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Yes, Jeffrey Friedl's Mastering Regular Expressions, 3rd Edition.

Here it is on Amazon

So you want to match the blocks in the encapsulated area as well?

I've edited the regex and added the underscore as separator: https://regex101.com/r/2hLzaQ/2 You might need to clean the ">" in your matches. It also works in ruby: http://rubular.com/r/is6igkXXsl

I'm not sure how it will turn out with nested replies. Keep me posted :_)

I'm still learning regex, but the one thing I think I see is that there's a problem with the braces. It looks you're asking for: OpenBrace, 0 or 1 commas, OpenBrace, DoubleQuote, and then "Name".

I could be misreading it though.

edit: I don't think you need the 8th capture group (,??) as long as you're using the global modifier. I think that and the 3rd capture group are always going to capture nothing.

edit 2: This is what I came up with to match name. But it renders 6th and 7th capture groups empty on the example you used. ({(.*?(,??)"(Name)":"(.*?)"|"(.*?)":"(.*?)")})
Regex101 link

Cygwin comes with a tool called txt2regex. Which is a command line wizard for creating regex and will output it in perl, php, postgres, python, sed, and vim formats. While it's probably available on native linux distros, I don't think it would support conversion of one format to another. Someone who is adventurous could probably write a converter based on the above codebase...

"^[^Name].*" happens to work for the sample text because it matches lines that do not start with any of the characters "N", "a", "m" or "e".

It seems that the regex engine implemented in notepad++ is this: http://www.scintilla.org/SciTERegEx.html (according to its online help) which supports very minimal regular expression syntax.

Cheers.

You haven't specified a language (nobody seems to bother with this extremely important detail) so I'll give you a solution in Perl.

First, as always, let's define a set of example data we want to test our regular expression against:

"log": true blah "write": false blah blah "log": false blah "log: true blah "write": true blah blah "log": true blah blah "write": false

Now we want any "log":true ... "write": false section to turn into a "log":true ... "write": true section.

To do this we will try the following regular expression and test it at http://regex101.com/ by setting the /g (global search), /x (extended so we can split the regex over several lines), \m (so ^ and $ mean start and end of lines), and s (so . also includes newlines) flags.

s/ ( # start capturing prefix ^\s*"log":\s*true # "log": true line .? # non-greedy anything until... ^\s"write":\s* # "write": ) # stop capturing prefix (.*?)$ # whatever comes after "write": on that line /\1true/smgx

Check out RegEx101.com, it will explain each part of a regex.

In this case the explanation is:
(?:\s|^) Non-capturing group

1st Alternative: \s \s match any white space character [\r\n\t\f ] 2nd Alternative: ^ ^ assert position at start of the string

Not as far as I know, in native JS. But there is one in NodeJs (https://nodejs.org/api/path.html#path_path_parse_path) which is what I targeted!

Unfortunately, it doesn't provide an answer to my splitting problem, which I think I will without an answer... :)

It works with the first example, but here is another one that should work in theory: http://rubular.com/r/DTa59pQru4

The second block doesn't match for some reason.

I'm getting these examples from my batch of emails, so I have to replace the actual text with placeholders for obvious reasons.

Well I would've helped faster if it was for php, but since it is for python, I tooks some time to remember the basics xD

Here's a solution you can rely on to move forward using lxml(to install): http://codepad.org/oFpXELhD

BS has big limitations, and the complexity here is to be good at xPath (which is not supported by BS), that's why I used lxml...

Have you selected "regular expression" mode in the find window? Have you selected "wrap around"? What exactly are you typing into the search box?

Here's an example of it working in an online tester.

Edit: wait, are you attempting to search for a file that contains those strings, or search a document for those strings? I am not certain if regex supports the former.

I believe this is what you're asking but the solution by /u/BashAtTheBeach96 should work unless you have other lines matching "rpg" outside the "<et>" tag

http://regex101.com/r/yQ5iM5/1

Sorry, that was the solution to a previous problem I encountered, but I just realized you are working with non-delimited strings.

The catch-22 here is that regex can't tell the difference between 30 03 and 3 00 3, because they can't calculate a given position as odd or even (zero length assertions cannot be repeated or contain back-references). The engine would need to consider the distance from the beginning recursively, which for this reason would not work (especially for variable-length strings).

If I were you, I would run two regexes. The first would delimit the string into each binary value to prevent this. This adds the information of where one value ends and the other begins, which we know is true because hex strings are even-numbered.

First replace:

(\d\d)

with:

\1

notice the space on the end of the \1.

Then run

(?!00)(\d\d)

Seen here

The first bracket group is non-capturing, and it checks that the match is not 00. The second bracket group is capturing, and note my use of the g flag, so that all are captured.

Edit: I just noticed your string is null-delimited. If that's always so, use the solution in /u/Caek_ 's link.

Are you looking for these exact patterns? Or do some of the letters/numbers change?

I'm going to make a few assumptions.
1. JLX\ doesn't change
2. The numbers can change, but are always 5 numbers for the first match and 6 numbers for the second.
3. The letters B, V, A, and E can change. If they don't, the following expression could be tightened up a bit.

If that's the case, then this expression should work for you:

/JLX\(?:(?:\w(\w{2}\d{5}))|(\w\d{6}))/i

Here are examples on regexe101.com

No, using \S doesnt give the "more efficient" way as you claim... In your first try you were on the right path, but you seem to have limited regex knowledge (we can tell from your pattern :o)

You can try <code>this</code> instead... A performace difference will be noticed as long as the searched text gets bigger...

To learn more about REGEX, I'll advice you <code>this wonderful tutorial</code>

Well, why complicating things... You want to remove the dollar sign, then just replace it with an empty string...

Something like this(it removes also multiple spaces): http://regex101.com/r/nR7yR2/5

This should work without problem as long as you input is what you said

Good one... That's what I wanted to do :P

Since the 2nd part of the regex is almost identical to the 1st part, you could use recursion for a shorter pattern (must check if it's supported where it's going to be used)

http://regex101.com/r/oS8jH1/5

This does not validate the time, so something like 5:90 pm would break it.

/(?P<time>1?\d(?::\d\d)?(?: [ap].m.)?(?:-1?\d(?::\d\d)?(?: [ap].m.))?)/gi

Try it out

Hello,

would something like this do the trick ?

You can use \1 in a pattern to reference the content of the first capturing parenthesis : <([a-z]+)></\1>

/^(?!.(anime|manga|hentai|sex|xxx)).$/i

My regex will just pass/fail for single sentences that you run the test on. In this example I set up captures so you can see what passed and what failed.

I'm not familiar with what you are doing so I can adjust my regex if you answer my questions.

Are you trying to capture anything? or are you just trying to see if a string (sentence) passes/fails? It sounds like you want to test against sentences but your existing regex appears that it is trying to parse URLs. Can you provide some data to test against? That would really help.

One thing to point out is that this match is greedy. If you have more than one "<!-- End section -->", it will match up to that last one and everything in between, including other End Sections. See what it would match here.

That site (Regex101.com) will also explain what each part of the regex is doing.

I would use Perl for this.

perl -n -e 'print "$1","\n" if /(^(?:\s?[\w\d]+){8})/gm' your_file.txt

Check it here: http://regex101.com/r/eL1sG7

The "." after the "z" is a literal . (Its literal when it's the first or last in a character class.

{5,} Means five or more of the preceding.

This regex should work... ^([a-z0-9]{6})\.[a-z]{3}$

DEMO

> {6,}

Quantifier: Between 6 and unlimited times, as many times as possible, giving back as needed [greedy]

Then you can simply use /^The (\S+) ball is \1\.$/

\S+ will match all non space caracters (including words with - and ')...

\1 refers to the 1st captured group

<code>DEMO</code>

First of all, since it's a regex question. You should really specify what language you're using. Since there are different engines with different capabilities.

Let's suppose we're using PCRE and let's try to answer your first question:

• Suppose we define a word as

  \w{3,}\b

for simplicity

• What you basically want, is go through every "word" and check if that word exists previously in the string/line. You will need a backreference like \1

• There are several problems though :

  • You can't just infinitely lookbehind like this

  (?<!word.*?)word

in some flavors like .NET that would be possible.

• Let's say we could and you were using .NET, we said we need to use backreferences

  (?<!\1.*?)(word)

But how in the world will the regex engine know what "\1" is if it didn't even reach (word) ?

Conclusion : IMO, I think it's close to impossible* to do this in one go.

Workaround : there is a not-well known escape sequence in PCRE \K, it will forget what was matched before. What you could do is the following

(\w{3,}\b).*?\K\1

the problem with this is that it will consume the characters, so you will need to use the pattern several times until 0 replaces were done. See this demo http://regex101.com/r/mE4hN7

As for your second question. It doesn't make sense.

Q: "If I do not want to remove Duplicate words that occur together" -> A: well you do nothing ?

Even the output doesn't make sense, I mean CCCCCC and DDDDDD are alone, so why aren't they removed ?

* I don't want to say that it's impossible, I mean who knows some guru shows up with a solution ?

PS: I signed up just because of this question o_o

Is the data in record-per-line format?

If so want to cut and paste an example line of text, and perhaps I can give you a Perl one-liner you could use to blank out the matched text?

Use the following: (?:@[domain_regex]) This (?:) forms a non-capturing group.

\b[-a-zA-Z0-9.]+(?:@[-a-zA-Z0-9.]+.(?:de|es|cz|fr|uk))\b

I’m not sure where to view it on regex101 but if you test out the example in https://regexr.com/ it will show in the bottom Tools section what are the capture groups and what are just matching groups (they also have a super-handy cheatsheet on the sidebar).

I’d paste it myself but it seems regexr doesn’t have mobile browser support.

Doesn't make a difference...

So i have this line i believe i have it

testing the regex

testing website: https://regexr.com

1 Raw_Read_Error_Rate 0x000b 100 100 016 Pre-fail Always - 0

I used the following regex to get the last digit

\d$

Might post to zabbix since that's the environment I'm using for smartctl. Just using this forum as a platform to share

Are u on apple or amdroid. It was this one here.

You just add pdf or other format books and it reads them but the issue is it reads more than you want it too like header footer page numbers and headings, citations if its an academic book etc. If it's a normal book might be ok. This app had the most natural voices out of the ones i tested.

Yeah, I 100% agree with /u/Splitshadow. You'd sure have a lot of results with your asterisk. If you took the quantifiers out to where you just had a fixed amount of options, you could conceivably do something like that by parsing the string and looping through an array of each ending.

Not sure how useful this would actually be, but I have cobbled together an example to show you what I mean.

Here is a demo

No, it should be highlighting. I mean, it's matching it, even if it's not highlighting it. It looks like some sort of bug. If you see down in the replacement, you can see that it is actually replacing the newline portion, even though it is not highlighted above.

Also, I ran this through a different regex tester that it is highlighting the newline properly.

https://regexr.com/54se3

So, I guess there's just something wonky going on with regex101.

Do you want to strip single quotes that occur both at the beginning and end?

var tests = ["'footloose'", "Godfather", "'Inception", "Frozen'"];

tests.forEach(function(s) { var title = s.replace(/^'(.*)'$/, "$1");

console.log(title); });

Output:

footloose Godfather 'Inception Frozen'

Try it online!

Hmm, what about https://regex101.com/r/lW2gD5/1

It's one regex that covers all 4 cases... Using preg_match_all you'll get an array (of arrays) of matches $matches:

• $matches[1] = is array containing all 1 cases
• $matches[2] = is array containing all 2 cases
• ...

I kept it simple so it'll work as long as your fileNames doesnt contain a dot in them like cgr.bat.SS... So if it's the case, it will need additional tweaks...

Yes... You replace the slaches with an empty string (or use some functions to strip the slaches like in php if there is one), then you parse it...

<strong>This</strong> is an online tool that helps you reading and understanding it's structure in order to parse it...

I'm not really familiar with MarcEdit, but this slide deck seems to indicate that replacing with capture groups supported.

Solutions from https://stackoverflow.com/questions/30953603/an-atypical-password-validation-regexp-at-least-3-letters-3-numbers-no-double:

^(?=(?:[^a-zA-Z]*[a-zA-Z]){3})(?=(?:\D*\d){3})[^\x22]{8,32}$

^(?=(?:.*?[a-zA-Z]){3})(?=(?:\D*\d){3})[^\x22]{8,32}$

No problem. Keep in mind that regex is only a solution for parsing the contents of single HTML tags. Once you need to parse multiple degrees of nested tags, you need a DOM parser.

This is the famous answer that expresses the hatred behind using regex to parse complicated HTML.

Have a look at https://emacsfodder.github.io/blog/easy-regexp-generation-with-emacs/ - in particular the <code>regexp-opt</code> function. It takes a list of words to match and builds a more efficient expression.

e.g.

(regexp-opt '("one" "two" "three"))

outputs:

"\(?:one\|t\(?:hree\|wo\)\)"

Taking away all those extra slashes gives you:

(?:one|t(?:hree|wo))

In order to do that you are going to have to write a little more complicated logic above. Since the text could have multiple images you are going to have to use preg_match_all and loop over the array of matches. Regex is great though. It will help you do a ton of things.

By numerical evaluation, I mean like this

element_contents = "1.44x" if float(element_contents[:-1]) < 1.45: ...

[:-1] removes the x, and float() converts the text number to a float number, so it can be compared to 1.45 using arithmetic. Math is far easier and faster for computers to do, so it's better to use math when you can. Obviously I don't know the whole story, so it might make more sense to do it with regex, but

Side note: It would be better if you used a DOM parser like xml.etree.ElementTree. It's generally ill-advised to use regex to parse xml/html.

Edit: obligatory regex-html rant

Looks like you can

&#x200B;

>Groups and backreferences
>
>You can use parentheses (…) to group expressions, and then insert the matched expression using a dollar sign $followed by the number of the group.
>
>Example: Search for “(Renamer [0-9]*)(Mac)”, the text is “I frequently use Renamer 6 on my Mac”:
>
>$1 = Renamer 6 = 1st group: “(Renamer [0-9]*)” $2 = Mac = 2nd group: “(Mac)”

&#x200B;

Source: https://renamer.com/help/English.lproj/regular_expressions.html

There's no need to use regex. Try this:

tbody:nth-child(2) > tr:nth-child(index) > td:nth-child(4) > input

Substitute index by 1, 2, 3 and so forth.

&#x200B;

Take a look at https://www.browserstack.com/guide/css-selectors-in-selenium if your use case varies.

this is a fantastic article on Regex. https://dev.to/codechunker/introduction-to-regex-expressions-for-java-developers-11jn

Here's a simple way to go: http://regex101.com/r/rD4mW7/1

But if the order isnt fixed, then, that is a problem... A workarround would be to use the operator OR "|" and add all the possible combinations wich will give a huuuuuuuuuuge regex xD

1. Check THIS and try to find out what's wrong :)

2. From your answer on Q.1, there's an easy fix :)

If you still stuck and you just want a solution, click HERE

This may not be right, but I'm going to assume that the file is some sort of MBox email file. If that's the case, then it will sort of go like this (though I could be wrong):

Each message starts with "From ", then a bunch of headers, including the Subject line. Then there will be two new lines, then the body of the message, then two more new lines followed by either the end of file or another "From " to start a new message. There is supposed to be some sort of escape for occurrences of "From " within the message.

Take a look at this regex (regex101.com). I'm ignoring the part where you mention "TECH" since it's grabbing the entire body based upon the subject. Hopefully you can use this as a starting point.

Assuming that all the src tags start with "/storage/" and have the question mark, you could just ignore everything else in the HTML tag and grab the filename.

Take a look at this and see how it works for you.

How about using look arounds for this!

Make sure we have a letter preceding the comma and NO space after the comma.

s/(?<=\pL),(?!\s)/, /g

http://regex101.com/r/oZ5iL3/1

>the VIN numbers will never have a Zero/Zed, but will occasionally have O's.

This is different than what the wiki link mk5p describes, which says there will never be an I, O, or Q. It also seems to be different than what I see on that site, for example this link has several zeroes in it.

Anywhere, here's (Rexex101.com link) what I came up with, following the rules on the wiki. I couldn't find any examples of cars that didn't have VIN numbers to test exclusion, though.

(Just notice that you had two different websites there, I didn't test for other links for that second site)

That regex doesn't make sense. It looks like you have a character class and are trying to perform an OR inside. This one has what you need.

/^(?\$([1-9][,\d]*(?:.\d\d)?))?$/gm

Try it out

I'd like to ask a few follow up questions if I may.

• First, Is the is the <time> string necessary?
  
• Second, why have two colons in this line (?::?
  
• Third, the characters (? appear to be the most important components to doing what I wanted to do. What is that called?
• Fourth, there is always an a.m. or p.m. at the end of the time string, so I don't think the (? characters are necessary. I would remove them myself but I don't want to screw anything up. Would you please help.
• Fifth, if it's fairly easy to do, can you explain how I would "control for" standard time and avoid military time? I suspect it involves using 1?\d[1-5](?::\d[0-5]\d[1-9]... characters to tell the regex that we don't want "20:00" hours. Is that the general idea? *Finally, what does the /gi mean a the end of the line?

ps: I use python btw. I don't know if that would change anything that you wrote. It worked for me here (and I adjusted it to only get times with a hyphen present).

pps: I am enormously grateful for your help. Sometimes working in regex (having just started learning any programming at all) can be enormously frustrating. Thank you!!!

Put the cases with similarities into groups, then build your regex...

You have two groups here, so you can try: thank (yo)?u|th(anks|x)

DEMO

BTW, this is the best tutorial I used myself to learn regex => http://www.regular-expressions.info/tutorial.html

After that, you'll just need to practice, a looooot and like forever xD

The problem was the character classes you were making.

/^(?:Copy of )?(?P<showName>.*?)(?: - )?(?P<episode>S\d\dE\d\d)$/gm

Try it out

EDIT: This one is better: /^(?:Copy of )?(?P<showName>.+?)(?: -)? (?P<episode>S\d\dE\d\d)$/gm

/(?:^|[_\s])(SK|PL|TESTCAT|FOO)(?:$|[\s_.])/g

This is pretty much the same as the last one. I just turned my OR statements into character classes. This one also accounts for file extensions.

(?:^|[_\s])

• beginning of string, whitespace, or an underscore

(SK|PL|TESTCAT|FOO)

• category list you are wanting to match. Each parameter is separated in an OR.

(?:$|[\s_.])

• End of string, whitespace, or a period (file extension beginning).

EDIT: Try it out http://regex101.com/r/tB5gA5/1

I'm a PHP guy. Couldn't figure out why the regex was invalid in the Ruby tool, but it works with PHP.

/(?:[a-z]+_and_)?(?P<superhero>[a-z]+man)(?:_and_[a-z]+)?(?:\s|$)/gi

http://regex101.com/r/pH5wO2

What do you mean by: > what I would really like is the text up to the second to the last colon

Do you mean that you want all the text from the 2nd : to the last : ?

If so, try <code>THIS</code>... If not please explain clearly what you want ? what is the expected result you want that .+: nor my example dont give ?

Weird question :>

What about /(?=(\d{1}))(?=(\d{2})?)(?=(\d{3})?)(?=(\d{4})?)(?=(\d{5})?)/g

<code>DEMO</code>

It uses lookaheads, depending on the language you're using (you didnt say) it may or may not be supported!

This case is easy... Try this

> ^([^-]+-)([^(]+)( [^.]+)(.{4})$

and replace with: $1$3$2$4

It will work for the examples you gave above, so be sure there's no special cases where it wont work before renaming !!

For example if the 1st group between parentheses contains a dot, the patern wont work, and it should be modified to the following for example:

> ^([^-]+-)([^(]+)( .+)(.{4})$

I hope it will work directly in TC since im too lazy to test it xD

> re.search(r"J+Neu","Jenny Neu")

J+Neu will match one J or more followed by the string "Neu"... In other words it will match JNeu, or JJNeu, or JJJNeu, or JJJJJJJJNeu...

But if you want to match "Jenny Neu", you can use "J" followed by any number of chars ie ".+"(the dot matches any chararcter) or "[a-z\s]+". Then followed by the string "Neu"... To sum up, you must try "J.*Neu" or "J[a-z\s]+Neu"

Try it here: http://regex101.com/r/uT2sZ2

Notice the use of the flags:

• g: to get all the matches (not the 1st only)
• m: activate multilines (each line is treated as separate string) => not needed in your case if you test only one string at a time
• i: activate the case insensitive match

You can also learn more about regex here: http://www.regular-expressions.info/tutorial.html

Both are correct...

To see clearer check <code>THIS</code>... You'll notice the "a" is in blue meaning there's a match, but if you replace it with "b" it wont be blue meaning there's no match... Well that's what you explained above and that's wrong !

Check the tab on the right named Match groups, when you have a "b" it says there's a match of one empty character, now if you replace it with an "a", it will show two matches: an "a" and an empty character...

This is due to your pattern wich says :

• Match an "a" at the begining of the string ^(a)

OR

• Match an empty character at the end of the string ()$

So, as you see, in both cases there's at least one match, wich explains why your code shows a match all the time...

Now, if you have any real test case, i'll be glad to help with :)

I couldn't make it work, but it taught me about negative symbols.

https://regexr.com/4gk1q

This does the first step above, but I want to explore this a bit more and try to avoid a programtic approach.

Here you go, search for the following and replace with an empty string to get rid of what you don't want: ".*"\K|[^"\n]+ ==> <code>DEMO</code>

But is there a reason you're using notepad++ ? I'll advise you using Sublime Text instead... It's waaaaaaaaaaaaay better

>^.+(?:\n.+)+\n\n^.+(?:\n.+)+

I like how it showed as one match without 2 separate groups. Unfortunately, it doesn't work in docparser.com. Their regex feature seems iffy with quantifiers or backreferencing. @mfb-'s solution seems to be the only that work though. Thanks so much for the help!

Hi u/Phantom569,
Thanks for the suggestion, I'll have a try and come back with the feedback.

The code that I'm trying is posted on codeshare.io if you want to give it a look :)

This question is not just to get the answer but to understand the concepts behind, just trying to learn a bit about the topic, that's all :)

howdy binvius,

i presume you are talking about this app ...

Tasker for Android
— https://tasker.joaoapps.com/

since the regex posted here has failed to work as expected, you may want to use their forum.

take care,
lee