I have data that looks like this:
#1
(1) This is a test.
a) This is a subtest one.
b) And another one.
(2) A really cool test.
(3) Here is the problem, text for each numbered line is
supposed to be on a single line like in (1) and (2), but
the text often spans multiple lines of text.
(4) How can I match the multi-line entries and unwrap them to single lines?
#2
(1) This is a test.
a) This is a subtest one.
b) And another one.
(2) A really cool test.
(3) Here is the problem, text for each numbered line is
supposed to be on a single line like in (1) and (2), but
the text often spans multiple lines of text.
(4) How can I match the multi-line entries and unwrap them to single lines?
#3
(1) This is a test.
a) This is a subtest one.
b) And another one.
(2) A really cool test.
(3) Here is the problem, text for each numbered line is
supposed to be on a single line like in (1) and (2), but
the text often spans multiple lines of text.
(4) How can I match the multi-line entries and unwrap them to single lines?
I need a Regular Expression that matches multiple multi-line text entries so I can unwrap them to single lines.
I've tried this:
$pattern = '/^(\(?[a-z0-9]+\) )([\s\S]+?(?!#))(^\(?[a-z0-9]+\))/mS';
$text = preg_replace_callback ($pattern, function ($grp) {
return $grp[1] . unwrap ($grp[2]) . PHP_EOL . $grp[3];
}, $text);
I feel like this should be a simple regex to write, but I'm having trouble for some reason.
You can match every entry using lookahead with the following regex and unwrap the whole match:
'^\(\d+\)[^#]*?(?=\n\(\d\)|\Z|#)'
See Demo
EDIT: from your question it's not clear how you want to handle sub-entries like a) and b). In this case they will be recognized as normal text.
EDIT2: in order to match a) and b) as entries as well:
'^(?:[a-z]\)|\(\d+\))[^#]*?(?=\n\(\d\)|\Z|#|\n[a-z]\))'
Demo
I would use preg_split with this pattern:
~(\R+|^#\d+$)+(?=\(\d+\)|[a-z]\)|\z)~m
The idea consists to describe what are the delimiters between the target parts:
at least one or more newlines followed with a target item or the end of the string
an eventual line with this format: #number
$res = preg_split('~(\R+|^#\d+$)+(?=\(\d+\)|[a-z]\)|\z)~m', $str, -1, PREG_SPLIT_NO_EMPTY);
demo
This is a bit shorter and more efficient than a preg_match_all approach and if you want to make it more flexible, you can add optional horizontal spaces in the pattern.
Note also that the pattern contains a capture group. The capture is totally useless, if you want you can change it to a non-capture group, but if you feel like a rebel and an adventurer, you can also begin the pattern with (?n) starting with PHP 7.3. With this modifier, capture groups are seen as non-capturing groups.
Related
I am trying to extract part of a long text, such as information about caring for a plant. The text contains paragraphs and blank lines. I am not able to capture the specific text I want, the second problem is that the last word isn't showing in the extracted text, and the last problem is when my search starts at the beginning of the line.
I tried searching for the text I want to extract by using a word that isn't at the beginning of the line, it worked except that the end of the desired text is missing a word, and if that word is on new line, it won't show any results at all.
I was using https://scriptun.com/tools/php/preg_match for testing
//The first word to start the search is 'How to'. And I want to capture it as well
// The second word where the text I want ends is '(optional):'
'/(?=How to).*?\s(?=\(optional\):)/'
The sample text I am using to test is:
//Text comes before this..
How to care for Split Leaf Plant
The Split leaf philodendron, also called monstera deliciosa or swiss
cheese plant, is a large, popular, easy- care houseplant that is not
really in the philodendron family. There is a great deal of confusion
about what to call this plant; the various names have become
inter-changeable over the years.
Here is more info (optional):
//And more text goes here
I want to extract all the text from the word 'How to' ending with '(optional):'. Regardless of how many lines or paragraphs are in between
The expected extracted text:
How to care for Split Leaf Plant
The Split leaf philodendron, also called monstera deliciosa or swiss
cheese plant, is a large, popular, easy- care houseplant that is not
really in the philodendron family. There is a great deal of confusion
about what to call this plant; the various names have become
inter-changeable over the years.
Here is more info (optional):
Thank you
That's pretty easy. You can use the following pattern:
https://regex101.com/r/TjE2x8/2
Pattern: ^How to[\w\W]+?\(optional\):$
Pattern: ^How to(?:.|\R)*optional\):$
demo on regex101
Explanation:
^ match the first instance where How to appears at the beginning of the line
(?: ) non capturing group. We need it because of the following OR instruction which is the pipe |. But we don't need to capture the contents. That's why we use ?: after the first parenthesis.
. every character
| or
\R every kind of new line
* make sure to capture zero to every instance of the group
optional\):$ match the word optional with parenthesis (escaped, because it is not an instruction) \) and a colon : at the very end of the text $
Pattern 2: /^How to.*optional\):$/ms
demo on regex101
This pattern is even simpler, but requires the m and s flag to be set in order to match multiline and the . character class to match new lines.
I have a peculiar use case where I need to detect paragraphs that end in !!. Normal occurrences of ! (a single one) is fine in the paragraph, but the block ends when !! is found.
For example:
test foo bar !!
longer paragraph this time!
goes on and on
and then stops !!
Should be detected as two separate matches, one covering the first line, and another (separate) covering lines 2, 3 and 4. This brings it to a total of 2 matches.
(Preferably it should work with multiline-mode, as it's part of a larger regex that employs this mode.)
How would I accomplish this? I tried [^!!]* which to me says, find as many non-!! characters as possible, but I'm not sure how to leverage that, and worse yet it still finds single occurrences of !.
There is a common idiom in regular expressions that is used for escape sequences. (Like "\n" in a string.) You can use the same concept here.
The trick is to match either NOT the first character, or the first character followed by a valid second character.
In your case, that would be:
(?: # this is a package, either A or B, choose one
[^!] # Not a bang
| # or
![^!] # Bang, followed by not-a-bang
)
This pair of alternatives describes all the characters in your paragraph. So you can repeat it either 0 times (*) or one-or-more times (+) depending on what you are doing in the rest of your pattern.
# All together:
(?:[^!]|![^!])* # zero or more
(?:[^!]|![^!])+ # one or more
(Obviously, you can match '!!' at the end if you like...)
^([!]?[^!]+[!]?[^!]+)*[!]{2}$/gm
This regex worked for me. It ensures any single ! characters are separated by non-! characters, but there don't have to be any single ! characters. It worked on multiline mode. This also has the added benefit of extracting the text that comes before an occurrence of "!!" since I assume you want to work with it.
/^([!]?[^!]+[!]?[^!]+)*.?[!]{2}$|^([!]?[^!]+[!]?[^!]+)*[^!]?[!]?$/gm
This slightly longer regex captures text that occurs after the final !! (ie, if the file has text between !! and EOF). I wouldn't recommend using the capturing groups though as on my regex checker, they didn't seem to work properly (that may have just been an implementation glitch, however, as the capturing groups look like they should work properly).
Try this:
([\w\s!]+?\!{2})
DEMO
Output:
MATCH 1
1. [0-15] `test foo bar !!`
MATCH 2
1. [15-76] `
longer paragraph this time!
goes on and on
and then stops !!`
or
(?:\n?([\w\s!]+?)\s?\!{2})
DEMO
Output:
MATCH 1
1. [0-12] `test foo bar`
MATCH 2
1. [16-73] `longer paragraph this time!
goes on and on
and then stops`
Try following regex using lookahead
VERSION #1
/(?<=!!|^).*?(?=!!)/gms
Please see https://regex101.com/r/cQ0wC0/2
Result should be
OUTPUT:
test foo bar
longer paragraph this time!
goes on and on
and then stops
VERSION #2
Since OP want to capture last paragraph of text after !! even it's not ending with bang signs.
/(?<=!!|^).*?(?=!!)|(?<=!!).*$/gms
Please see demo https://regex101.com/r/cQ0wC0/4
INPUT:
test foo bar !!
longer paragraph this time!
goes on and on
and then stops !!
longer paragraph this time!
goes on and on
OUTPUT:
test foo bar
longer paragraph this time!
goes on and on
and then stops
longer paragraph this time!
goes on and on
I am trying to retrieve matches from a comma separated list that is located inside parenthesis using regular expression. (I also retrieve the version number in the first capture group, though that's not important to this question)
What's worth noting is that the expression should ideally handle all possible cases, where the list could be empty or could have more than 3 entries = 0 or more matches in the second capture group.
The expression I have right now looks like this:
SomeText\/(.*)\s\(((,\s)?([\w\s\.]+))*\)
The string I am testing this on looks like this:
SomeText/1.0.4 (debug, OS X 10.11.2, Macbook Pro Retina)
Result of this is:
1. [6-11] `1.0.4`
2. [32-52] `, Macbook Pro Retina`
3. [32-34] `, `
4. [34-52] `Macbook Pro Retina`
The desired result would look like this:
1. [6-11] `1.0.4`
2. [32-52] `debug`
3. [32-34] `OS X 10.11.2`
4. [34-52] `Macbook Pro Retina`
According to the image above (as far as I can see), the expression should work on the test string. What is the cause of the weird results and how could I improve the expression?
I know there are other ways of solving this problem, but I would like to use a single regular expression if possible. Please don't suggest other options.
When dealing with a varying number of groups, regex ain't the best. Solve it in two steps.
First, break down the statement using a simple regex:
SomeText\/([\d.]*) \(([^)]*)\)
1. [9-14] `1.0.4`
2. [16-55] `debug, OS X 10.11.2, Macbook Pro Retina`
Then just explode the second result by ',' to get your groups.
Probably the \G anchor works best here for binding the match to an entry point. This regex is designed for input that is always similar to the sample that is provided in your question.
(?<=SomeText\/|\G(?!^))[(,]? *\K[^,)(]+
(?<=SomeText\/|\G) the lookbehind is the part where matches should be glued to
\G matches where the previous match ended (?!^) but don't match start
[(,]? *\ matches optional opening parenthesis or comma followed by any amount of space
\K resets beginning of the reported match
[^,)(]+ matches the wanted characters, that are none of ( ) ,
Demo at regex101 (grab matches of $0)
Another idea with use of capture groups.
SomeText\/([^(]*)\(|\G(?!^),? *([^,)]+)
This one without lookbehind is a bit more accurate (it also requires the opening parenthesis), of better performance (needs fewer steps) and probably easier to understand and maintain.
SomeText\/([^(]*)\( the entry anchor and version is captured here to $1
|\G(?!^),? *([^,)]+) or glued to previous match: capture to $2 one or more characters, that are not , ) preceded by optional space or comma.
Another demo at regex101
Actually, stribizhev was close:
(?:SomeText\/([^() ]*)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\))
Just had to make that one class expect at least one match
(?:SomeText\/([0-9.]+)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\)) is a little more clear as long as the version number is always numbers and periods.
I wanted to come up with something more elegant than this (though this does actually work):
SomeText\/(.*)\s\(([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?\)
Obviously, the
([^\,]+)?\,?\s?
is repeated 6 times.
(It can be repeated any number of times and it will work for any number of comma-separated items equal to or below that number of times).
I tried to shorten the long, repetitive list of ([^\,]+)?\,?\s? above to
(?:([^\,]+)\,?\s?)*
but it doesn't work and my knowledge of regex is not currently good enough to say why not.
This should solve your problem. Use the code you already have and add something like this. It will determine where commas are in your string and delete them.
Use trim() to delete white spaces at the start or the end.
$a = strpos($line, ",");
$line = trim(substr($line, 55-$a));
I hope, this helps you!
I'm going nuts trying to get a regex to detect spam of keywords in the user inputs. Usually there is some normal text at the start and the keyword spam at the end, separated by commas or other chars.
What I need is a regex to count the number of keywords to flag the text for a human to check it.
The text is usually like this:
[random text, with commas, dots and all]
keyword1, keyword2, keyword3, keyword4, keyword5,
Keyword6, keyword7, keyword8...
I've tried several regex to count the matches:
-This only gets one out of two keywords
[,-](\w|\s)+[,-]
-This also matches the random text
(?:([^,-]*)(?:[^,-]|$))
Can anyone tell me a regex to do this? Or should I take a different approach?
Thanks!
Pr your answer to my question, here is a regexp to match a string that occurs between two commas.
(?<=,)[^,]+(?=,)
This regexp does not match, and hence do not consume, the delimiting commas.
This regexp would match " and hence do not consume" in the previous sentence.
The fact that your regexp matched and consumed the commas was the reason why your attempted regexp only matched every other candidate.
Also if the whole input is a single string you will want to prevent linebreaks. In that case you will want to use;
(?<=,)[^,\n]+(?=,)
http://www.phpliveregex.com/p/1DJ
As others have said this is potentially a very tricky thing to do... It suffers from all of the same failures as general "word filtering" (e.g. people will "mask" the input). It is made even more difficult without plenty of example posts to test against...
Solution
Anyway, assuming that keywords will be on separate lines to the rest of the input and separated by commas you can match the lines with keywords in like:
Regex
#(?:^)((?:(?:[\w\.]+)(?:, ?|$))+)#m
Input
Taken from your question above:
[random text, with commas, dots and all]
keyword1, keyword2, keyword3, keyword4, keyword5,
Keyword6, keyword7, keyword8
Output
// preg_match_all('#(?:^)((?:(?:[\w]+)(?:, ?|$))+)#m', $string, $matches);
// var_dump($matches);
array(2) {
[0]=>
array(2) {
[0]=>
string(49) "keyword1, keyword2, keyword3, keyword4, keyword5,"
[1]=>
string(31) "Keyword6, keyword7, keyword8..."
}
[1]=>
array(2) {
[0]=>
string(49) "keyword1, keyword2, keyword3, keyword4, keyword5,"
[1]=>
string(31) "Keyword6, keyword7, keyword8"
}
}
Explanation
#(?:^)((?:(?:[\w]+)(?:, ?|$))+)#m
# => Starting delimiter
(?:^) => Matches start of line in a non-capturing group (you could just use ^ I was using |\n originally and didn't update)
( => Start a capturing group
(?: => Start a non-capturing group
(?:[\w]+) => A non-capturing group to match one or more word characters a-zA-Z0-9_ (Using a character class so that you can add to it if you need to....)
(?:, ?|$) => A non-capturing group to match either a comma (with an optional space) or the end of the string/line
)+ => End the non-capturing group (4) and repeat 5/6 to find multiple matches in the line
) => Close the capture group 3
# => Ending delimiter
m => Multi-line modifier
Follow up from number 2:
#^((?:(?:[\w]+)(?:, ?|$))+)#m
Counting keywords
Having now returned an array of lines only containing key words you can count the number of commas and thus get the number of keywords
$key_words = implode(', ', $matches[1]); // Join lines returned by preg_match_all
echo substr_count($key_words, ','); // 8
N.B. In most circumstances this will return NUMBER_OF_KEY_WORDS - 1 (i.e. in your case 7); it returns 8 because you have a comma at the end of your first line of key words.
Links
http://php.net/manual/en/reference.pcre.pattern.modifiers.php
http://www.regular-expressions.info/
http://php.net/substr_count
Why not just use explode and trim?
$keywords = array_map ('trim', explode (',', $keywordstring));
Then do a count() on $keywords.
If you think keywords with spaces in are spam, then you can iterate of the $keywords array and look for any that contain whitespace. There might be legitimate reasons for having spaces in a keyword though. If you're talking about superheroes on your system, for example, someone might enter The Tick or Iron Man as a keyword
I don't think counting keywords and looking for spaces in keywords are really very good strategies for detecting spam though. You might want to look into other bot protection strategies instead, or even use manual moderation.
How to match on the String of text between the commas?
This SO Post was marked as a duplicate to my posted question however since it is NOT a duplicate and there were no answers in THIS SO Post that answered my question on how to also match on the strings between the commas see below on how to take this a step further.
How to Match on single digit values in a CSV String
For example if the task is to search the string within the commas for a single 7, 8 or a single 9 but not match on combinations such as 17 or 77 or 78 but only the single 7s, 8s, or 9s see below...
The answer is to Use look arounds and place your search pattern within the look arounds:
(?<=^|,)[789](?=,|$)
See live demo.
The above Pattern is more concise however I've pasted below the Two Patterns provided as solutions to THIS this question of matching on Strings within the commas and they are:
(?<=^|,)[789](?=,|$) Provided by #Bohemian and chosen as the Correct Answer
(?:(?<=^)|(?<=,))[789](?:(?=,)|(?=$)) Provided in comments by #Ouroborus
Demo: https://regex101.com/r/fd5GnD/1
Your first regexp doesn't need a preceding comma
[\w\s]+[,-]
A regex that will match strings between two commas or start or end of string is
(?<=,|^)[^,]*(?=,|$)
Or, a bit more efficient:
(?<![^,])[^,]*(?![^,])
See the regex demo #1 and demo #2.
Details:
(?<=,|^) / (?<![^,]) - start of string or a position immediately preceded with a comma
[^,]* - zero or more chars other than a comma
(?=,|$) / (?![^,]) - end of string or a position immediately followed with a comma
If people still search for this in 2021
([^,\n])+
Match anything except new line and comma
regexr.com/60eme
I think the difficulty is that the random text can also contain commas.
If the keywords are all on one line and it is the last line of the text as a whole, trim the whole text removing new line characters from the end. Then take the text from the last new line character to the end. This should be your string containing the keywords. Once you have this part singled out, you can explode the string on comma and count the parts.
<?php
$string = " some gibberish, some more gibberish, and random text
keyword1, keyword2, keyword3
";
$lastEOL = strrpos(trim($string), PHP_EOL);
$keywordLine = substr($string, $lastEOL);
$keywords = explode(',', $keywordLine);
echo "Number of keywords: " . count($keywords);
I know it is not a regex, but I hope it helps nevertheless.
The only way to find a solution, is to find something that separates the random text and the keywords that is not present in the keywords. If a new line is present in the keywords, you can not use it. But are 2 consecutive new lines? Or any other characters.
$string = " some gibberish, some more gibberish, and random text
keyword1, keyword2, keyword3,
keyword4, keyword5, keyword6,
keyword7, keyword8, keyword9
";
$lastEOL = strrpos(trim($string), PHP_EOL . PHP_EOL); // 2 end of lines after random text
$keywordLine = substr($string, $lastEOL);
$keywords = explode(',', $keywordLine);
echo "Number of keywords: " . count($keywords);
(edit: added example for more new lines - long shot)
I'd like to capture up to four groups of text between <p> and </p>. I can do that using the following regex:
<h5>Trivia<\/h5><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p>
The text to match on:
<h5>Trivia</h5><p>Was discovered by a freelance photographer while sunbathing on Bournemouth Beach in August 2003.</p><p>Supports Southampton FC.</p><p>She has 11 GCSEs and 2 'A' Levels.</p><p>Listens to soul, R&B, Stevie Wonder, Aretha Franklin, Usher Raymond, Michael Jackson and George Michael.</p>
It outputs the four lines of text. It also works as intended if there are more trivia items or <p> occurrences.
But if there are less than 4 trivia items or <p> groups, it outputs nothing since it cannot find the fourth group. How do I make that group optional?
I've tried: <h5>Trivia<\/h5><p>(.*?)<\/p>(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)? and that works according to http://gskinner.com/RegExr/ but it doesn't work if I put it inside PHP code. It only detects one group and puts everything in it.
The magic word is either 'escaping' or 'delimiters', read on.
The first regex:
<h5>Trivia<\/h5><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p>
worked because you escaped the / characters in tags like </h5> to <\/h5>.
But in your second regex (correctly enclosing each paragraph in a optional non-capturing group, fetching 1 to 5 paragraphs):
<h5>Trivia</h5><p>(.*?)</p>(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?
you forgot to escape those / characters.
It should then have been:
$pattern = '/<h5>Trivia<\/h5><p>(.*?)<\/p>(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?/';
The above is assuming you were putting your regex between two / "delimiters" characters (out of conventional habit).
To dive a little deeper into the rabbit-hole, one should note that in php the first and last character of a regular expression is usually a "delimiter", so one can add modifiers at the end (like case-insensitive etc).
So instead of escaping your regex, you could also use a ~ character (or #, etc) as a delimiter.
Thus you could also use the same identical (second) regex that you posted and enclose for example like this:
$pattern = '~<h5>Trivia</h5><p>(.*?)</p>(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?~';
Here is a working (web-based) example of that, using # as delimiter (just because we can).
You can use the question mark to make each <p>...</p> optional:
$pattern = '~<h5>Trivia</h5>(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?~';
Use the Dom is a good option too.