I am looking for a way to extract X number of words on either side of a given word in a search.
For example, if a user enters "inmate" as a search word and the MySQL query finds a post that contains "inmate" in the content of the post, I would like to return not the entire contents of the post but just x number of words on either side of it to give the user the gist of the post and then they can decide if they want to continue on to the post and read it in full.
I am using PHP.
Thanks!
You might not be able to fully solve this problem with regex. There are too many possibilities of other characters between the words...
But you can try this regex:
((?:\S+\s*){0,5}\S*inmate\S*(?:\s*\S+){0,5})
See here : rubular
You might also want to exclude certain characters as they are not counted as words. Right now the regex counts any sequence of non space characters that are surrounded by spaces as word.
To match only real words:
((?:\w+\s*){0,5}<search word>(?:\s*\w+){0,5})
But here any non word character (,". etc.) brakes the matching.
So you can go on...
((?:[\w"',.-]+\s*){0,5}["',.-]?<search word>["',.-]?(?:\s*[\w"',.-]+){0,5})
This would also match 5 words with one of "',.- around your search term.
To use it in php:
$sourcestring="For example, if a user enters \"inmate\" as a search word and the MySQL";
preg_match_all('/(?:\S+\s*){0,5}\S*inmate\S*(?:\s*\S+){0,5}/s',$sourcestring,$matches);
echo $matches[0][0]; // you might have more matches, they will be in $matches[0][x]
I would use this regex for php which also takes UTF8 characters into account
'~(?:[\p{L}\p{N}\']+[^\p{L}\p{N}\']+){0,5}<search word>(?:[^\p{L}\p{N}\']+[\p{L}\p{N}\']+){0,5}~u'
In this case '~' is the delimiter and the modificator 'u' at the end identifies the regex is UTF8 interpreted.
please see a documentation about the Unicode Regex identifiers here:
http://www.regular-expressions.info/refunicode.html
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