\\.gif\"[^>]*>\\xA0*<strong>([^(<]+)\\(([^)]+)\\)</strong>"
...and you'd have to use numbers instead of names to refer to the group captures.

REGEX is REGEX regardless of language. The REGEX you've posted will work on both Java and PHP. You do need to make some adjustments as both language don't take the pattern exactly the same (though the pattern itself will work in both languages).
Points to Consider
You should know that Java's Pattern object applies flags without having to specify them on the pattern string itself.
Delimiters should not be included as well. Only the pattern itself.