The data along with the regex pattern I'm using is linked here:
(?m)(?<=Note:)(\w+|\s+)*$
The sample text is:
Date:21
Month:03
Year:2017
Amount:50
Category:Test
Account:Testimg
Note:Tested
Date:21
Month:03
Year:2017
Amount:48
Category:Great
Account:Good
Note:Better
As you can imagine, I want all the text after the word "Note:" including the spaces and right up to the end of the line. I'm getting the results I need, but I'm not sure if this is a proper solution.
Is this the right way of going about it? Could it be made simpler?
Thank you.
Since your lines start with Note: and you need to use ^ anchor before it. You may use capturing as I suggested in my first comment:
preg_match_all('/^Note:(.+)/m', $s, $matches)
See this demo.
Here, ^Note:(.+) will assert the position at the start of the line, then Note: will get matched, and then any 1+ chars other than line break chars will get captured into Group 1, you will just need to access it using the right index.
Alternatively, use \K to drop the Note::
preg_match_all('/^Note:\K.+/m', $s, $matches)
See another regex demo
Here, ^Note:\K.+ will also match the Note: at the start of the line, and then the text will be dropped due to \K match reset operator, and then 1+ chars other than line break chars will get consumed and placed into the match buffer.
Note the $ anchor is not even necessary here, since .+ will only match greedily up to the end of line on its own.
You can simplify this to just /Note:(.*)$/gm, I've updated your regex101 example. But other than that yes you're going about it the right way.
Related
I can't capture what i want with capturing parantheses...
I'm searching in /hodsakers/marsh-zwartArray/d and i want to capture marsh-zwartArray but sometimes the last / is not present in what i'm searching.
I search and try many things =/ like :
(marshall[\s\S]*)\/
it work but if the last backslash is not present it doesn't.
I also try
(marsh[\s\S]*)(\/)?
in this case that's the opposite, it work without the last backslash but not anymore if there is one, it will get all the string and capture nothing =/
So i don't know how i can capture in both cases =/
Thanks for your help
You may use a [^\/]* negated character class to match 0+ chars other than /:
/marsh[^\/]*/
See the regex demo
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";
How do you inverse a Regex expression in PHP?
This is my code:
preg_match("!<div class=\"foo\">.*?</div>!is", $source, $matches);
This is checking the $source String for everything within the Container and stores it in the $matches variable.
But what I want to do is reversing the expression i.e. I want to get everything that is NOT inside the container.
I know there is something called negative lookahead, but I am really bad with Regular expressions and didn't manage to come up with a working solution.
Simply using ?!
preg_match("?!<div class=\"foo\">.*?</div>!is", $source, $matches);
Does not seem to work.
Thanks!
New solution
Since your goal is to remove the matching divs, as mentioned in the comment, using the original regex with preg_split, plus implode would be the simpler solution:
implode('', preg_split('~<div class="foo">.*?</div>~is', $text))
Demo on ideone
Old solution
I'm not sure whether this is a good idea, but here is my solution:
~(.*?)(?:<div class="foo">.*?</div>|$)~is
Demo on regex101
The result can be picked out from capturing group 1 of each matches.
Note that the last match is always an empty string, and there can be empty string match between 2 matching divs or if the string starts with matching div. However, you need to concatenate them anyway, so it seems to be a non-issue.
The idea is to rely on the fact that lazy quantifier .*? will always try the sequel (whatever comes after it) first before advancing itself, resulting in something similar to look-ahead assertion that makes sure that whatever matched by .*? will not be inside <div class="foo">.*?</div>.
The div tag is matched along in each match in order to advance the cursor past the closing tag. $ is used to match the text after the last matching div.
The s flag makes . matches any character, including line separators.
Revision: I had to change .+? to .*?, since .+? handle strings with 2 matching div next to each other and strings start with matching div.
Anyway, it's not a good idea to modify HTML with regular expression. Use a parser instead.
<div class=\"foo\">.*?</div>\K|.
You can simply do this by using \K.
\K resets the starting point of the reported match. Any previously consumed characters are no longer included in the final match
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.
Having a little trouble with regex. I'm trying to test for a match but only if nothing follows it. So in the below example if I go to test/create/1/2 - it still matches. I only want to match if it's explicitally test/create/1 (but the one is dynamic).
if(preg_match('^test/create/(.*)^', 'test/create/1')):
// do something...
endif;
I've found some answers that suggest using $ before my delimiter but it doesn't appear to do anything. Or a combination of ^ and $ but I can't quite figure it out. Regex confuses the hell out of me!
EDIT:
I didn't really explain this well enough so just to clarify:
I need the if statement to return true if a URL is test/create/{id} - the {id} being dynamic (and of any length). If the {id} is followed by a forward slash the if statement should fail. So that if someone types in test/create/1/2 - it will fail because of the forward slash after the 1.
Solution
I went for thedarkwinter's answer in the end as it's what worked best for me, although other answers did work as well.
I also had to add an little extra in the regex to make sure that it would work with hyphens as well so the final code looked like this:
if(preg_match('^test/create/[\w-]*$^', 'test/create/1')):
// do something...
endif;
/w matches word characters, and $ matches end of string
if(preg_match('^test/create/\w*$^', 'test/create/1'))
will match test/create/[word/num] and nothing following.
I think thats what you are after.
edit added * in \w*
Here you go:
"/^test\\/create\\/([^\\/]*)$/"
This says:
The string that starts with "test" followed by a forward slash (remember the first backslash escapes the second so PHP puts a letter backslash in the input, which escapes the / to regex) followed by create followed by a forward slash followed by and capture everything that isn't a slash which is then the end of the string.
Comment if you need more detail
I prefer my expressions to always start with / because it has no meaning as a regex character, I've seen # used, I believe some other answer uses ^, this means "start of string" so I wouldn't use it as my regex delimiters.
Use following regular expression (use $ to denote end of the input):
'|test/create/[^/]+$|'
If you want only match digits, use folloiwng instead (\d match digit character):
'^test/create/\d+$^'
The ^ is an anchor for the beginning of the line, i.e. no characters occurring before the ^ . Use a $ to designate the end of the string, or end of the line.
EDIT: wanted to add a suggestion as well:
Your solution is fine and works, but in terms of style I'd advise against using the carat (^) as a delimiter -- especially because it has special meaning as either negation or as a start of line anchor so it's a bit confusing to read it that way. You can legally use most special characters as long as they don't occur (or are escaped) in the regex itself. Just talking about a matter of style/maintainability here.
Of course nearly every potential delimiter has some special meaning, but you also often tend to see the ^ at the beginning of a regex so I might chose another alternative. For example # is a good choice here :
if(preg_match('#test/create/[\w-]*$#', $mystring)) {
//etc
}
The regex abc$ will match abc only when it's the last string.
abcd # no match
dabc # match
abc # match