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 need to capture numbers and dots between brackets on lines containing the string 0020,000d, for example:
I: (0020,000d) UI [1.2.410.200001.1104.20160720104648421 ] # 38, 1 StudyInstanceUID
Using this regexp 0020,000d.*\[([\.0-9]+)\] I can match the needed value only if it doesn't have a space inside the brackets. How can I match the needed value ignoring any other character?.
Edit
If I use this regexp 0020,000d.*\[([\.0-9(\s|^\s))]+)\] I can capture numbers and dots and/or spaces, now if the string contains a space how can I capture in a group everything but the space?.
To clarify, I want to extract the 1.2.410.200001.1104.20160720104648421 string.
Codifying my (apparently helpful) answer from the comments:
You just need to allow zero or more spaces after the numbers-and-dots sequence before the closing bracket:
0020,000d.*\[([.0-9]+) *\]
Also, please note that you don't need to escape a dot in a character class.
Try this
let regex = /(?!\[)[.\d]+(?=[(\s)*\]])/g
let str = 'I: (0020,000d) UI [1.2.410.200001.1104.20160720104648421 ]'
let result = str.match(regex);
console.log(result);
My job is to create a regular expression that matches any of the variations on the name: vic, Victor, Victoria, victor, Victoria, VICKY, VICTOR. VICTORIA, Vic, VIC
and so I wrote this
(Vic[a-z])|(vic[a-z])|(VIC[A-Z]*)
I have a problem though. For example if I typed "Hello aVictoria", it would match the string "Victoria"...How do I make it to match such that the first character has to be V or v?
You failed to add *,
(Vic[a-z]*)|(vic[a-z]*)|(VIC[A-Z]*)
In the below regex, you failed to add + or * on the first two regexes,
(Vic[a-z])|(vic[a-z])|(VIC[A-Z]*)
(Vic[a-z]) - It matches any single alphabetical character after Vic but it wont match more than one character after that word Vic.
If you need start of the word, you can use \b (word boundary 0-width matcher). If you need start of the line, you should use ^ (start of the line 0-width matcher). Regexp engines that also have multi-line capabilities will also define \A (start of the string 0-width matcher). Depending on what you want, stick one of those in front of your regexp, like this:
^((Vic[a-z])|(vic[a-z])|(VIC[A-Z]*))
(You also need extra parentheses, because otherwise you would match "Victor" only at start, but "victor" and "VICTOR" anywhere at all.)
That said, this won't match "vic", since you are requiring another [a-z] after it. This would be the correct regexp:
^((Vic[a-z]*)|(vic[a-z]*)|(VIC[A-Z]*))
This will match "Victor", but not "aVictor". If you write \b instead of ^, you will additionally match "I am not a Victor" (while still disallowing "aVictor").
^ character to indicate start of string, like
^(Vic[a-z])|(vic[a-z])|(VIC[A-Z]*)
Try this
/\bvic[a-zA-Z]{0,}/ig
for Explanation
Sample Example
'vic, Victor, Victoria, victor, Victoria, VICKY, VICTOR. VICTORIA, Vic, VIC, Hello aVictoriaX'.match(/\bvic[a-zA-Z]{0,}/ig)
//output
["vic", "Victor", "Victoria", "victor", "Victoria", "VICKY", "VICTOR", "VICTORIA", "Vic", "VIC"]
//Hello aVictoriaX isn't matched
I want to split a text into it's single words using regular expressions. The obvious solution would be to use the regex \\b unfortunately this one does split words also on the hyphen.
So I am searching an expression doing exactly the same as the \\b but does not split on hyphens.
Thanks for your help.
Example:
String s = "This is my text! It uses some odd words like user-generated and need therefore a special regex.";
String [] b = s.split("\\b+");
for (int i = 0; i < b.length; i++){
System.out.println(b[i]);
}
Output:
This
is
my
text
!
It
uses
some
odd
words
like
user
-
generated
and
need
therefore
a
special
regex
.
Expected output:
...
like
user-generated
and
....
#Matmarbon solution is already quite close, but not 100% fitting it gives me
...
like
user-
generated
and
....
This should do the trick, even if lookaheads are not available:
[^\w\-]+
Also not you but somebody who needs this for another purpose (i.e. inserting something) this is more of an equivalent to the \b-solutions:
([^\w\-]|$|^)+
because:
There are three different positions that qualify as word boundaries:
Before the first character in the string, if the first character is a word character.
After the last character in the string, if the last character is a word character.
Between two characters in the string, where one is a word character and the other is not a word character.
--- http://www.regular-expressions.info/wordboundaries.html
You can use this:
(?<!-)\\b(?!-)
If I have a string like this
Newsflash: The Big(!) Brown Dog's Brother (T.J.) Ate The Small Blue Egg
how would I convert that into the following using regex:
newsflash-the-big-brown-dogs-brother-tj-ate-the-small-blue-egg
In other words, punctuation is discarded and spaces are replaced with hyphens.
It sounds like you want to create a "URL plug" -- a URL-friendly version of an article's title, for example. That means you'll want to make sure you remove all possible non-URL-friendly characters, not just a few. You might do it this way (in order):
Remove all non-letter non-number non-space characters by:
Replacing regex [^A-Za-z0-9 ] with the empty string "".
Replace all spaces with a dash by:
Replacing regex \s+ with the string "-".
Lower-case the string by:
Java s = s.toLowerCase();
JavaScript s = s.toLowerCase();
C# s = s.ToLowerCase();
Perl $s = lc($s);
Python s = s.lower()
PHP $s = strtolower($s);
Ruby s = s.downcase
Replace the regex [\s-]+ with "-", then replace [^\w-] with "".
Then, call ToLowerCase or equivalent.
In Javascript:
var s = "Newsflash: The Big(!) Brown Dog's Brother (T.J.) Ate The Small Blue Egg";
alert(s.replace(/[\s+-]/g, '-').replace(/[^\w-]/g, '').toLowerCase());
Replace /\W+/ with '-', that will replace all non-word characters with a dash.
Then, collapse dashes by replacing /-+/ with '-'.
Then, lowercase the string - pure regex solutions cannot do that. You didn't say which language you are using, so I cannot give you an example, but your language might have String.toLowercase() or a tr/// call (tr/A-Z/a-z/, for example, in Perl).