My question is about fixing spacing issues in text, specifically sentences that lack spaces after a dot. For example:
See also vadding.Constructions on this term abound.
I also have URLs within the text, such as:
See also vadding.Constructions on this term abound. http://example.com/foo/bar
Locate and extract a URL while also identifying all other occurrences of dots and converting them to a dot followed by a space:
var re = /((?:https?|ftps?):\/\/\S+)|\.(?!\s)/g;
var str = 'See also vadding.Constructions on this term abound.\nSee also vadding.Constructions on this term abound. http://example.com/foo/bar';
var result = str.replace(re, function(m, g1) {
return g1 ? g1 : ". ";
});
document.body.innerHTML = "<pre>" + result + "</pre>";The regular expression for the URL - (?:https?|ftps?):\/\/\S+ - identifies patterns starting with http, https, ftp, or ftps, followed by :// and one or more non-whitespace characters (\S+). For more complex URL matching expressions, resources like Stack Overflow can provide useful insights. Check out What is a good regular expression to match a URL?.
Explanation of the process in more depth:
The regex
((?:https?|ftps?):\/\/\S+)|\.(?!\s)|) recognizing a dot that is not followed by whitespace (\.(?!\s)).
IMPORTANT: The use of (?!\s) serves as a negative lookahead assertion to locate a dot not succeeded by a space.
When utilizing string.replace(), it's possible to specify an anonymous callback function as the second argument to handle matches and captured groups. In this case, there's one match value (m) and one capture group value g1 (representing the identified URL). If the URL is found, g1 won't be null. Therefore, return g1 ? g1 : ". "; preserves group 1 if matched, replacing standalone dots with . otherwise.
If you want to target the period character that is not followed by two or three lowercase letters or a space, consider using RegExp
/(\.)(?!=[a-z]{2}\/|[a-z]{3}\/|\s+|$)/g"To learn more about this subject, check out myriad resources available online at http://example.com/foo/bar"
.replace(/(\.)(?!=[a-z]{2}\/|[a-z]{3}\/|\s+|$)/g, "$1 ")
Using inspiration from @MarcelKohls
var phrase = "Explore more: Resources about this topic are widely available. http://sample.com/foo/bar";
var url_pattern = /(\bhttps?:\/\/(?:(?:(?!&[^;]+;)|(?=&))[^\s"'<>\]\[)])+\b)/gi;
phrase = phrase.split(url_pattern).map(function(phrase) {
if (phrase.match(url_pattern)) {
return phrase;
} else {
return phrase.replace(/\.([^ ])/g, '. $1');
}
}).join('');
document.body.innerHTML = '<pre>' + phrase + '</pre>';Try implementing this regular expression:
/\.(?! )((?:ftp|http)[^ ]+)?/g
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