Hippolyte | HTTP Stubbing in Swift | Mock library

I have this two following classes in my model:

I am playing bandit from overthewire.org; getting to level 10 requires me to find strings preceded with several "=" characters (equal sign) (I interpreted "several" as "two or more") in a text file.

The target lines look like this:

========== passwordhere123

i.e. ten equal signs, one space, and a string of letters and numbers, followed by line break (not sure which exact type).

These lines should be excluded:

c========== EqualSignDidNotStartLine

= only-one-equal-sign

equalsign=somewhereElse

No equal signs at all

The original data did not contain any lines preceded by less than ten but more than one ='s; there are some +'s (plus signs) littered in the text, but +'s and ='s are never in the same line.

The bandit server runs some kind of linux @ 4.18.12 (uname -r), GNU bash 4.4 (from man page), and GNU grep 2.27 (from man page).

The raw data contains non-readable parts, so it is fed through strings first to leave only human-readable strings fro grep to process.

From what I learned, grep's default regex engine (BRE, thanks Casimir) should not be too different from PCRE's. * is still a quantifier (match the preceding pattern zero times or more), not as a standalone pattern meaning "anything, zero times or more". This confuses me in grep's behavior below.

Edit: per this chart, "+" needs to be escaped (i.e.\+) in BRE. It does not help though. I will make some more testing strings to try to decipher what's going on.

Here's the command I tried:

I've created a regex for checking a date format ( 01-01-0000 to 31-12-9999). I tried an example regex, and it works, so there is something wrong with my regex, but when I try it in a debugger (regexr) it works just fine.

What am I missing?

I have been at this on and off for a few days, but my RexEx mastery is not great. Yes I understand that RegEx is not for parsing HTML. I am doing server side "cleaning" of CKEditor input, which already does this, but only client side.

After striping none white-listed tags...

First: $html = preg_replace(' on\w+=(["\'])[^\1]*?\1', '', $html); remove all event attributes properly quoted with either ' or " quotes

Second: $html = preg_replace(' on\w+=\S+', '', $html); *remove the ones that have no quotes but still can fire, ex. onclick=blowUpTheBase()

What I would like to do is ensure the onEvent is between < & > but I can only get it to work if the onEvent attribute is the first one after a tag. Everything I try ends up capturing most of the code. I just cant get it lazy enough.

ex. $html = preg_replace('<([\s\S]?)( on\w+=\S+) ([\s\S]*?)>', '<$1 $3>', $html);

EDIT: I am going to select @colburton's answer because RegEx is what I asked for. I will also use it for my particular situation because it will due the trick. (it is an internal application anyhow)

BUT

I want to thank @Casimir et Hippolyte for his answer because it gives a great example and explanation about how to do this the "right way". I will in short order write up a function using DOMDocument and it will become my goto way of handling RTE/WYSIWYG/HTML input.

Hey there, you Regex Lovers !

I'm quite in Regex, these times and had a purely theorical problem. To put it simple, I will present it as a game.

The game :
Let's say you have a list of words separated by spaces.
What I call a word is as they are defined by regular expressions : [a-zA-Z_0-9]+ (There is no empty word here)
Example of list :
Horse Banana Joker RoXx0r A_Long_Word Joker 1337

What I want you to do is replace each word except Joker by a number of $ equal to the number of character of the matched word.
With our previous list we would obtain :
$$$$$ $$$$$$ Joker $$$$$$ $$$$$$$$$$$ Joker $$$$

In fewer words : I want a regex that matches each character that does not belong to the word "Joker" (In the string, I mean, not that compose the word Joker)

While it is not easy, it's not impossible (I have my own regex for that). That's why I will set some rules.

The rules :

• It must be done with only 1 regex
• I will not accept any regex that works only in specific languages
• I will still accept most common features like Conditionals, Lookarounds, etc... even if some languages can't read them
• No recursion allowed (but if you have a working recursive one, post it, just for the beauty of the regex ^^)
• The regex must be optimized for performance
• If your regex matches (get it ? ;) ) these rules but does not satisfy me, I will feel free to add some more rules
  

Added rules :

• None

To help you out, here are some strings on which the regex must work :
Horse Banana Joker RoXx0r A_Long_Word Joker 1337 Joke Poker Joker Jokers
Must return after replacement :
$$$$$ $$$$$$ Joker $$$$$$ $$$$$$$$$$$ Joker $$$$ $$$$ $$$$$ Joker $$$$$$

Joker Joker Joker
Must return after replacement :
Joker Joker Joker

Again, solving the problem is not the goal here, I want to see different solutions, and more importantly I want to see the best ones !

Solutions :

A very elegant one by Casimir et Hippolyte :
(?:\G(?!^)|(? (replace : $)
See the post
However the \G take the fun out of the problem and does not work in every language, so I can't accept it unless is is possible to create a custom delimiter that is equivalent to \G

Almost accepted answer also by Casimir et Hippolyte :
((?:\s+|\bJoker\b)*)\S((?:\s+Joker)*\s*$)? (replace : $1$$2)
See the post
Does not work when there are only Joker words in the string

A similar solution by ClasG :
(\bJoker[^\w]+)\w|\w([^\w]+Joker\b)|\w (replace : $1$$2)
See the post
Does not work when there are only Joker words in the string

Another one by ClasG :
[^Joker\s]|(? (replace : $)
See the post
Not very efficient, though, but it's another way of seeing things ;)

I came up with a similar regex after reading the comment of Rahul below :
(?(?<=\b|\bJ|\bJo|\bJok|\bJoke|\bJoker)(?!(?:Joke|oke|ke|e|)r\b)\w|\w) (replace $)
Regex101
It is also inefficient, but use the same lookaround list thing :)