I am trying to grab the text after the first hyphen in a pattern
<title>.*?-(.*?)(-|<\/title>)
which then grabs DesiredText from the pattern below:
<title>Stuff - DesiredText - Other Stuff</title>
However in this pattern:
<title>Stuff - Unwanted - DesiredText - Otherstuff</title>
I want it to skip the 'Unwanted' text and match the text after the next hyphen instead (DesiredText). I made a regex101 with both patterns and need to modify my basic regex so that if a word or words I don't want to match are present in that capture group it then matches the second hyphen text instead:
https://regex101.com/r/veSqH3/1
I believe this is what you are looking for. The key is in using the caret (^) character within the square-bracket character list ([]). Using the caret and brackets together indicate a blacklist. It will only match things that are NOT in the list.
https://regex101.com/r/alAZhj/3
Pattern: <title>.*?-\s*([^-\s]*)\s*- End<\/title>
This matches anything in between the middle hyphens that is not a hyphen or space. You can of course modify the pattern to include such characters by using the following pattern.
Pattern: <title>.*?-\s*([^-]*)\s*- End<\/title>
This will match anything in between the middle hyphens that is not a hyphen, so that you can have less restricted text in there.
This will use a negative lookahead to disqualify Note. There may be ways to optimize the pattern, but I cannot do so with confidence because I don't know how variable your inputs strings are.
Pattern: /<title>.*?- (?P<title>(?!Note).*?)(?= -|<])/
Demo
I am using a positive lookahead to ensure the captured match doesn't have any unwanted trailing characters.
If you just want the second last delimited value, you could do something like this to return the value as the fullstring match:
~- \K[^-]*(?= - [^-]*?</title>)~
Or faster with a capture group:
~- ([^-]*) - [^-]*?</title>~
This assumes there are no hyphens in the value.
I took a different approach and focused on returning the capture prior to the last word, rather than any sort of negation. In this way it's highly generic.
This pattern will match what you want in the capture group:
\s-\s([a-zA-Z]+)\s-\s[a-zA-Z]+<\/title>
If you are concerned that this only match between title tags, then you can add:
<title>.*?\s-\s([a-zA-Z]+)\s-\s[a-zA-Z]+<\/title>
Here's a link to the Test
The only limitation to this I see, is that it uses words and whitespace, so if your desired match is "- Some phrase -" then this won't work with it, but that was not indicated in your example. It's a bit unclear because you used "other stuff" and then "otherstuff".
Related
Is it possible to have a RegEx fall back to the beginning of the string and begin matching again?
Here's why I ask. Given the below string, I'd like to capture the sub strings black, red, blue, and green in that order, regardless of the order of occurrence in the subject string and only if all substrings are present in the subject string.
$str ='blue-ka93-red-kdke3-green-weifk-black'
So, for all of the below strings, the RegEx should capture black, red, blue, and green (in that order)
'blue-ka93-red-kdke3-green-weifk-black'
'green-ka93-red-kdke3-blue-weifk-black'
'blue-ka93-black-kdke3-green-weifk-red'
'green-ka93-black-kdke3-blue-weifk-red'
I wonder if there isn't a way to match a capture group then fall back to the start of the string and find the next capture group. I was hoping that something like ^.*(?=(black))^.*(?=(red))^.*(?=(blue))^.*(?=(green)) would work but of course the ^ and lookaheads do not behave this way.
Is it possible to construct such a RegEx?
For context, I'll be using the RegEx in PHP.
You can use
^(?=.*(black))(?=.*(red))(?=.*(blue))(?=.*(green))
Note: This will require all these keywords to be in the string.
See demo
There is no way to reset RegEx index when matching, so, you can only use capturing mechanism inside a positive lookahead anchored at the start. The lookahead will match an empty location at the start of the string (due to ^) and each of tose lookaheads in the RegEx above will be executed one after another if the previous one returned true (found a string of text meeting its pattern).
Your RegEx did not work the same way because you matched, consumed the text with.* (this subpattern was outside the lookaheads) and repeated the start of string anchor that automatically fails a RegEx if you do not use a multiline modifier.
Why not just use capture groups for maintaining the order.
^(?:(black)|(red)|(blue)|(green)|.)+$
This will match any string, all colors are optional.
See demo at regex101 or php demo at eval.in
Given a text string (a markdown document) I need to achieve one of this two options:
to replace all the matches of a particular expression ((\W)(theWord)(\W)) all across the document EXCEPT the matches that are inside a markdown image syntax .
to replace all the matches of a particular expression ({{([^}}]+)}}\[\[[^\]\]]+\]\]) ONLY inside the markdown images, ie.: ![Blah {{theWord}}[[1234]] blah](url).
Both expressions are currently matching everything, no matter if inside the markdown image syntax or not, and I've already tried everything I could think.
Here is an example of the first option
And here is an example of the second option
Any help and/or clue will be highly appreciated.
Thanks in advance!
Well I modified first expression a little bit as I thought there are some extra capturing groups then made them by adding a lookahead trick:
-First one (Live demo):
\b(vitae)\b(?![^[]*]\s*\()
-Second one (Live demo):
{{([^}}]+)}}\[\[[^\]\]]+\]\](?=[^[]*]\s*\()
Lookahead part explanations:
(?! # Starting a negative lookahead
[^[]*] # Everything that's between brackets
\s* # Any whitespace
\( # Check if it's followed by an opening parentheses
) # End of lookahead which confirms the whole expression doesn't match between brackets
(?= means a positive lookahead
You can leverage the discard technique that it really useful for this cases. It consists of having below pattern:
patternToSkip1 (*SKIP)(*FAIL)|patternToSkip2 (*SKIP)(*FAIL)| MATCH THIS PATTERN
So, according you needs:
to replace all the matches of a particular expression ((\W)(theWord)(\W)) all across the document EXCEPT the matches that are inside a markdown image syntax
You can easily achieve this in pcre through (*SKIP)(*FAIL) flags, so for you case you can use a regex like this:
\[.*?\](*SKIP)(*FAIL)|\bTheWord\b
Or using your pattern:
\[.*?\](*SKIP)(*FAIL)|(\W)(theWord)(\W)
The idea behind this regex is tell regex engine to skip the content within [...]
Working demo
The first regex is easily fixed with a SKIP-FAIL trick:
\!\[.*?\]\(http[^)]*\)(*SKIP)(*FAIL)|\bvitae\b
To replace with the word of your choice. It is a totally valid way in PHP (PCRE) regex to match something outside some markers.
See Demo 1
As for the second one, it is harder, but acheivable with \G that ensures we match consecutively inside some markers:
(\!\[.*?|(?<!^)\G)((?>(?!\]\(http).)*?){{([^}]+?)}}\[{2}[^]]+?\]{2}(?=.*?\]\(http[^)]*?\))
To replace with $1$2{{NEW_REPLACED_TEXT}}[[NEW_DIGITS]]
See Demo 2
PHP:
$re1 = "#\!\[.*?\]\(http[^)]*\)(*SKIP)(*FAIL)|\bvitae\b#i";
$re2 = "#(\!\[.*?|(?<!^)\G)((?>(?!\]\(http).)*?){{([^}]+?)}}\[{2}[^]]+?\]{2}(?=.*?\]\(http[^)]*?\))#i";
I have a string. An example might be "Contact /u/someone on reddit, or visit /r/subreddit or /r/subreddit2"
I want to replace any instance of "/r/x" and "/u/x" with "[/r/x](http://reddit.com/r/x)" and "[/u/x](http://reddit.com/u/x)" basically.
So I'm not sure how to 1) find "/r/" and then expand that to the rest of the word (until there's a space), then 2) take that full "/r/x" and replace with my pattern, and most importantly 3) do this for all "/r/" and "/u/" matches in a single go...
The only way I know to do this would be to write a function to walk the string, character by character, until I found "/", then look for "r" and "/" to follow; then keep going until I found a space. That would give me the beginning and ending characters, so I could do a string replacement; then calculate the new end point, and continue walking the string.
This feels... dumb. I have a feeling there's a relatively simple way to do this, and I just don't know how to google to get all the relevant parts.
A simple preg_replace will do what you want.
Try:
$string = preg_replace('#(/(?:u|r)/[a-zA-Z0-9_-]+)#', '[\1](http://reddit.com\1)', $string);
Here is an example: http://ideone.com/dvz2zB
You should see if you can discover what characters are valid in a Reddit name or in a Reddit username and modify the [a-zA-Z0-9_-] charset accordingly.
You are looking for a regular expression.
A basic pattern starts out as a fixed string. /u/ or /r/ which would match those exactly. This can be simplified to match one or another with /(?:u|r)/ which would match the same as those two patterns. Next you would want to match everything from that point up to a space. You would use a negative character group [^ ] which will match any character that is not a space, and apply a modifier, *, to match as many characters as possible that match that group. /(?:u|r)/[^ ]*
You can take that pattern further and add a lookbehind, (?<= ) to ensure your match is preceded by a space so you're not matching a partial which results in (?<= )/(?:u|r)/[^ ]*. You wrap all of that to make a capturing group ((?<= )/(?:u|r)/[^ ]*). This will capture the contents within the parenthesis to allow for a replacement pattern. You can express your chosen replacement using the \1 reference to the first captured group as [\1](http://reddit.com\1).
In php you would pass the matching pattern, replacement pattern, and subject string to the preg_replace function.
In my opinion regex would be an overkill for such a simple operation. If you just want to replace instance of "/r/x" with "[r/x](http://reddit.com/r/x)" and "/u/x" with "[/u/x](http://reddit.com/u/x)" you should use str_replace although with preg_replace it'll lessen the code.
str_replace("/r/x","[/r/x](http://reddit.com/r/x)","whatever_string");
use regex for intricate search string and replace. you can also use http://www.jslab.dk/tools.regex.php regular expression generator if you have something complex to capture in the string.
im looking for a regex that matches words that repeat a letter(s) more than once and that are next to each other.
Here's an example:
This is an exxxmaple oooonnnnllllyyyyy!
By far I havent found anything that can exactly match:
exxxmaple and oooonnnnllllyyyyy
I need to find it and place them in an array, like this:
preg_match_all('/\b(???)\b/', $str, $arr) );
Can somebody explain what regexp i have to use?
You can use a very simple regex like
\S*(\w)(?=\1+)\S*
See how the regex matches at http://regex101.com/r/rF3pR7/3
\S matches anything other than a space
* quantifier, zero or more occurance of \S
(\w) matches a single character, captures in \1
(?=\1+) postive look ahead. Asserts that the captrued character is followed by itsef \1
+ quantifiers, one or more occurence of the repeated character
\S* matches anything other than space
EDIT
If the repeating must be more than once, a slight modification of the regex would do the trick
\S*(\w)(?=\1{2,})\S*
for example http://regex101.com/r/rF3pR7/5
Use this if you want discard words like apple etc .
\b\w*(\w)(?=\1\1+)\w*\b
or
\b(?=[^\s]*(\w)\1\1+)\w+\b
Try this.See demo.
http://regex101.com/r/kP8uF5/20
http://regex101.com/r/kP8uF5/21
You can use this pattern:
\b\w*?(\w)\1{2}\w*
The \w class and the word-boundary \b limit the search to words. Note that the word boundary can be removed, however, it reduces the number of steps to obtain a match (as the lazy quantifier). Note too, that if you are looking for words (in the common meaning), you need to remove the word boundary and to use [a-zA-Z] instead of \w.
(\w)\1{2} checks if a repeated character is present. A word character is captured in group 1 and must be followed with the content of the capture group (the backreference \1).
So the user may search for "10 mbit" after which I want to capture the "10" so I can use it in a speed-search rather than a string-search. This isn't a problem, the below regexp does this fine:
if (preg_match("/(\d+)\smbit/", $string)){ ... }
But, the user may search for something like "10/10 mbit" or "10-100 mbit". I don't want to match those with the above regexp - they should be handled in another fashion. So I would like a regexp that matches "10 mbit" if the number is all-numeric as a whole word (i.e. contained by whitespace, newline or lineend/linestart)
Using lookbehind, I did this:
if (preg_match("#(?<!/)(\d+)\s+mbit#i", $string)){
Just to catch those that doesn't have "/" before them, but this matched true for this string: "10/10 mbit" so I'm obviously doing something wrong here, but what?
If the slash or hyphen is the only thing you care about, this should do it:
'#(?<![\d/-])(\d+)\s+mbit#i`
The problem with your regex is that \d+ is only required to match one digit. It can't match the 10 in 10/10 mbit because it's preceded by a slash, but the 0 isn't. To make sure it matches from the beginning of the number, you have to include \d in the list of things it can't be preceded by.
You lookback assertion is negative. It tells the string should not be preceded by /
So the / is matched inside the string (as the regex cannot match only "10" : you forbid it explicitely with the assertion). Maybe you wanted a positive lookbehind?